# 1. Solving Quadratic Equations by Factoring

The general form of a quadratic equation is

ax^{2}+bx+c= 0

where * x* is the variable and *a*, *b* & *c* are constants

### Examples of Quadratic Equations

(a) 5*x*^{2} − 3*x* − 1 = 0 is a
quadratic equation in quadratic form where

`a = 5`, `b = -3`, `c = -1`

(b) 5 + 3*t* − 4.9*t*^{2} = 0 is a
quadratic equation in quadratic form.

Here, `a = -4.9`, `b = 3`, `c = 5`

[This equation arose from finding the time when a projectile, being acted on by gravity, hits the ground.]

(c)_{ } (*x* + 1)^{2} = 4 is a quadratic
equation but **not** in quadratic form.

It has to be expanded and simplified to:

x^{2}+ 2x− 3 = 0

### Summary

In general, a quadratic equation:

- must contain an
*x*^{2}term - must NOT contain terms with degrees higher than
*x*^{2}eg.*x*^{3},*x*^{4}etc

### Examples of NON-quadratic Equations

*bx*− 6 = 0 is NOT a quadratic equation because there is no*x*^{2}term.*x*^{3}−*x*^{2}− 5 = 0 is NOT a quadratic equation because there is an*x*^{3}term (not allowed in quadratic equations).

## Solutions of a Quadratic Equation

The solution of an equation consists of all numbers (roots) which **make the equation true**.

All quadratic equations have 2 solutions (ie. 2 roots). They can be:

- real and distinct
- real and equal
- imaginary (complex)

### Example 1

The quadratic equation *x*^{2} − 7*x* + 10 = 0 has roots of

`x = 2` and `x = 5`. (We'll show below how to find these roots.)

This can be seen by substituting in the equation:

When *x* = 2,

*x*^{2} − 7*x* + 10

= (2)

^{2}− 7(2)+ 10= 4 − 14 + 10

= 0

(This can be shown similarly for *x* = 5). In this example, the roots are **real **and **distinct**.

### Example 2

The quadratic equation *x*^{2} − 6 *x* + 9 = 0 has **double roots** of *x* = 3 (both roots are the same)

This can be seen by substituting *x* = 3 in the
equation:

*x*^{2} − 6*x* + 9

= (3)

^{2}− 6(3) + 9= 9 − 18 + 9

= 0

### Example 3

The quadratic equation

x^{2}+ 9 = 0

has **imaginary roots** of

`x=sqrt(-9)` or `-sqrt(-9)`

Learn more about imaginary numbers.

## Solving a Quadratic Equation by Factoring

For the time being, we shall deal only with quadratic equations that can be factored (factorised).

If you need a reminder on how to factor, go back to the section on:

Factoring Trinomials.

Using the fact that a product is zero if any of its factors is zero we follow these steps:

(i) Bring all terms to the left and simplify, leaving zero on the right side.

(ii) Factorise the quadratic expression

(iii) Set each factor equal to zero

(iv) Solve the resulting linear equations

(v) Check the solutions in the original equation

### Example 4

Solve *x*^{2} − 2*x* − 15 = 0

Answer

*x*^{2} − 2*x* − 15 = 0

Factoring gives:

(*x* − 5)(*x* + 3) = 0

Now, if either of the terms (*x* − 5) or (*x* + 3) is 0, the product is zero. So we conclude:

(

x− 5) = 0, therefore

x= 5

or

(

x+ 3) = 0, therefore

x= − 3

Hence the roots are * x* = 5 and *x* = − 3.

Are we correct?

We check the roots in the original equation by substitution.

When *x* = 5:

*x*^{2} − 2*x* − 15

= (5)

^{2}− 10 − 15= 25 − 10 − 15

= 0

(Similarly, when we substitute `x = -3`, we also get `0`.)

### Alternate method (Po-Shen Loh's approach)

We could have proceded as follows to solve this quadratic equation. The following approach takes the guesswork out of the factoring step, and is similar to what we'll be doing next, in Completing the Square.

