Trigonometric ratios express the sides of a right-angled triangle. There are six trigonometric ratios:
Sine, cosine, and tangent ratios are the ratios of the two lengths of a right-angled triangle. The ratios represent the angles formed by the right-angled triangle hypotenuse and legs. The sine ratio is the length of a side opposite the angle it represents over the hypotenuse.
Similarly, right-angled triangles have ratios that represent their base angles. Along with sine and tangent ratios, cosine ratios represent two different sides of a right-angled triangle. Specifically, cosine ratios are a triangle's side base angle over the hypotenuse.
Alongside cosine and sine ratios, the tangent ratios are the ratios of two different sides of a right-angled triangle. The tangent ratios refer to the side's ratios opposite to the length adjacent to the angle they represent.
The other trigonometric ratios, cosecant, secant and cotangent are reciprocals to the basic trigonometric ratios: sine, cosine and tangent. The secant ratio describes the hypotenuse ratio to any side opposite to a given angle of a right-angled triangle. Cosecant represents the hypotenuse ratio to the right of a right-angled triangle, whereas the cotangent is the ratio adjacent to the opposite of a right-angled triangle.
If you have a right-angled triangle, the trigonometric ratios of each of the angles that are not 90 degrees can be solved using different formulas. However, we will limit our discussion to finding sine, abbreviated as sin in trigonometric ratios.
To find sine:
Sin θ = Length of the leg opposite to the angle (O) / Length of the Hypotenuse (H)
Please note that the symbol "θ" is the Greek letter for "theta," representing angles in trigonometric ratios.
Solving a triangle entails finding the length and angles of all the sides. The sine rule applies when you are given a triangle with two angles and a side; alternatively, a triangle two sides and a non-included angle. Note that the sine rule applies to any triangle as long as one side and opposite angles are provided.
To best understand the sine rule ensure you have one of the following items:
Let's go over the actual rule. In a triangle ABC, the sides are a, b and c. So you need this information to get the sine rule formula:
a/SinA=b/SinB=c/SinC
As a result, the sine rule can be reciprocated this way:
Sin/a=SinB/b=SinC/c
In a triangle, ABC, C= 102.3°
, B= 28.7°
, and b= 27.4
feet, find the remaining two sides.
A= 180°-B-C
A= 180°-28.7°-102.3°
A=49.0°
Use the sine rule to get the following:
a/SinA=b/SinB=c/SinC
Now use b= 27.4 to get this:
a=b/SinB(SinA)
a= 27.4/Sin28.7°(49.0)
a= 43.06 feet
And C:
C = b/SinB(SinC)
C= 27.4/Sin28.7(Sin102.3°)
C=55.75 feet
If you want to find the length of a side of a triangle, use the formula:
a/sin (A) = b/ sin (B)
In this formula, the lengths are on the top. You only need two parts of the formula and also one pair of the side with its opposite angle for the formula to apply.
Determine the length of x in a triangle where one of the sides is 7cm, and the two are 60 and 80 degrees, respectively.
Start by writing the sine rule formula.
a/sin (A) = b/ sin (B)
Provide the known and unknown values in the sine rule formula.
x/sin(80°)=7/sin(60°)
Solve the resulting equation to find the unknown side. Give your answer to 3 significant figures.
x/sin(80°)=7/sin(60°)
Multiply both sides by sin(80°)
x=7/sin(60°) × sin(80°)
x=7.96( written in three significant figures)
Find the missing side in a triangle with two angles given as 32° and 95° with one of the sides measuring 21cm.
a/Sin(A)= b/Sin(B)
p/Sin(32°) = 21/Sin(95°)
Multiply both sides by Sin(32°)
p= 21/Sin(95° * Sin(32°)
P= 11.2 cm
To find an angle's size, use the sine rule formula where the angles are on the top.
Sin(A)/a= Sin(B)/b
As mentioned earlier, you only need two parts to use the sine rule, one side, and an opposite angle.
Work out angle n° in a triangle where one of the angles is 75°, and the two sides are 8 and 10cm.
Write out the sine rule formula for finding the angles.
Sin A/a =Sin B/b
Fill in the known and unknown values.
Sin(n°)/8=Sin(75°)/10
Multiply both sides by 8
Sin(n°) = sin(75°)/10 * 8
Sin(n°) = 0.773( three significant figures)
Use the inverse-sine function to find the angle n°.
n° = Sin-1(0.773)
=50.6°
Find the missing angle in a triangle with two sides measuring 5.1cm and 3.6 cm if one of the angles is 100°.
Sin(A)/a = Sin(B)/b
Sin(b)/3.6= Sin(100°)/5.1
Multiply both sides by 3.6
Sin(b)= Sin(100)/5.1 *3.6
Sin(b) =0.695(3 significant figures)
b = Sin-1(0.695)
b = 44.0°
There are three practical applications where the sine ratio is applied. First, the sine ratio is essential in measuring the height of a building or a mountain. The distance of a building from the ground and the elevation angle determines the building's actual height using trigonometric functions.
Second, trigonometric ratios are applied in aviation. Aviation takes into account speed, direction, distance, and direction and speed of the wind. These different variables use the sine ratio to find solutions.
Third, trigonometric ratios are useful in oceanography. Looking at waves as large right-triangles, scientists can determine the height of waves in the ocean.
Knowing the sine trigonometric ratios is an integral part of understanding mathematics and the world around you.
]]>This brings us to the meaning of shifting functions. Shifting functions don't change the size and shape of the graph but rather its position.
A shift is an addition or subtraction to the x
or f(x)
component. When you shift a function, you're basically changing the position of the graph of the function. A vertical shift raises or lowers the function as it adds or subtracts a constant to each y
coordinate, while the x
coordinate remains the same. A horizontal shift moves the function right or left since it adds or subtracts a constant to each x
coordinate while keeping the Y coordinate unchanged. You can combine vertical and horizontal shifts in a single expression. If the constants are grouped with x
, then the shift is horizontal; otherwise, it is vertical.
Common functions include:
y = c
y = x
y = x^2
y = x^3
y = |x|
y = sqrt(x)
Consider quadratic functions and their associated parabolas. When you first graph quadratic functions, you start with the basic equation.
f(x) = x^2
Then, you make some related graphs such as
g(x) = -x^2 - 4x + 5
h(x) = x^2 - 3x - 4
k(x) = (x + 4)^2
In each case, the basic parabolic shape is the same. The only difference is where the vertex is and whether it is right-side up or upside down. If you've been doing hands-on graphing, you've probably started to notice some relationships between the equation and the graph. The topic of function transformations makes these relationships more explicit.
Let's start by looking at the function symbols for the basic quadratic equation.
f(x) = x^2
A function transformation, or translation, is a fancy way of saying that you change the equation a bit so that the graph moves.
To move the function up, add to the function: f(x) + b
is f(x)
moved up by b
units. The same is true for shifting the function downward; f(x) - b
is f(x)
shifted downward by b
units.
We can add a 3 to the basic quadratic equation f(x) = x^2
, going from the basic quadratic function x^2
to the transformed function x^2 + 3
.
This moves the function up three units.
To move a function left or right, a constant is added or subtracted from x
, respectively.
Let’s look at the equation y = (x + 3)^2
In this graph, f(x)
shifts three units to the left. Now, instead of graphing f(x)
we are graphing f(x + 3)
. This means the equation is now y = (x + 3)^2
, and the original graph is shifted three units to the left.
When moving a function to the left, always add to the function's argument: f(x + b)
shifts f(x)
b units to the left. The reverse is true for moving to the right, f(x - b)
is always f(x)
shifted b
units to the right.
Functions flipped over the x-axis and mirrored across the y-axis are called reflections. You find the equation for functions reflected over the x-axis by taking the original function's negative value.
If you reflect f(x) = x^2 + 2x - 3
over the x-axis, it becomes f(x) = -(x^2) - 2x + 3
. This always works for flipping a function upside down.
To illustrate how this transformation works, remember that f(x)
is the same as y
. By adding a minus sign to everything, you change all positive (upper axis) y
values to negative (lower axis) y
values, and vice versa. Any points on the x-axis will stay as they are, and only the off-axis points will move.
To reflect a function across the y-axis, let's consider the cubic function g(x) = x^3 + x^2 - 3x 1
.
If you replace the x
from the original function with -x
, you get
g(-x) = (-x)^3 + (-x)^2 - 3(-x) – 1
g(-x) = -x^3 + x^2 - (-3x) – 1
g(-x) = -x^3 + x^2 + 3x – 1
This transformation will flip the original graph across the y-axis. Any point on the y-axis stays on the y-axis; only off-axis points change sides.
f(x) + b
shifts the function up by b
units.
f(x) – b
shifts the function down by b
units.
f(x + b)
shifts the function to the left by b
units.
f(x - b)
shifts the function to the right by b
units.
-f(x)
reflects the position of the function on the x-axis (i.e., inverted).
f(-x)
reflects the position of the function on the y-axis (i.e., left and right swapped).
If the sun is shining, water tends to reflect everything. If you look at the mountains behind the lake from a distance, you will see that the mountains are reflected downward. In mathematical terms is reflected over the x-axis creating the same image, only flipped.
Changes in mirrors. When you look into a mirror, you see an accurate reflection of yourself. Most of us look in the mirror every day and are not even aware of this reflection.
A moving car. The car demonstrates transformation because it is moving, but it does not change in size or shape. It also does not flip sideways or upside down.
]]>A = (1/2)b*h
, where height is measured as a vertical line from the base to the opposite vertex. This formula makes calculating the area of a triangle relatively easy, but it is quite difficult to naturally find a triangle that has at least one known side (base) and a known height. Obtaining the length of all three sides of the triangle is often much easier. Fortunately, a formula does exist for calculating the area of a triangle when all three sides are known.
Heron's formula allows us to find the area of a triangle when we know the length of all three sides of that triangle. It turns out that we can calculate its area without proving any additional theorems or using trigonometry. The following equation is Heron’s formula.
A = sqrt(s(s-a)(s-b)(s-c))
Where s = (a+b+c)/2
or 1/2
the perimeter, and A
stands for area
The triangle DeltaABC
has sides a
, b
, and c
. We can create a formula for the area of triangle ABC using these three sides.
First, we'll recall the standard equation to find the area of a triangle - it is the base multiplied by the height of the triangle, all divided by two (1/2)bh
. We know all three side lengths, so we can use any of them as the base - in this case, we choose side b
. Since we don't have the height of the triangle, we draw a line for the height, h
, at a right angle to side b
up to the vertex of a
and c
. Drawing line h
creates two right triangles - one on the right and one on the left of h
.
Labeling the section of side b
from h
to the vertex of b
and c
as x
, we now have two right triangles. The left-hand triangle has sides h
, (b-x)
, and a
, and the right-hand triangle has sides h
, x
, and c
.
Using the Pythagorean theorem, we can derive expressions for x
and h
by simply using a
, b
, and c
.
The solution for x
:
From the right triangle on the right side, we get
h^2 + x^2 = c^2
, or h = sqrt(c^2 - x^2)
It is also possible to start with the left side of the right triangle.
h^2 + (b-x)^2 = a^2
, or h = sqrt(a^2 - (b-x)^2) = sqrt(a^2 - b^2 + 2bx - x^2)
Since h
is the same in both these triangle we can set the two equations for h
equal to each other.
sqrt(c^2-x^2) = sqrt(a^2-b^2+2bx-x^2)
Since the lengths here are all positive, we can square both sides and get c^2-x^2 = a^2-b^2+2bx-x^2
.
By simplifying and rearranging, we can solve for x
and get
x = (-a^2+b^2+c^2)/(2b)
.
The solution for h
:
Now we can substitute the equation for x
into one of our original right triangle equations above to solve for h
. Using the right-hand side triangle, h^2 + x^2 = c^2
, so h^2 = c^2 - x^2
. Plugging in the equation for x
, we get h^2 = c^2-((-a^2 + b^2 + c^2)/(2b))^2
.
