Quadratic equation

The standard form of a Quadratic equation is ax2 + 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 x2 , 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:

The Solution of a Quadratic Equation by Factorization

Determine 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. For Example: solve for x, x2 + 4x = 0

solution

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

So, x2 + 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

Determine the solution of a quadratic equation by factorization

first rearrange the equation in its usual form.

that is:

3x2 = - 6x – 3

3x2 + 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 3x2 + 6x + 3 = 0 can be written as:

3x2 + 3x + 3x + 3 = 0

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

(3x + 3)(x + 1) (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

Determine the solution of a quadratic equation by factorization

Solution

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

Then, we can write the equation 10y2 – 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

Determine the solution of a quadratic equation by factorization

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:

4x2 -10x - 10x + 25 = 0

2x(2x - 5) – 5(2x -5) = 0

(2x – 5)(2x - 5) (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

Determine the solution of a quadratic equation by factorization

Solution

We can write the equation as x2 - 42 = 0. This is a difference of two squares. The difference of two squares is an identity of the form:

a2 - b2 = (a - b)(a + b).

So, x2 - 42 = (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

Find the solution of a quadratic equation by completing the square

Completing the square.

Example

Find the solution of a quadratic equation by completing the square

find a term that must be added to make the following expression a perfect square: x2 + 10x

Example

Find the solution of a quadratic equation by completing the square

Example

Find the solution of a quadratic equation by completing the square

General Solution of Quadratic Equations

The Quadratic Formula

Derive the quadratic formula

The special quadratic formula used for solving quadratic equation is:

Quadratic Equations using Quadratic Formula

Solve quadratic equations using quadratic formula

Example

Solve quadratic equations using quadratic formula

Example

Solve quadratic equations using quadratic formula

Word problems leading to quadratic equations

Given a word problem; the following steps are to be used to recognize the type of equation.

Step1: choose the variables to represent the information

Step 2: formulate the equation according to the information given

Step 3: solve the equation by using any of the method you know

In order to be sure with your answers, check if the solution you obtained is correct.

Example

Solve quadratic equations using quadratic formula

Solution

Let the width be x

The length of a plot is 8 more than the width, so the length of a plot be x + 8

We are given the area of a plot = 240cm2and the area of a rectangle is given by length ×width

then (x + 8) × x = 240

x2 + 8x = 240

rearrange the equation

x2 + 8x – 240 = 0

then solve the equation to find the value of x

Solving by splitting the middle term, two numbers whose product is -240 and their sum is 8, the number

are -12 and 20

our equation becomes; x2 + 20x – 12x – 240 = 0

x(x + 20) – 12(x + 20) = 0

either (x - 12) = 0 or (x + 20) = 0

x = 12 or x = -20

since we don’t have negative dimensions, then the width is 12cm and the length is 12 + 8 = 20cm

Therefore the rectangular plot has the length of 20cm and the width of 12cm.

Example

Solve quadratic equations using quadratic formula

Exercise

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1. Solve each of the following quadratic equations by using factorization method:

1. -6x2+ 23x – 20 = 0
2. X2– x -12 = 0

2. Solve these equations by completing the square: