Quadratic equations are where the degree of the polynomial is $2$.  The standard form of the quadratic equation is $ax^2 + bx + c$ = $0$ where $a \neq 0$.  Solving the quadratic equation we can get $1, 2$ or $0$ real solutions.  There are many methods of solving quadratic equations like

• Solving by Graphing

• Solving by factorization

• Solving by square root method

• Solve by completing the square.

This lesson we are going to concentrate on solving the quadratic equation by factorization and by completing the square.

## Solve the Quadratic Equation by Factorization

Let us take few examples on solving quadratic equation by factoring. These examples are arranged in an order for easy understanding and slowly moving up a level.
Example 1:

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

Solution:

As we can see this quadratic equation is already in factored form, we can use the zero product property or the null product property which states is $a \times b$ = $0$ then either $a$ = $0$ or $b$ = $0$.

We have $(x - 2)(2x - 5)$ = $0$

Therefore, $x - 2$ = $0$ or $2x - 5$ = $0$

We can solve for $x$ and get it as $x$ = $2$ or $x$ = $\frac{5}{2}$

Note in solving the quadratic equations by factoring the zero product rule plays a great part and should be used.
Example 2 :

$x^2 – 2x$ = $0$

Solution:

Step 1: Factorize by taking out the Greatest Common Factor.

We get $x(x - 2)$ = $0$
Step 2: Apply zero product rule

$x$ = $0$ or $x - 2$ = $0$
Step 3: Solve for $x$

$x$ = $0$ or $x$ = $2$
Example 3:

Solve $3x^2$ = $15x$

Solution:

This problem is to first written by bringing all the variables to one side

$3x^2$ = $15x$

$3x^2 - 15x$ = $0$

Factor out GCF here it is $3x$

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

Now the Zero product rule

$3x$ = $0$ or $x - 5$ = $0$

Solve for $x$

$x$ = $0$ or $x$ = $5$
Example 4:

Solve $3x^2 - 27x$ = $0$

Solution:

Factor out the GCF here it is $3$.

We get $3(x^2 - 9)$ = $0$

Here we can apply the difference of squares rule which states $(a^2 - b^2)$ = $(a - b)(a + b)$

$3(x^2 - 3^2)$ = $0$

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

$(x - 3)(x + 3)$ = $0$ (divide by $3$ on either side)

Now the zero product property

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

solve for $x$

$x$ = $3$, or $x$ = $-3$
Example 5:

Solve $x^2 - 13x + 36$ = $0$

Solution:

Find two factors of $36$ which add up to $-13$. We can see that $-9 \times -4$ = $36$ and $-9 + (-4)$ = $-13$.

Therefore,

$x^2 – 13x + 36$ = $0$

$(x-4)(x-9)$ = $0$

Zero product rule

$x - 4$ = $0$ or $x - 9$ = $0$

Solve for $x$

$x$ = $4$ or $x$ = $9$
Example 6:

Solve $6x^2 - 5x + 1$ = $0$

Solution:

$6x^2 - 5x + 1$ = $0$

First find the product of $6 \times 1$ = $6$. $-2 \times -3$ = $+6$ and $(-2) + (-3)$ = $-5$

$(6x -2)(6x-3)$

Factor out and eliminate the common factors from each set of parantheses or the factorization would turn up incorrect. Factoring out and canceling $2$ from the first parenthesis and $3$ from the second we get

$(3x - 1)(2x - 1)$ = $0$

Now apply the Zero product rule

$3x - 1$ = $0$ or $2x - 1$ = $0$

Solve for $x$

$x$ = $\frac{1}{3}$ or $x$ = $\frac{1}{2}$

Note if this method of factorization is not comfortable one can try $AC$ Method, splitting the middle term and factor by grouping

$6x^2 - 5x + 1$ = $0$

$6x^2 - 3x - 2x + 1$ = $0$

$(6x^2 - 3x) + (-2x + 1)$ = $0$

$3x(2x - 1) -1 (2x - 1)$ = $0$

$(2x - 1)(3x - 1)$ = $0$

Then we have to follow the same procedure as the above.

## Solve Quadratic Equation by Completing The Square

Before starting this one needs to understand the binomial square

$(x + b)^2$ = $x^2 + 2bx + b^2$

$(x - b)^2$ = $x^2 - 2bx + b^2$

We need to get a part of the given quadratic equation to the binomial square form $x^2 + 2bx + b^2$ or $x^2 - 2bx + b^2$ this would help us write them as a square term.

Let us learn this with few examples:
Example 7:

Solve $x^2 + 4x - 10$ = $0$

Solution:

Step 1: Move the constant to the other side of the equation

$x^2 + 4x$ = $10$
Step 2: Divide the coefficient of $x$ by $2$ and square it

$\frac{4}{2}$ = $2$ and we square it to $2^2$.
Step 3: Add the result to both sides of the equation

$x^2 + 4x + 2^2$ = $10 + 2^2$
Step 4: The left side would be a perfect square because we introduce the missing square term.

$(x + 2)^2$ = $10 + 4$

$(x + 2)^2$ = $14$

Step 5: Take square root on both sides, not to forget to put ± in front of the square root.

$x + 2$ = $\pm \sqrt{(14)}$
Step 6: Solve for $x$

$x$ = $-2 \pm \sqrt{(14)}$

Solution can be written as $x$ = $-2\ -\ \sqrt{(14)}$ or $x$ = $- 2\ +\ \sqrt{(14)}$

Note: This method works fine when the coefficient of $x^2$ is $1$.  When the coefficient of $x^2$ is not equal to $1$, we should divide through out by the coefficient of $x^2$ and make the coefficient as $1$ and we can solve as we did.
Example 8:

Solve $2x^2 - 5x - 7$ = $0$

Solution:

Step 1: Since the coefficient of $x^2$ is not $1$, we divide throughout by the  coefficient of $x^2$

$x^2\ -$ $\frac{5}{2}$ $x\ -$ $\frac{7}{2}$ = $0$
Step 2: Move the constant to the other side

$x^2\ -$ $\frac{5}{2}$ $x$ = $\frac{7}{2}$
Step 3: Divide $x$ coefficient by $2v$ and square it

$x^2\ -$ $\frac{5}{2}$ $x\ +$ $(\frac{5}{4})^2$ = $\frac{7}{2}$ $+$ $(\frac{5}{4})^2$

Step 4: $(\frac{x - 5}{4})^2$ = $\frac{7}{2}$ $+$ $\frac{25}{16}$
$(\frac{x - 5}{4})^2$ = $\frac{(56 + 25)}{16}$
$(\frac{x - 5}{4})^2$ = $\frac{81}{16}$
Step 5: Take square root on both the sides

$\frac{x - 5}{4}$ = $\pm$ $\sqrt{(\frac{81}{16})}$
$\frac{x - 5}{4}$ = $\pm$ $\frac{9}{4}$

Step 6: Solve for $x$

$x$ = $(\frac{5}{4})$ $\pm$ $(\frac{9}{4})$
$x$ = $\frac{(5 + 9)}{4}$ or $x$ = $\frac{(5 - 9)}{4}$
$x$ = $\frac{7}{2}$ or $x$ = $-1$