# BASIC MATHEMATICS-TOPIC 3: QUADRATIC EQUATIONS

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767 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 EquationsThe standard form of a Quadratic equation is ax2 + bx + c =0whereby abc 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:

1. by Factorization
2. by completing the square

The Solution of a Quadratic Equation by FactorizationDetermine the solution of a quadratic equation by factorizationWe
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 = 0solutionSince 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 1Solve the equation: 3x2 =- 6x – 3.first rearrange the equation in its usual form.that is:3x2 = – 6x – 33x2 + 6x + 3 = 0now, factorize the equation by splitting the middle term. Let us find two numbers whose product is 9and 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 = 03x(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) = 0therefore 3x = -3 or x = -1x = -1 (divide by 3 both sides) or x = -1Therefore, since the values of x are identical then x = -1.Example 2solve the equation 10 – 3y – 1 = 0 by factorization.SolutionTwo 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) = 0Therefore, either 2y – 1 = 0 or 5y + 1 = 0

Example 3solve the following quadratic equation by factorization: 4x2 – 20x + 25 = 0.SolutionWe 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 = 02x(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 = 02x = 5 then, divide by 2 both sides.Therefore

Example 4solve the equation x2 – 16 = 0.SolutionWe 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) = 0Now, either x– 4 = 0 or x + 4 = 0Therefore x = 4 or x = -4The Solution of a Quadratic Equation by Completing the SquareFind the solution of a quadratic equation by completing the squareCompleting the square.

Example 5Add a term that will make the following expression a perfect square: x2 – 8x

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

Example 6solve the following quadratic equation by completing the square: x2 + 4x + 1 = 0

Example 7solve by completing the square: 3x2 + 7x – 6 = 0

Example 9solve this quadratic equation by using quadratic formula: 3x2 = – 7x – 4

Word problems leading to quadratic equationsGiven a word problem; the following steps are to be used to recognize the type of equation.Step1: choose the variables to represent the informationStep 2: formulate the equation according to the information givenStep 3: solve the equation by using any of the method you knowIn order to be sure with your answers, check if the solution you obtained is correct.Example 10the length of a rectangular plot is 8 centimeters more than the width. If the area of a plot is 240cm2, find the dimensions of length and width.SolutionLet the width be xThe length of a plot is 8 more than the width, so the length of a plot be x + 8We are given the area of a plot = 240cm2 and the area of a rectangle is given by length ×widththen (x + 8) × x = 240x2 + 8x = 240rearrange the equationx2 + 8x – 240 = 0then solve the equation to find the value of xSolving by splitting the middle term, two numbers whose product is -240 and their sum is 8, the numberare -12 and 20our equation becomes; x2 + 20x – 12x – 240 = 0x(x + 20) – 12(x + 20) = 0either (x – 12) = 0 or (x + 20) = 0x = 12 or x = -20since we don’t have negative dimensions, then the width is 12cm and the length is 12 + 8 = 20cmTherefore the rectangular plot has the length of 20cm and the width of 12cm.Example 11A
piece of wire 40cm long is cut into two parts and each part is then
bent into a square. If the sum of the areas of these squares is 68
square centimeters, find the lengths of the two pieces of wire.

Exercise 11. 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: