The term \"quadratic\" traces from the Latin word \"quad,\" which means \"square.\"<\/p>\n
This is because the variable gets squared(X\u00b2).<\/p>\n
A quadratic equation is, thus, sometimes referred to as Equation of Degree 2 since the greatest power is 2 (having one or more variables raised to the second power).<\/p>\n
The standard form of any quadratic equation must be expressed as AX\u00b2+ BX + C\u22600, where A, B, and C are values, except that A can't be equal to zero, and X is unknown (yet to be solved).<\/p>\n
There are, basically, three methods of solving Quadratic Equations by Factoring:<\/p>\n
This method is mainly used by students who find it challenging to use the guessing method, (or the trial and error method). Unlike the trial and error method, the Product Sum Method is generally easier to apply since it identifies an equation that cannot be factored.<\/p>\n
This method takes various forms, i.e:<\/p>\n
Case 1: X\u00b2 + BX + C= 0 ( A= 1).<\/p>\n
Case 2: AX\u00b2 + BX + C = 0 ( A \u22601).<\/p>\n
Simply follow these steps when solving an equation using the product sum method:<\/p>\n
Step One: Find two integers whose product is C.<\/p>\n
Step Two: Give the integers any characters of your choice, for example, M and N.<\/p>\n
Step Three: Make one factor ( X + M ) and the other ( X + N).<\/p>\n
Find the value of x by factorization.<\/p>\n
X\u00b2 + 16X + 55= 0<\/p>\n
Solution:<\/p>\n
Find two integers whose product is 15. The table below shows the numbers.<\/p>\n
Below are the pairs of the numbers<\/p>\n
1 and 5<\/p>\n
55 and 11<\/p>\n
-1 and -5<\/p>\n
-55 and -11<\/p>\n
<\/p>\n
You can then select the pair that has the sum of 16 and product 55.<\/p>\n
That pair is 5 and 11.<\/p>\n
Therefore, the factors are X\u00b2 + 16X +55 = (X +5)(X +11) = 0,<\/p>\n
Thus;<\/p>\n
X= - 5 or X = - 11.<\/p>\n
<\/p>\n
Find the value of X in X\u00b2 -16X +60 =0<\/p>\n
Solution<\/p>\n
Identify a duo of integers whose product is 60; the pairs are listed below.<\/p>\n
You can have a table with different values for different pairs.<\/p>\n
You should then select a pair that has sum -16.<\/p>\n
Thus, the pair is:<\/p>\n
-6 and \u2013 10.<\/p>\n
Therefore;<\/p>\n
X\u00b2 -16X + 60 =(X- 6)(X-10) = 0<\/p>\n
Hence X = 6 or X = 10.<\/p>\n
If the equation AX\u00b2+BX + C =0 and A\u22601 you only need a little extra effort to find the factors using the product sum method.<\/p>\n
Here are the steps to follow:<\/p>\n
To illustrate this case, let's consider the following examples.<\/p>\n
Find the value of X given that 2X\u00b2+ X -10=0<\/p>\n
Find two integers whose product AC= (2)\u00d7(-10)=-20. You should then draw a table on your working paper to come up with several pairs.<\/p>\n
You can now select the pair that has the sum of B = 1. This pair is - 4 and 5.<\/p>\n
Rewrite the expression as 2X\u00b2- 4X + 5x - 10 = 0<\/p>\n
Taking out GCF, we get 2X(X-2)5(x-2).<\/p>\n
Now we have the common parenthesis, which is X-2.<\/p>\n
Therefore,(2X +5)(X-2)= 0 where X = 2 or X = -5\/2<\/p>\n
Calculate the value of X given that 3X\u00b2+X-2 =0<\/p>\n
Find two integers whose product is AC= 3 \u00d7-2=-6<\/p>\n
Next, list the pairs in a tabular form.<\/p>\n
Select the pair that has the sum of B=1. This pair is 3 and -2<\/p>\n
You should then rewrite the function as 3X\u00b2+3X-2X-2= 0.<\/p>\n
Taking out GCF, we get:<\/p>\n
3X( X+1)-2 ( X+1)=0<\/p>\n
Now the common parenthesis is X+1;<\/p>\n