The squared difference identity states that x^2-y^2 = (x+y)(x-y)
. Replacing x
and y
in this identity with b
and (-a^2+b^2+c^2)/(2b)
, respectively, we get
h^2 = (c+(-a^2+b^2+c^2)/(2b))(c-(-a^2+b^2+c^2)/(2b))
= ((2bc-a^2+b^2+c^2)/(2b))((2bc+a^2-b^2-c^2)/(2b))
. Now let's simplify and rearrange it a bit. Multiplying across and factoring out the denominator, we get
h^2 = (1/(4b^2))(2bc-a^2+b^2+c^2)(2bc+a^2-b^2-c^2)
.
From here we can factor and then apply the square difference identity again to get the following.
h^2 = (1/(4b^2))((b+c)^2-a^2)(a^2-(b-c)^2)
= (1/(4b^2)) ((b+c)+a)((b+c)-a)(a+(b-c))(a-(b-c))
It follows that h = (1/(2b))sqrt((b+c+a)(b+c-a)(a+b-c)(a+c-b)
.
Since the area of the triangle is (1/2)bh
, we can simply multiply them together to get
A = (1/4)sqrt((b+c+a)(b+c-a)(a+b-c)(a+c-b)
.
Simplifying Heron's Formula to its Final Form
So, let's play with the lengths to get a simpler formula and an interesting result.
Since a
, b
, and c
are the sides of a triangle, a+b+c
is its perimeter. Defining s
as half of the perimeter we get s = (a+b+c)/2
, or 2s = b+c+a
.
From this we can find that b+c-a = 2(s-a)
; a+b-c = 2(s-c)
, and a+c-b = 2(s-b)
. Plugging these into the equation for the area of DeltaABC
above, we get
A = (1/4)sqrt((2s)(2(s-a))(2(s-b))(2(s-c))
.
Then, with some simplification and a little rearranging we get
A = sqrt(s(s-a)(s-b)(s-c))
.
Now we have reached the final state of Heron’s formula, and we can use it to find the area of a triangle if the height is unknown, but all three sides are.
While Heron's formula is used to find the area of a triangle from just the lengths of three sides, it can also be used to find the height of a triangle.
If you have all three sides, use Herron's formula. Herron's formula has two parts. First, you must find the variable s
, which is equal to half the triangle’s perimeter. This formula is: s = (a+b+c)/2
.
For a triangle with side lengths a = 3
, b = 4
, and c = 5
,
s = (3+4+5)/2
s = (12)/2
s = 6
Then use the second part of Heron's equation, A = sqrt(s(s-a)(s-b)(s-c))
. A = sqrt(6(6-3)(6-4)(6-5))
A = sqrt(6(3)(2)(1)) = 6
The area of the triangle is 6 square units.
In addition to finding the area of a triangle when only the three sides are known, this formula also allows us to calculate the height of a triangle. To accomplish this, set the area to 1/2bh
and then solving for the height. This gives us the equation 1/2bh = sqrt(s(s-a)(s-b)(s-c))
or h = (2sqrt(s(s-a)(s-b)(s-c)))/b
For our example triangle with side lengths 3, 4, and 5, it is solved as follows.
(1/2)(4)h = sqrt(6(6-3)(6-4)(6-5)
2h = 6
h = 3
So, with side b
as the base, the height is 3.
Before getting into the theory of exponents, let's first look at a few notations and related terminology.
Let's take the term x^n
.
We read this as "x raised to the power of n," which means we multiply x n number of times.
x^n
=x⋅x⋅x...n times.
Here, x is any real number and n is a positive integer. We refer to x as "base," and n as "power" or "exponent."
Another example, we read 9^5
as "9 raised to the power of 5" and it is equal to 9⋅9⋅9⋅9⋅9.
9^5
= 9⋅9⋅9⋅9⋅9 = 59,049
There are eight fundamental rules of exponents.
A number to the 1 power is equal to the number itself.
x^n
= n where n = 1
We multiply x one time, which is equal to the number itself.
x^1
= x
Examples:
9^1
= 9
01 = 0
-1^1 = -1
Any number raised to the power of 0 (except zero) is equal to 1.
x^0 = 1
Examples:
9^0 = 1
1^0 = 1
Note: 0^0
is undefined.
When multiplying, you can add two exponents with the same base.
x^a⋅x^b = x^a+b
As the bases of both the terms are the same, we can add their powers.
Examples:
9^5⋅9^7 = 9^(5+7) = 9^12
8^-2⋅8^10 = 8^(-2+10) = 8^8
In special cases, we can apply this rule even when there are the same bases but with different signs (positive/negative). We just need to rewrite the terms in accordance with our rule.
For example:
Simplify: (-9)^5⋅(9)^7
We can write (-9)^5
as (-1⋅9)^5
which is equal to (-1)^5⋅(9)^5
.
This implies, (-9)^5⋅(9)^7 = (-1)^5⋅(9)^5⋅(9)^7 = (-1)^5⋅(9)^5+7 = (-1)^5⋅(9)^12 = -9^12
.
By subtracting the exponents, we can divide two powers with the same base.
x^a/x^b = x^(a-b)
Here the bases of both the terms under division are the same, so we subtract the exponents.
Examples:
9^5/9^7 = 9^(5-7) = 9^-2
2^2/2^2 = 2^(2-2) = 2^0 = 1
To raise a power to another power, we multiply the exponents.
(x^m)^n = x^(m⋅n)
We raise the inner term to an exponent and so we multiply both the exponents.
Examples:
(9^5)^7 = 9^(5⋅7) = 9^35
(8^2)^2 = 8^(2⋅2) = 8^4
Any non-zero number raised to a negative power equals its reciprocal raised to the opposite positive power.
x^-n = 1/x^n
x is raised to a negative exponent and we can write it as x's reciprocal with the opposite sign of n.
Examples:
9^-5 = 1/9^5
1^-1 = 1/1^1 = 1
When there is a power to a fraction, we can rewrite the term as both the numerator and denominator raised to the power of the exponent.
