**TOPIC 3: QUADRATIC EQUATIONS ~ MATHEMATICS FORM 2**

Solving quadratic equations can be difficult, but luckily there are several different methods that we can use depending on what type of quadratic that we are trying to solve.

The four methods of solving a quadratic equation are factoring, using the square roots, completing the square and the quadratic formula.

**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

*x*,

*x*

^{2}+ 4

*x*= 0

*x*as a common factor.

So, *x*^{2} + 4*x* = *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

*x*

^{2}=- 6

*x*– 3.

*x*

^{2}= – 6

*x*– 3

*x*

^{2}+ 6

*x*+ 3 = 0

*x*

^{2}+ 6

*x*+ 3 = 0 can be written as:

^{2}+ 3

*x*+ 3

*x*+ 3 = 0

Example 2

*y*– 1 = 0 by factorization.

*y*

^{2}– 3

*y*– 1 = 0 as:

*y*(5

*y*+ 1) – 1(5

*y*+ 1) = 0

(2*y* – 1)(5*y* + 1) = 0

*y*– 1 = 0 or 5

*y*+ 1 = 0

Example 3

*x*

^{2}– 20

*x*+ 25 = 0.

*x*

^{2}-10x – 10

*x*+ 25 = 0

*x*(2

*x*– 5) – 5(2

*x*-5) = 0

*x*– 5)(2

*x*– 5) (take out common factor. The resulting factors are identical. This is a perfect square)

*x*– 5 = 0

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

Therefore

Example 4

*x*

^{2}– 16 = 0.

*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*).

*x*

^{2}– 4

^{2}= (

*x*– 4)(

*x*+ 4) = 0

*x*– 4 = 0 or

*x*+ 4 = 0

*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.**

^{2}– 8

*x*

*x*

^{2}+ 10

*x*

*x*

^{2}+ 4x + 1 = 0

*x*

^{2}+ 7

*x*– 6 = 0

**General Solution of Quadratic Equations**

The special quadratic formula used for solving quadratic equation is:

Quadratic Equations using Quadratic Formula

Solve quadratic equations using quadratic formula

Example 8

solve 5x^{2} – 8x + 3 = 0 by using quadratic formula.

**Example 9**

^{2}= – 7x – 4

**Word problems leading to quadratic equations**

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

Example 10

^{2}, find the dimensions of length and width.

^{2}and the area of a rectangle is given by length ×width

**×**x = 240

^{2}+ 8x = 240

^{2}+ 8x – 240 = 0

are -12 and 20

^{2}+ 20x – 12x – 240 = 0

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

Example 11

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