Solving equations

Solving Equations

The standard form of a Quadratic equation is $ax^2 + bx + c = 0$ whereby $a$, $b$, $c$ are known values and $a$ can't be 0. $x$ is a variable (we don't know it yet). $a$ is the coefficient of $x^2$, $b$ is the coefficient of $x$ and $c$ is a constant term. Quadratic equation is also called an equation of degree 2 (because of the 2 on $x$). There are several methods which are used to find the value of $x$. These methods are:

• by Factorization
• by completing the square
• by using quadratic formula

The Solution of a Quadratic Equation by Factorization

We can use any of the methods of factorization we learnt in previous chapter. But for simplest we will factorize by splitting the middle term.

Example: Solve for $x$, $x^2 + 4x = 0$

Solution

Since the constant term is 0 we can take out $x$ as a common factor.

So, $x^2 + 4x = x(x + 4) = 0$. This means the product of $x$ and $(x + 4)$ is 0. Then, either $x = 0$ or $x + 4 = 0$. If $x + 4 = 0$ that is $x = -4$. Therefore the solution is $x = 0$ or $x = -4$.

Example 1

Solve the equation: $3x^2 = -6x - 3$.

First rearrange the equation in its usual form. That is:

$3x^2 = -6x - 3$

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

Now, factorize the equation by splitting the middle term. Let us find two numbers whose product is 9 and their sum is 6. The numbers are 3 and 3. Hence the equation $3x^2 + 6x + 3 = 0$ can be written as:

$3x^2 + 3x + 3x + 3 = 0$

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

$(3x + 3)(x + 1) = 0$ (take out common factor which is $(x + 1)$)

Either $(3x + 3) = 0$ or $(x + 1) = 0$

Therefore $3x = -3$ or $x = -1$

$x = -1$ (divide by 3 both sides) or $x = -1$

Therefore, since the values of x are identical then $x = -1$.

Example 2

Solve the equation $10y^2 - 3y - 1 = 0$ by factorization.

Solution

Two numbers whose product is -10 and their sum is -3 are 2 and -5.

Then, we can write the equation $10y^2 - 3y - 1 = 0$ as:

$2y(5y + 1) - 1(5y + 1) = 0$

$(2y - 1)(5y + 1) = 0$

Therefore, either $2y - 1 = 0$ or $5y + 1 = 0$

Example 3

Solve the following quadratic equation by factorization: $4x^2 - 20x + 25 = 0$.

Solution

We need to split the middle term by the two numbers whose product is 100 and their sum is -20. The numbers are -10 and -10.

The equation can be written as:

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

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

$(2x - 5)(2x - 5) = 0$ (take out common factor. The resulting factors are identical. This is a perfect square)

Since it is a perfect square, then we take one factor and equate it to 0. That is:

$2x - 5 = 0$

$2x = 5$ then, divide by 2 both sides.

Therefore

Example 4

Solve the equation $x^2 - 16 = 0$.

Solution

We can write the equation as $x^2 - 4^2 = 0$. This is a difference of two squares. The difference of two squares is an identity of the form:

$a^2 - b^2 = (a - b)(a + b)$.

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

Now, either $x - 4 = 0$ or $x + 4 = 0$

Therefore $x = 4$ or $x = -4$

The Solution of a Quadratic Equation by Completing the Square

Completing the square

Example 1

Add a term that will make the following expression a perfect square: $x^2 - 8x$

Find a term that must be added to make the following expression a perfect square: $x^2 + 10x$

Example 2

Solve the following quadratic equation by completing the square: $x^2 + 4x + 1 = 0$

Example 3

Solve by completing the square: $3x^2 + 7x - 6 = 0$