(x/y)^n = x^n/x^y
Examples:
(2/5)^5 = 2^5/5^5
(1/2)^2 = 1^2/2^2 = 1/2^2
When you have a fractional exponent, the numerator is the power and the denominator is the root (square root for denominator = 2, cube root for denominator = 3, and so on).
b^n/m = (^m √b)^n
As we studied the fundamental laws of exponents, we shall now implement these rules in the following examples. Observe the statement or equation carefully before rushing in to apply the rules. You may be able to solve what seems like a five-step problem in just two steps. This is especially true when dealing with exponents.
Simplify: ((x/y)^n)^(1/n)
Solution
From the power rule, we get (x/y)^(n⋅(1/n))
which is equal to (x/y)^1 = x/y
.
Simplify: ((x^a/x^b)x^c)^-d
Solution
From the negative exponent rule, we get 1/((x^a/x^b)x^c)^d
.
From the quotient rule, we get 1/((x^a-b)x^c)^d
.
From the product rule, we get 1/(x^a-b+c)^d
.
From the power rule, we get 1/x^((a-b+c)^d)
which equals to 1/x^(ad-bd+cd)
.
Hence, 1/((x^a/x^b)x^c)^-d = 1/(x^(ad-bd+cd))
Simplify: (((x⋅x⋅x⋅x⋅x)^y)^-1)⋅(((x⋅x⋅x⋅x⋅x)^y)^1)
Solution
Look carefully. We can simplify this in two steps without even simplifying the individual terms under multiplication. Upon observation, we find that the bases of the powers are equal and we can apply the rule of product.
(((x⋅x⋅x⋅x⋅x)^y)^-1)⋅(((x⋅x⋅x⋅x⋅x)^y)^1) = ((x⋅x⋅x⋅x⋅x)^y)^-1+1 = ((x⋅x⋅x⋅x⋅x)^y)^0
Applying the zero rule we get, ((x⋅x⋅x⋅x⋅x)^y)^0 = 1
Our aim is not just to solve problems or equations. Our aim is to solve them efficiently and effectively. It is true that we can solve or handle the statements without using these rules but that process is highly ineffective, complex, and time-consuming. So, we use these rules of exponents to solve and simplify equations and statements efficiently.
]]>When we define the trigonometric ratios, let us define a right-angled triangle with one of the angles named x. This angle is 90°. You define the sides of a triangle as a, b, and c where a is the side adjacent to x and b is the side opposite x. c is the hypotenuse or the side opposite the right angle. There are six fundamental trigonometric functions.
sin x = (opposite) / (hypotenuse) = b / c
cos x = (adjacent) / (hypotenuse) = a / c
tan x = (opposite) / (adjacent) = b / a
(b / c) / (a / c)
, you will get b/a
which is tan x. So tan x can be expressed as the ratio of sin to cos. tan x = sin x / cos x
.csc x = 1 / sin x
sec x = 1 / cos x
cot x = 1 / tan x
Out of the six fundamental trigonometric functions, you will mostly be concerned with sin, cos, and tan.
You can define a cosine function using a right-angled triangle as defined above. However, you can use cosine in several other applications.
You can use the cosine using differential equations. The cos and sin are the two differentiable trig functions and they have a special relationship.
cos x = ( d / dx ) sin x
and
-sin x = ( d / dx ) cos x
The above definitions are useful when solving differential equations. Both of the above expressions are solutions to the differential equation:
y” + y = 0
Trigonometric functions are also defined using power series. By applying the Taylor series to cosine, you can obtain another definition.
cos x = 1 – ( x2 / 2! ) + ( x4 / 4! ) – ( x6 / 6! )
…..
Euler had related the sine and cosine functions by the expression:
ejx = cos x + j sin x
e-jx = cos x – j sinx
The j in the above expressions refers to the imaginary unit, which is equivalent to the square root of (-1). Euler's expression or relationship is true for all complex values. This means that the formula is true for all real values of x.
If we add the above equations, we can find a concise expression for cos x in the complex domain as:
cos x = ( ejx + e-jx ) / 2
If the value of x is real, you can write the expression as:
cos x = Re( ejx )
Since a full circle is 360°, you can express the cosine in different parts of a circle starting at 0° up to 360°. In the first quadrant of a circle, angles from 0° to 90°, the value of cos is positive. In the second quadrant with a range of angles from 90° to 180°, the value of cos is negative. In the third quadrant with a range of angles from 180° to 270°, the value of cos is still negative. In the fourth quadrant, with the range of angles from 270° to 360°, the value of cos is positive.
Before I proceed, let me introduce a trigonometric identity. Trigonometric identities are relationships between the trigonometric functions which are true at all conditions. One of them is cos2 x + sin2 x = 1
. Let's look at a few examples and apply this trigonometric identity.
A right triangle has a sin of 0.866. Find the cosine of the angle.
Taking our trigonometric identity, we can rearrange the expression.
cos2 x = 1 – sin2 x
cos x = ( 1 – sin2 x )1/2
Since we know the value of sin x, let us substitute it for sin2 x in the expression.
cos x = ( 1 – sin2 x )1/2
cos x = (1 – 0.8662 )1/2
cos x = 0.5
A right triangle (ABC) has a right angle at B. The length of the hypotenuse, AC, is 5cm and the side BC is 3 cm. Find the angle at C.
To refresh your memory, the cosine of an angle is adjacent/hypotenuse. Let the angle at C be x.
cos x = 3 / 5
x = cos-1 ( 3 / 5 )
x = 53°
The angle at C is 53°.
The expression cos-1 means the cos inverse. It is the inverse of the cos function. If the cosine of an angle is x, then cos-1 x is the original angle.
cos 60° = 0.5
cos-1 0.5 = 60°
Find the cosine of the following angles using our circle quadrants.
660° is bigger than a circle, which is 360°. But since an angle is a degree of turning, it means that the point has moved a full circle and then some. The full circle will not count since the angle of interest is the amount it turned from the starting point to the final point.
So cos 660° = cos ( 660 – 360 )° = cos 300°
Since 300° falls within the fourth quadrant, it means that the value of cos is positive.
cos 300° = 0.5
For the second question, 234° is less than 360° so the moving point has not moved a full circle. Also, 234° falls within the third quadrant. Therefore, the value of cos is negative.
cos 234° = -0.588
The third problem has a negative angle of -60°. Negative angles mean that the direction of movement is clockwise instead of the normal anticlockwise. So if you move clockwise 60° you will end up in the fourth quadrant.
