## Solving Quadratic Equations by Factoring

In the previous post, we have learned how to solve quadratic equations by extracting the roots. In this post, we are going to learn how to solve quadratic equations by factoring.

To solve quadratic equations by factoring, we need to use the zero property of real numbers. It states that the product of two real numbers is zero if at least one of the two real numbers is zero. In effect, we need to transfer all the terms to the lefth hand side, let the right hand side be 0, equate factored form zero and find the value of x.

Example 1: $2x(x + 5) = 0$

Solution
$2x = 0, x = 0$
$x + 5 = 0, x = -5$.

This means the solutions of $2x(x+5) =0$ are $0$ or $-5$.

Example 2: $x^2 - x = 6$

Solution
Subtracting $6$ from both sides, we have
$x^2 - x - 6 = 0$.

Factoring, we have
$(x - 3)(x + 2) = 0$
$x - 3 = 0, x = 3$
$x + 2 = 0, x = -2$.

This means the solutions of $x^2 - x = 6$ are $3$ or $-2$.

Example 3: $x^2 = x \sqrt{3}$

Solution

Subtracting $\sqrt{3}$ from both sides,
$x^2 - x \sqrt{3} = 0$.

Factoring out $x$, we have
$x(x - \sqrt{3}) = 0$.

Equating both expression to 0, we have
$x = 0$
$x - \sqrt{3} = 0, x = \pm \sqrt{3}$.

This means the solutions of $x^2 - x = 6$ are $0$ or $\pm \sqrt{3}$.

Example 4: Solve $x^2 = 16$.

Solution

$x^2 - 16 = 0$
$(x + 4)(x - 4) = 0$
$x = -4, x = 4$.

This means the solutions of $x^2 = 16$ are $-4$ or $4$.

Example 5: Solve $x^2 + 2x + 1 = 0$.

Solution
$x^2 + 2x + 1 = 0$
$(x + 1)(x + 1) = 0$
$x + 1 = 0, x = -1$
$x + 1 = 0, x = -1$
$x = -1, x = -1$.

This means the solutions of $x^2 + 2x + 1 = 0$ are $katex -1$ or $-1$.

Example 6: Solve $(x - 3)^2 = 4x$

Solution
$(x - 3)^2 = 4x$
$x^2 - 6x + 9 = 4x$
$x^2 - 6x - 4x + 9 = 0$
$x^2 - 10x + 9 = 0$
$(x - 1)(x - 9) = 0$
$x = 1, x = 9$.

This means the solutions of $(x - 3)^2 = 4x$ are $1$ or $9$.

Example 7: $\frac{2x - 3}{x + 3} = \frac{x + 3}{x - 3}$
Solution

Cross multiplying, we have
$(2x - 3)(x - 3) = (x + 3)(x + 3)$.

Expanding, we have
$2x^2 - 9x + 9 = x^2 + 6x + 9$.

Transposing all the terms to the left hand side, we have
$x^2 - 15x = 0$
$x(x - 15) = 0$
$x = 0$
$x - 15 = 0$
$x = 15$.

This means the solutions are $0$ or $15$.

## Solving Quadratic Equations by Extracting the Square Root

In the previous post, we have learned about quadratic equations or equations of the form $ax^2 + bx + c = 0$, where a is not equal to 0. In this equation, we want to find the value of x which we call the root or the solution to the equation.

There are three strategies in finding the root of the equation: by extracting the roots, by completing the square, and by the quadratic formula. In this example, we will discuss, how to find the root of the quadratic equation by extracting the root.

Just like in solving equations, if we want to find the value of x, we put all the numbers on one side, and all the x’s on one side. Since quadratic equations contain the term $x^2$, we can find the value of x by extracting the square roots. Below are five examples on how to do this.

Example 1: $2x^2 = 0$

Solution

Dividing both sides by 2, we have

$\frac{2x^2}{2} = \frac{0}{2}$.

This gives us

$x^2 = 0$.

Extracting the square root of both sides, we have

$x = 0$.

Therefore, the root $x = 0$.

Example 2: $x^2 - 36 = 0$

Solution

$x^2 - 36 = 0$.

Adding 36 to both side, we have

$x^2 - 36 + 36 = 0 + 36$.

$x^2 = 36$.

Extracting the square root of both sides, we have

$\sqrt{x^2} = \sqrt{36}$.

$x = \pm 6$.

In this example, x has two roots: x = 6 and x = -6.

Example 3: $x^2 + 81 = 0$

Solution

Subtracting 81 from both sides, we have

$x^2 + 81 - 81 = 0 - 81$

$x^2 = -81$

$\sqrt{x^2} = - 81$

$x = \sqrt{-81}$.

In this case, there is no number that when multiplied by itself is negative. For example, negative times negative is equal to positive, and positive times positive is equal to positive. Therefore, there is no real root. There is, however, what we call a complex root as shown in the video below.

Example 4: $5x^2 = 12$

Solution

$5x^2 = 12$

Dividing both sides by 5, we have

$x^2 = \frac{12}{5}$.

Extracting the square root of both sides, we have

$\sqrt{x^2} = \sqrt{\frac{12}{5}} = \frac{\sqrt{12}}{\sqrt{5}}$

$x = \frac{\sqrt{4}\sqrt{3}}{\sqrt{5}}$

$x = \frac{2 \sqrt{3}}{\sqrt{5}}$.

Example 5: $3x^2 = - 4$

$3x^2 = -4$

Dividing both sides by 3, we have

$x^2 = \frac{-4}{3}$.

Extracting the square root of both sides, we have

$\sqrt{x^2} = \sqrt{\frac{-4}{3}}$.

Again, the sign of the number inside the radical is negative, so there is no real root. To know how to compute for the complex root, watch the video below.

The length of a rectangle is 3 cm more than its width. Its area is equal to 54 square centimeters. What is its length and width?

Solution

Let

x = width of rectangle
x + 3 = length of rectangle

The area of a rectangle is the product of the length and width, so we have

Area= x(x + 3)

which is equal to 54.

Therefore, we can form the following equation:

x(x + 3) = 54.

By the distributive property, we have

$x^2 + 3x = 54$

Finding the value of x

In the equation, we want to find the value of x that makes the equation true. Without algebraic manipulation, we can find the value of x by assigning various values to x. The equation $x^2 + 3x = 54$ indicates that one number is greater than the other by 3 and their product is 54. Examining the numbers with product as 54, we have,

1 and 54
2 and 27
3 and 18
6 and 9.

Note: We have excluded the negative (e.g. (-1)(-54) = 54) numbers since a side length cannot be negative.

Now, 9-6 = 3 which means that the side lengths of the rectangle are 6 and 9. Yes, their product is 54 and one is 3 greater than the other.

In the equation above, subtracting both sides by 54, we have

$x^2 + 3x - 54 = 54 - 54$

$x^2 + 3x - 54 = 0$.

The equation that we formed above is an example of a quadratic equation.

A quadratic equation is of the form $ax^2 + bx + c = 0$, where a, b, and c are real numbers and a not equal to 0. In the example above, a = 1, b = 3, and c = -54.

In the problem above, we got the value of x by testing several values, however, there are more systematic methods. In the next post, we will be discussing one of these methods. These methods are factoring, completing the square, and quadratic formula.

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