**Step 1:** Take 1/2 of the absolute value of the *x* coefficient. In this case, `1/2 × |−2| = 1`

**Step 2:** Expand `(1 − u)(1 + u) = 1 − u^2`

**Step 3:** Set that expansion equal to the constant term: `1 - u^2 = -15`

**Step 4:** Solve for `u`:

`1 - u^2 = -15`

`u^2 = 16`

`u= +-4`

**Step 5:** Substitute either value (we'll use `+4`) into the `u` bracket expressions, giving us the same roots of the quadratic equation that we found above:

`x=(1-u)=1-4 = -3,` or

`x=(1+u)=1+4 = 5`

For more on this approach, see: A Different Way to Solve Quadratic Equations (video by Po-Shen Loh).

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### Example 5

Solve

`9x^2+ 6x + 1 = 0`

Answer

9*x*^{2} + 6*x* + 1 = 0

Factoring gives:

(3*x* + 1)(3*x* + 1) = 0

So we conclude:

(3

x+ 1) = 0,

therefore

`x=-1/3`

We say there is a **double root** of `x = -1/3`.

### Alternate method (Po-Shen Loh's approach)

We first need to re-express the quadratic with a `1` as the coefficient of `x^2.`

`x^2+ 2/3x + 1/9 = 0`

**Step 1:** Take 1/2 of the absolute value of the *x* coefficient. In this case, `1/2 × |2/3| = 1/3`

**Step 2:** Expand `(1/3 − u)(1/3 + u) = 1/9 − u^2`

**Step 3:** Set that expansion equal to the constant term: `1/9 - u^2 = 1/9`

**Step 4:** Solve for `u`:

`1/9 - u^2 = 1/9`

`u^2 = 0`

`u= 0`

**Step 5:** Substitute `u=0` into the `u` bracket expressions, giving us the same (repeated) root for the quadratic equation that we found above:

`x=1/3-0 = 1/3,` or `x=1/3+0 = 1/3`

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### Example 6 (involving fractions)

Solve

`2-1/x=3/(x+2)`

Answer

`2-1/x=3/(x+2)`

Multiply throughout by `x(x+2)` to remove the denominators (bottoms) of the fractions:

`2x(x+2)-(x(x+2))/x=(3(x)(x+2))/(x+2)`

Cancelling gives:

`2x(x+2)-(x+2)=3x`

Expanding the brackets:

`2x^2+4x-x-2=3x`

`2x^2-2=0`

`x^2-1=0`

Factoring gives:

`(x+1)(x-1)=0`

So `x = -1` or `x = 1`.

CHECK: Substituting `x = -1` into both the left hand side and right hand side of the question gives:

`"LHS"=2-1/x=2-1/-1=3`

`"RHS"=3/(x+2)=3/(-1+2)=3="LHS" `

Likewise, for `x = +1`,

LHS `= 2 - 1 = 1`

RHS `= 3/3 = 1 =` LHS

### Exercises

- Determine if the following are quadratic equations. If so,
determine
*a*,*b*, and*c.*

a. 5

x^{2}= 9 −xb. (3

x− 2)^{2}= 2

Answer

**Q1a** `5x^2= 9 − x`

`5x^2+ x − 9 = 0`

So, yes, it is a quadratic equation with

`a = 5`, `b = 1`, `c = -9`

**Q1b** `(3x − 2)^2= 2`

`9x^2− 12x + 4 = 2`

`9x^2− 12x + 2 = 0`

So, yes, it is a quadratic equation with

`a = 9`, `b = -12`, `c = 2`

- Solve for
*x*:

2

x^{2}− 7x+ 6 = 3

Answer

2*x*^{2} − 7*x* + 6 = 3

2

x^{2}− 7x+ 3 = 0(2

x− 1)(x− 3) = 0

So

`x=1/2` or `x=3`.

### Alternate method (Po-Shen Loh's approach)

Once again, we first need to re-express the quadratic with a `1` as the coefficient of `x^2.`

`x^2- 7/2x + 3/2 = 0`

**Step 1:** Take 1/2 of the absolute value of the *x* coefficient. In this case, `1/2 × |-7/2| = 7/4`

**Step 2:** Expand `(7/4 − u)(7/4 + u) = 49/16 − u^2`

**Step 3:** Set that expansion equal to the constant term: `49/16 - u^2 = 3/2`

**Step 4:** Solve for `u`:

`49/16 - u^2 = 3/2`

`u^2 = 25/16`

`u= +-5/4`

**Step 5:** Substitute either value (we'll use `+5/4`) into the `u` bracket expressions, giving us the same roots of the quadratic equation that we found above:

`x=7/4-5/4 = 1/2,` or `x=7/4+5/4 = 3`

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