-60° = ( 360 – 60 )° = 300°
cos 300° = 0.5
You can easily find the cos of an angle by looking it up in a cos table or by pressing cos and the angle on a scientific calculator. On most scientific calculators, the cos-1 inverse function is a second function of the cos, usually on the same key. For such calculators, to use the cos inverse function, press SHIFT and COS on the calculator.
]]>ax² + bx + c = 0
Some quick examples of quadratic equations include:
2x² + 5x – 8 = 0
7x² + 9 = 0
xx² – 26 = 3x
For this particular article, I will show you how the discriminant affects the solutions to quadratic equations. The discriminant of a quadratic formula is the part of the quadratic formula that determines the root type in a quadratic equation (imaginary, real, singular).
Solutions to a quadratic equation are values of an unknown variable that make the equation true. There are four standard ways of finding the roots of a quadratic equation.
This method is applicable if you can factorize the coefficients of the quadratic equation as av + bx + c = a( rx + n )( px + m ) = 0
. Where n and m are the roots of the quadratic equation.
This method is useful when you cannot factorize the coefficients of the quadratic equation as shown above. In completing the square method, the quadratic equation is expressed in the form
ax² + bx + c = x2 + ( b / a )x + ( c / a ) = 0
x² + ( b / a )x + ( c / a ) = ( x +½b)2 + ( c / a ) – ( b² / 4 ) = 0
( x +½b )2 = ( b² / 4 ) – ( c / a )
Solving for x gives the roots of the quadratic equation.
You get the quadratic formula by completing the square method. If a quadratic equation is given as ax² + bx + c
then the roots of the quadratic equation are given by x = (-b+-(b²–4ac )1/2 )/2a
.
In this method, you plot the quadratic equation, and the points where the graph cuts the x-axis are the roots of the equation.
For the purpose of this topic, however, we will focus on the quadratic formula.
You can solve all quadratic equations using the quadratic formula method. Because of its versatility, we call it the almighty formula. You can find the roots of a quadratic equation using x = ( -b +- ( b² – 4ac )1/2 ) / 2a
.
The term b² – 4ac
under the square root determines the quadratic equation's roots and is the quadratic equation's discriminant. There are three possible outcomes for the discriminant.
b² - 4ac > 0
This happens when b²
is greater than 4ac. If this is the case, you'll get two real roots of the quadratic equation. This is true because the square root of any positive number is a positive number. If you plot the graph of the quadratic equation, it will cut the x-axis at two points.
b² – 4ac = 0
This happens when b²
is equal to 4ac. There is only one real root to the quadratic equation when this is your outcome. The square root of zero is zero. If you plot the quadratic equation graph, it will touch the x-axis at only one point.
(b² - 4ac) = 0
This happens when b²
is less than 4ac. This is a job for imaginary roots. The roots are imaginary since the square root of a negative number is an imaginary number. The graph of such a quadratic equation will not touch the x-axis.
Let's illustrate the different cases where the discriminant determines the roots of quadratic equations.
Find the following quadratic equations' roots:
x² + 7x + 3 = 0
3x² – 13x – 12 = 0
6y² + 10y = 0
Since we want to demonstrate how the discriminant affects a quadratic equation's roots, we will use the formula method to solve the problems above.
The quadratic formula is x = (-b + -(b² - 4ac)1 / 2) / 2a
x² + 7x + 3 = 0
a = 1, b = 7 and c = 3
Substitute the coefficient values of a, b, and c in the quadratic formula.
( -7 +- ( 72 – 4 * 1 * 3 )1/2 ) / ( 2 * 1 )
The discriminant here is ( 72 – 4 * 1 * 3 )
and comes to 37. Since 37 is greater than 0, it means that we have two real roots. Let us solve and get the roots!
( -7 +- ( 72 – 4 * 1 * 3 )1/2 ) / ( 2 * 1 )
( -7 +- 371/2 ) / ( 2 * 1 )
( -7 +- 6.08 ) / ( 2 * 1 )
The roots are
( -7 + 6.08 ) / ( 2 * 1 ) and ( -7 – 6.08 ) / ( 2 * 1 )
-0.46 and -6.54
The roots of x² + 7x + 3 = 0
are -0.46 and -6.54
3x² – 13x – 12 = 0
a = 3
, b = -13
and c = -12
After substituting the values for a, b and c in the formula, we have
( -(-13) +- ( -132 – 4 * 3 * -12 )1/2 ) / ( 2 * 3 )
( 13 +- ( 313 )1/2 ) / ( 2 * 3 )
( 13 +- 17.69 ) / ( 2 * 3 )
The roots are
( 13 + 17.69 ) / ( 2 * 3 ) and ( 13 – 17.69 ) / ( 2 * 3 )
The roots of 3x2 – 13x – 12 = 0
are 5.11 and -0.78
6y² + 10y = 0
a = 6
, b = 10
and c = 0
Coefficient c is zero, which is why it did not appear in the question.
After substituting the values for a, b, and c in the quadratic formula, we have this:
( -10 +- ( 102 – 4 * 6 * 0)1/2 ) / ( 2 * 6 )
( -10 +- ( 102 )1/2 ) / ( 2 * 6 )
The roots are as follows:
( -10 + 10 ) / 12 and ( -10 – 10 ) / 12
0 and -1.67
For all the questions, the discriminant was greater than 0. All the roots are real and in pairs.
Your quadratic equation is 2x2 + 4x + 2 = 0
.
a = 2
, b = 4
and c = 2
Substitute the coefficient values for a, b, and c in the quadratic equation.
( -4 +- ( 42 – 4 * 2 * 2 )1/2 ) / 2 * 2
( -4 +- 0 ) / 4
The roots are
( -4 + 0 ) / 4 or ( -4 – 0 ) / 4
The root of the quadratic equation is -1. In this example, the discriminant is equal to 0, and we arrived at only one root.
Find the roots of 3x2 + 2x + 7 = 0
.
a = 3
, b = 2
and c = 7
Input your values for a, b, and c into the quadratic formula.