Therefore ( 3X-2)(X+1)=0<\/p>\n
Thus X = 1 or X =2\/3<\/p>\n
This method involves arranging the terms into smaller groupings with common factors. Use the factoring by grouping method if you can't find the common factor for all the terms.<\/p>\n
Further, by taking two terms at the same time, you can get something to divide the terms.<\/p>\n
Use this easy procedure in solving the equation by factorizing using the grouping method.<\/p>\n
Suppose you are given a general equation AX\u00b2 +BX + C;<\/p>\n
First Pair \ud83d\ude41 AX\u00b2+QX)+(PX+C)<\/p>\n
Second Pair: X(AX+Q)+(PX+C)<\/p>\n
Depending on your selection of P and Q, you will factor out a constant on the second parenthesis, remaining with two identical expressions as shown in the example below:<\/p>\n
Find the value of X given 5X\u00b2 + 11X +2= 0<\/p>\n
First find the product AC;<\/p>\n
5 \u00d72=10<\/p>\n
Then think of two factors of 10 that can add up to 11<\/p>\n
Next, write 11X in the product of 10 and 1.<\/p>\n
Hence 5X\u00b2+1X + 10X +2.<\/p>\n
You should now group the pairs into two.<\/p>\n
(5X\u00b2+ 1X) + (10 X +2)<\/p>\n
After grouping, take out the common factor.<\/p>\n
X (5X+1)+2(5X +1)<\/p>\n
Thus, (X +2)(5X+1)=0<\/p>\n
Therefore X =-2 or X= -1\/5<\/p>\n
<\/p>\n
Compute the value of X given that X\u00b2+2X-24=0<\/p>\n
First, find the product AC = 1\u00d7-24=-24<\/p>\n
Think of two factors, such that their product is -24 and their sum is 2.<\/p>\n
Let the factors be -4 and +6<\/p>\n
Next, write +2X in the form -4X and 6X<\/p>\n
Therefore, the expression becomes X\u00b2- 4X +6X-24=0<\/p>\n
Pair the equation into 2 terms, thus:<\/p>\n
(X\u00b2 -4X)+(6X-24)<\/p>\n
Next, take out the common factor.<\/p>\n
X(X-4)+6(X-4)<\/p>\n
Now, X-4 becomes the common parenthesis.<\/p>\n
Therefore (X+ 6)(X-4)=0<\/p>\n
Thus, X=4 or X= -6<\/p>\n
The Special Product Method requires special cases that can be factored quicker. You can do this using two special quadratics:<\/p>\n
Identify the value of X given that X\u00b2+ 14X +49=0<\/p>\n
The first term is X\u00b2 and the last term is 49; both X\u00b2 and 49 are perfect squares whose roots are X and 7 respectively.<\/p>\n
The middle term, 14X is two times the roots of the other terms.<\/p>\n
Thus,<\/p>\n
X\u00b2 +14X+49=(X+7)\u00b2 = 0<\/p>\n
Therefore (X+7) (X+7) =0<\/p>\n
This X =-7<\/p>\n
Special Product Method is used here since:<\/p>\n
Identify the value of X given that X\u00b2-16=0<\/p>\n
Note that in the quadratic equation above:<\/p>\n
Thus:<\/p>\n
X\u00b2-16=(X-4)(X+4)=0<\/p>\n
Therefore X= 4 or X=-4<\/p>\n
Find the value of X given that X2-64=0<\/p>\n
The terms X\u00b2 and 64 are perfect squares and they are subtracted.<\/p>\n
Thus X2-64= (X-8)(X+8)<\/p>\n
Therefore X = 8 or X-8<\/p>\n
Quadratics are considered to be among the most challenging concepts in Mathematics.<\/p>\n
Regardless, getting the correct methods and learning how to apply the concepts can make teaching and learning Mathematics fun!<\/p>\n
Consequently, knowledge of quadratics is important in everyday life.<\/p>\n
For instance, quadratics is used in the determination of profits or even in formulating the speed and velocity of an object.<\/p>\n
Further, quadratics have been applied to athletic endeavours like shotput and javelin. Take time to learn the various methods to solve quadratic equations using the Factorizing method and you'll come to love it!<\/p>\n