A quadratic equation is a second-order polynomial. This means that the highest exponent of x is 2.

A typical quadratic equation is in the form:

$$Ax^2+Bx+C=0$$

They are three ways of solving the quadratic equation:

1. Factorization method

2. Completing the square method

Today, we shall be solving quadratic equations using the factorization method.

Example 1

Solve $x^2+4x-12=0$ by factorization

Solution:

By factorization

$x^2+4x-12=0$

$x^2+6x-2x-12=0$

$x(x+6)-2(x+6)=0$

$(x-2)(x+6)=0$

Now, we equate each factor to zero

$x-2=0$ or $x+6=0$

$x=2$ or $x=-6$

Example 2

$5x^2+7x-6=0$

Solution

By factorization

$5x^2+7x-6=0$

$5x^2+10x-3x-6=0$

$5x(x+2)-3(x+2)=0$

$(5x-3)(x+2)=0$

This translates to:

$5x=3$ or $x=-2$

$x=\frac{3}{5}$ or $x=-2$

In some cases, you need to substitute before solving the equation quadratically.

Here is an example:

Example 3

Solve $x^4+4x^2-12=0$

Solution.$x^4+4x^2-12=0$
$x^2(x^2)+4(x^2)-12=0$

Let $x^2$ be y

$y(y)+4(y)-12=0$

$y^2+4y-12=0$

By factorization

$y^2+4y-12=0$

$y^2+6y-2y-12=0$

$y(y+6)-2(y+6)=0$

$(y-2)(y+6)=0$

This translate to

$y-2=0$ or $y+6=0$

$y=2$ or $y=-6$

Recalled that $x^2$ is equal to y
So, when $x^2=2$ $$\sqrt{x^2}=\sqrt{2}$$
$$x=\sqrt{2}$$

Note: the square of a whole number cannot be negative, thus, $-6$ if extraneous
In other cases(as you will see below), the quadratic equation may be re-arranged.

Example 4

Solve $-25=4x^2+20x$ by factorization

Solution:

Re-arranging

$0=4x^2+20x+25$

$4x^2+20x+25=0$

$4x^2+20x+25=0$

$4x^2+10x+10x+25=0$

$2x(2x+5)+5(2x+5)=0$

$(2x+5)(2x+5)=0$

This translate to

$2x=-5$

Divide both sides by 2

$\frac{\require{cancel}\bcancel{2}x}{ \require{cancel}\bcancel{2}}=\frac{-5}{2}$
$x=\frac{-5}{2}$

In some cases, you may be given an incomplete quadratic equation
The next example will illustrate an incomplete quadratic equation

Example 5
Solve $4x^2-25=0$ by factorization

Solution:

This is a difference of two states $(2x)^2$, $(5)$, hence

$(2x+5)(2x-5)=0$

Hence, either

$(2x+5)=0$ or $(2x-5)=0$

$2x=-5$   or $2x=5$

This translate to

$x=\frac{-5}{2}$ or $x=\frac{5}{2}$
Related post

Example 6

Find the solution to $x^2+\frac{14x}{4}+\frac{49}{16}=0$

Solution:
First, let take L.C.M

$x^2+\frac{14x}{4}+\frac{49}{16}=0$

$\frac{16x^2+56x+49}{16}=0$

Cross multiply

$16x^2+56x+49=0\times16$

$16x^2+56x+49=0$

By factorization

$16x^2+28x+28x+49=0$

$4x(4x+7)+7(4x+7)=0$

$(4x+7)^2$=0

This translates to

$4x=-7$

$x=\frac{-7}{4}$

Example 7

Solve this: $9x^2+18x+19=26$

Solution:

First, let's rearrange

$9x^2+18x+19-26=0$

$9x^2+18x-7=0$

By factorization

$9x^2-3x+21x-7=0$$3x(3x-1)+7(3x-1)=0$

$(3x+7)(3x-1)=0$

$3x+7=0$ or $3x-1=0$

$3x=-7$ or $3x=1$

$\frac{\require{cancel}\bcancel{3}x}{\require{cancel}\bcancel{3}}=\frac{-7}{3}$ or  $\frac{\require{cancel}\bcancel{3}x}{\require{cancel}\bcancel{3}}=\frac{1}{3}$

$x=\frac{-7}{3}$ or $\frac{1}{3}$.

There you have it! You should be able to solve the quadratic equation by now. If not, then you can always go back to the examples above.

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