( -2 +- ( 22 – 4 * 3 * 7 )1/2 ) / 2 * 2
( -2 +- ( -80 )1/2 ) / 4
( -2 +- 8.9j ) / 4
The roots are
( -2 + 8.9j ) / 4 and ( -2 – 8.9j ) / 4
The roots here are imaginary. They contain the imaginary variable j, which we define as ( -1 )1/2
or the square root of -1. We have arrived at imaginary roots because the discriminant was less than zero.
Understanding how the discriminant affects the outcome of your quadratic equation solutions is as easy as memorizing a formula. If you are ever faced with this mathematical task, always choose the quadratic formula.
]]>y=mx+b
, can look odd and confusing due to its use of variables, it is easy to use with proper explanation and practice. Even more, you can extract vast amounts of information from a single equation.
As mentioned previously, slope-intercept form, much like other forms used to model functions, uses variables to describe the meanings behind values in an equation. In order to use the slope-intercept form effectively, it is crucial that you understand and can determine what these variables mean for the equation as a whole.
Arguably the most important variable in the slope-intercept form is x. This is the input of the equation. You could say that x is the value that goes into a "machine" (which is, in this case, the function) to produce another value. For example, say you have the equation 2x = ?
which is the "machine" in this case. When you plug x into the equation as a value, say, for the purpose of this example, 2, you would replace x with 2 and proceed to do 2 times 2, getting 4. The x value that you plugin can be any number, positive or negative.
Now, just as the function "machine" has an input, it also has output, modeled by the variable y. Unlike x, you typically do not decide what the y value is. Instead, you allow the x value to determine what the output will be. Say that someone presents you with another function, 3x + 1 = y
. Notice that there are now two variables with y replacing the question mark of the previous example. When you chose an x value, for example, 3, you would plug this in and do 3(3) + 1
to get 10, which is the output, y.
To show this input-output relationship between two values, you would write x and y values as something called an ordered pair. Ordered pairs simply put the x value (input) and y value (output) of a particular function together, formatted as (x, y). In the equation and example above, you would write the ordered pair as (3, 10) per the format.
The ordered pair changes with the x value, as changing the input, in turn, changes the output. This means that for every function, there are numerous unique ordered pairs. Ordered pairs are important when it comes to plotting a function on a plane, which is a surface of two intersecting perpendicular lines, one horizontal and one vertical. On this plane, ordered pairs act as directions for the person graphing the equation. The x telling them to go that many units right (or, if x is negative, left), and y telling them to go that many units up (or, if y is negative, down) when starting at the axis (where the two lines of a plane intersect).
The m variable, known as the "slope" in the slope-intercept form, is more complex compared to the other variables. Simply put, the slope is the rate at which the y values of a function grow or decrease in relation to the x values. In other words, m describes how a function's ordered pairs change. You can find a line's slope or function by examining the rate of rise (how much the y value changes between two points) to run (how much the x value changes between those same two points). You do this by using the equation (y2 - y1)/(x2-x1)=m
where (x1, y1) and (x2, y2) are ordered pairs of a function.
For example, if you were given two ordered pairs, (3, 5) and (5, 8), of a single function, the equation would look like (8-5)/(5-3)
. When this is simplified, 3/2 is produced as the answer and the slope, m, for the function. Later, we will discuss the slope's importance and how you can use it in the slope-intercept form.
Of course, there is still the question of the last variable, b. This variable is the constant in the slope-intercept form, meaning that it is an unchanging number in the equation compared to the variables x and y. The main job of the variable b is to tell you, at a glance, what y equals when x is 0, known as the y-intercept.
The y-intercept is found by simply plugging in a 0 for x in whatever function someone gives to you. For example, if the function is y = 4x + 3
, you plug in 0 to get y = 3
as the y-intercept (the ordered pair would be (0, 3)). Think of the b value as a shortcut to finding the y-intercept; instead of replacing x with 0, you just have to remember that (0, b) is the y-intercept.
With a basic understanding of slope-intercept form variables, you can now use and apply these concepts to find information and create graphs.
Although we mentioned that you wouldn't normally be the one to decide the value of y, there are cases where you will need to. Normally, you use this method when you want to find the x value given a particular y value. Using the slope-intercept form, this is a fairly easy thing to do if given the function and y-value. For instance, if you are presented with the function y = 6x - 1
and are told to find x when y is 11, you would plug in y, giving you 11 = 6x - 1. Then after adding 1 to each side of the equation and dividing by 6, you would get x = 2
.
As mentioned previously, ordered pairs (x, y) produced by inputting values for x and getting an output, y, are used as directions for points on a plane. When you have multiple ordered pairs (two or more), you can plot these points and connect them to present what is known as a linear function, which is a straight line going through all of the points in a function. Linear functions are capped on either side by arrows, which indicate that the line goes on forever in both directions.
The slope is a key component of using slope-intercept, specifically when trying to figure out how a graph will look plotted on a plane. If you understand slope and its meaning, with a brief interpretation of m, it is possible to determine the line of a function.
A general rule when examining the slope is that the larger m is, the steeper the line will be. This makes perfect sense. The output, y, increases faster, even when x grows at the same rate. For example, if you have two equations, y = 2x
and y = 6x
, and x is 1 for both, y in the first function would be 2 (ordered pair would be (1, 2)) but, for the second, the y value would be the greater value, 6 (ordered pair would be (1, 6)). If you graphed both of these functions, you would see that y = 6x is far steeper because the y values increase faster.
When m is positive, as seen in the examples used thus far, it is a general rule that the line produced from the function will appear to increase as the x value increases (gets larger). The opposite is generally true when m is negative: the line will appear to decrease as x increases. This information is extremely important when you are trying to quickly determine how your graph will look without finding ordered pairs and/or graphing the function.
Slope-intercept form not only gives you the ability to find ordered pairs but also allows you to form a general idea of what your graph will look like with minimal effort. As long as you understand the variables and what they mean in the context of a graph, you can do almost anything using the slope-intercept form.
]]>x/3
makes it an algebraic fraction. In algebraic fractions, you cannot divide by zero. As such, the denominator has certain restrictions. Take a look at the illustrations below.
5/x
, x cannot be equal to zero (x≠0
)2/(x-3)
, x cannot be equal to three (x≠3
)3/(a-b)
, a-b
cannot be equal to 0 (a-b≠0
); therefore, a cannot equal b (a≠b
)4/(a²b)
, neither a nor b can equal 0 (a≠0
, b≠0
)Simplifying an algebraic fraction means writing the fraction in its most compact and efficient form without necessarily interfering with the original fraction's value. Simplifying makes it easier for you to understand and solve the fraction. Here are some tips.
Now that you can simplify the basic equation, it will be easier to identify the equation's components. The way you write an algebraic equation is known as algebraic notation.
a/b + c/d
is an algebraic notation. The algebraic notation has several components, and they include operators, parentheses, coefficients, variables, and exponents. If you are given an equation like X + PX* V2 – (W/X)
, X is the variable, + is the operators, P is the coefficient, V2 is the exponent, and (W/X
) is the parentheses.
X, which we now know as the variable, represents the unknown number. It helps find the missing number in an equation, which is always the algebra goal. Any letter can represent a variable, and the variable is used together with coefficients, but it does not mean the answer will be the same for both variables.
Coefficients group the same variables into one to simplify work, and operators determine how we will solve the problem. If there are several operators in one problem, we use PEMDAS. Parentheses are the same as brackets. So, if you come across a problem with parentheses, solve it first, and if two parentheses are close to each other, for example (v) (w), you will multiply the two. On the other hand, exponents are numbers multiplied by themselves, V2 in this case, is the same as V*V
.
Solve the following algebraic fraction: a/b +c/d*(z/p - v/w)
We have addition, subtraction, multiplication, and brackets in the above equation. According to PEMDAS, we must start by solving the equations in the parentheses.
z/p – v/w = zw /pw – vp /pw
After parentheses, we will add the remaining equation then multiply the answer we got from the parentheses. In both cases, we apply the law of distribution.
a/b + c/d = ad/ bd + bc/ bd
(ad/bd + bc/bd)*(zw/pw – vp/pw) = adbc/bd x (zw/pw- vp/pw)
Remember, you are simplifying, not solving.
Vx – w / Qx -w
You'll find common factors in the numerator and denominator. Therefore we will group the common factors.
V (x – w)/Q(x-w)
You can cancel out similar factors. This results in the final answer being V/Q.
Simplify the equation x2/Vy*PY/QX
You will divide (simplify) x2 by QX resulting in X. You will also divide Y by Vy to get V. So, the final answer will be:
X/V*P/ Q
XP/VQ
The factor not only has to be common but can also be the highest factor. Therefore, if X is the highest common factor among all the values, you use X to divide each number.
Given a rectangle has four sides and the measurements include w+x/v
and w-p/x
, write an expression for its area.
Since we are finding the area, we will directly multiply the given figures. If it were an addition, we would have added all four sides.
W + X/V*W - P/X
(W + X)/VX and (W - P)/VX
W(W - P) +X(W-P)
W2 - WP + XW – XP
WP + WP
(WP)2/VX
The following two examples require you to use the factorization method and FOIL method, that is, multiplying two binomials. If the denominators are the same, you will use the examples above and if the denominators are not the same and the question is complex, use the two examples stated. In complex problems, use the following steps:
Simplify the following complex fraction: (P + p/x)/ (P – p/x2)
Our first step is to create a single fraction in both the numerator and denominator. We will add the fractions in the numerator and subtract the denominator's fractions.
{P(x/x) + p/x} / {P (x2/x2) – p/x2}
{(x/x + p/x)}/ {(x2 + x2) - p/x2}
{(x + p)/x} {/(x2-p)/x2}
((x + p)/x}. {(x2/x2 – p)}
{(x+p/p)}. {x2/ (x + p) (x – p)}
x / (x – p)
You can choose to go for a more straightforward method: the least common divisor. In this case, the least common divisor is x2. You will multiply x2 in both the numerator and denominator and quickly solve it.
Solve the complex algebraic equation {p/x – p/qx}/ {n/x – w/rx}
.
We will first create a single fraction in the denominator and numerator, and afterward, solve the equation.
(p/x – p/qx)/ (n/x – w/rx) = {p/x. (q/q) -p/qx}/ (n/x(r/r) – w/rx)
You can choose to leave the answer as it is or simplify depending on the given question.
The complex questions may seem hard, but once you understand the essential details, such as factorization, it becomes easy to solve. You can also use the least common divisor where applicable or the greatest common denominator. After multiplication, you can expand the problem to simplify the question.
]]>Simply put, an imaginary number is the square root of a negative number and does not have a tangible value. Although imaginary numbers are not real numbers and you cannot quantify them on a number line, these numbers are "real." We use them all the time in advanced Mathematics classes.
For a time, the belief held that you can't get the square root of a negative number. This resulted from the "non-existence" of numbers that were negative after you squared them. It was impossible to work backward by taking the square root since every number was positive after you squared them.
So you couldn't square root a negative number and expect to come up with anything practical. But you don't have to worry about working out the square roots of negative numbers. To solve this problem, you can use a new number.
This new number was invented during the Reformation period. During this time, nobody believed that you could use this number for any "real world" use. It was strictly for making it easy to work out the computations in solving certain equations. As such, the new number was generally viewed as "a pretend number" invented only for convenience.
The new number we are talking about is called "i," denoting "imaginary." People believed that this number—the square root of a negative number—wasn't real. You write this imaginary number as:
i = √-1
i² = (√-1)² = -1
i² = (√-1)² = √(-1)² = √1=1
To determine the square root of a negative number in terms of the imaginary unit "i," use the following property where a represents any non-negative real number:
√-a = √-1.a =√-1.√a= I√a
With this information, we can write:
√-9 = √-1.9 = √-1.√9= i.3 =3i
We would expect that 3i squared equals -9. So (3i)² =9i² =9(-1) = -9
. You can write the square root of any negative number in terms of the imaginary unit. Such numbers are called imaginary numbers.
Since multiplication is commutative, the imaginary numbers are equivalent and are often misinterpreted as part of the radicand. To deal with this confusion, place the imaginary number in front of the radical, then solve the problem. Let's consider the complex number 21-20i
.
Solve the equation 21-20i
.
21-20i=x2
21-20i=(a+bi)²
21-20i=a²+(2ab)i+(b2)i²
i²=-1
by definition of "i", rearrange the equation to give: 21-20i=(a²-b²)+(2ab)i
.Now that both sides of the equation are in the same form, compare the coefficients to obtain two equations in a and b. You have a²-b²=21
(call this equation 1). Next, compare the imaginary parts of the equation (the coefficients of i). You have 2ab=-20
(call this equation 2). You'll now have two equations with two unknowns. You can solve the simultaneous equations for a and b.
b=-10/a
.a²-(-10/a)²=21
.(a²+4)(a²-25)=0
a²=-4 or a²=25
Remember the assumption that a and b are real numbers. a²=-4
has no solutions of interest to you. This means your solutions are a=5
and a=-5
. Now substitute each a value into your earlier expression for b. This means that when a=5
, b=-2
and when a=-5
, b=2
. Lastly, put a and b into the context of the question and get the solution as 5-2i
and -5+2i
.
If you have an equation with a single radical, follow the procedure below:
Solve the equation √7x -2 = 5
√7x + 4 = 7
7x +4 = 7²
7x + 4 = 49
7x = 45
x = 45/7
You can now substitute x = 45/7
into the original equation √7x -2 =5
. Thus: 7-2 = 5
To divide imaginary numbers, you multiply the numerator and denominator by the complex conjugate a - bi
. In this case, assuming a - bi
is a complex number, then you will have:
(a + bi) (a - bi) = a²+ b²
Divide
x = (5 - 3i) / 4 + 2i
Multiply top and bottom by 4 - 2i
(5 - 3i)(4 - 2i) / (4 + 2i)(4 - 2i)
(20-10i-12i+6i²) / (16 - 8i + 8i - 4i2)
(20 - 22i + 6i²) / 16 - 4i2
(14 - 22i) / 20
Also called complex numbers, imaginary numbers are applicable in real life. For instance, in quadratic planes, these numbers show up in equations that do not touch the x-axis. Imaginary Numbers are especially very useful in advanced calculus.
Imaginary Numbers are also very essential in electricity, especially in alternating current (AC) electronic devices. Here, the AC electricity alternates between positive and negative in a sine wave. As such, combining AC currents can be extremely challenging. So using imaginary currents and real numbers has helped solve this problem by making it possible to do the calculations to avoid electrocution.
Lastly, imaginary numbers are essential in signal processing. This is especially so if what you're measuring relies on cosine or sine wave. Signal processing is vital in cellular and wireless technologies as well as radar and brain waves.
Initially, imaginary numbers were considered impossible to solve. From this discussion, however, it's evident that they're not as complex as they seem. You can actually solve problems involving these types of numbers. Knowledge of imaginary numbers has deep significance and profound importance to the understanding of Physics and Mathematics.
]]>If you're looking at a large group of data, these terms can help you identify important benchmarks in your information. We'll explain what they really mean and examine the differences between the terms.
The mean is the sum of all the values in a dataset, divided by the total number of values in the data.
Typically, there's a range of values in any given set of data. The average is one measure of the center of a set of data. We also call this the mean. It can also be referred to as the arithmetic mean. Imagine you are a teacher and all of your students score differently on their exams. When your supervisor asks about your class's performance, you tell them the average score of the class. This is the class's mean or the mean score of students.
As mean takes into account every value in the data set, it is affected by extreme values or outliers. For example, if two students do poorly on the test, earning F's, this would lower the class mean.
The mean of 12, 16, 9, 13, 21
Mean = (12 + 16 + 9 + 13 + 21) / 5
Mean = 14.2
The mean of 100, 99, 99, 95, 94, 90, 88, 86, 80, 75, 63
Mean = (100 + 99 + 99 + 95 + 94 + 90 + 88 + 86 + 80 + 75 + 63) / 11
Mean = 88.09
The median is the middle value or number in a given data set organized sequentially. In order to find the median, we need to arrange the data set in either ascending or descending order. The middle value is the median.
Unlike mean, we do not take into account every value in the dataset to calculate the median. So, the median is not affected by outliers or extreme values. The median is considered as a positional average while the mean is considered as an arithmetic average
Every now and then, we'll have two numbers in the middle (when the number of values is even). Find the average of those two numbers (a+b/2) to find the median. You can also use the median to separate the given data set into two halves, one half has all the higher values and the other half has all the lesser values.
Find the median of the data 20 10 10 40 57
5/2 = 2.5
-> 3rd value in the data set)Median = 20
Find the median of the data 20 10 10 25 40 57
Median = (10+15)/2 = 25/2
Median = 12.5
The mode is the most common number or the most repeated number in a given data set. While the mean represents the entire dataset, the mode doesn't. Mean takes into consideration each and every value while mode doesn't. It is simply the value that appears most frequently.
There is always a mean and a median for a given dataset—but this is not the case with mode. There may be no mode if no value appears more than any other.
Find the mode of 15, 15, 19, 23, 24, 19, 16, 19, 23, 30, 19, 22
Find the mode of 16, 19, 93, 45, 63, 24, 87, 33, 52, 23, 11, 01, 100
Most of the time, our aim is to calculate the central tendency of a dataset. When working on a dataset, to measure central tendency, the mean is preferred over the other two entities because of the fact that it takes into consideration every single value in the dataset.
However, if the dataset contains outliers or extreme values, the median is preferred over the mean since it isn't affected by those extreme values or outliers. In any case, mean and median are preferred over mode for calculating the central tendency of the data. There is a formula that shows the mathematical connection between the three terms:
Mode = 3 x Median - 2 x Mean
The above formula is derived from the observation that the difference between the mean and mode is almost equal to three times the difference between the mean and median.
Mean - Mode = 3(Mean - Median)
Depending on the situation, you can choose the term that will best represent the sections of data that you're identifying.
]]>