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subject: Quadratic Equation Exercises [print this page]

Quadratic equation is also called as second degree equation.

The basic form of a second degree equation is given by ay2 + b y + c = 0. Where, a = coefficient of y2, b = coefficient of y, c = constant and y is the variable.

Let us consider the equation 6y2 + 12y 9=0. In this equation, maximum power of the variable y is 2. So, it is a second degree equation.

Three methods are used to solve the quadratic equation.

1) Factoring method

2) Quadratic formula method

3) Completing the square method

Exercise Problems to Solve Quadratic Equation:

Exercise problem 1:

Determine the given equation is a quadratic equation or not.

(x - 10) (x - 20) = 0

Solution:

Multiply two expressions in the left hand side,

(x - 10) (x - 20) = 0

x(x-20) -10(x-20)=0

x2 -20x -10x + 10(20) = 0

x2 -30x +200 = 0

So, the highest exponent of variable x is 2.

So, the given equation is a quadratic equation.

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Exercise problem 2:

Solve the quadratic equation by factoring: x2 + 28x + 27 = 0

Solution:

x2 + 28x + 27 = 0

Here a = coefficient of x2 = 1

b = coefficient of x = 28

c = constant term = 27

We find a c = 1 27 = 27, a + c = 1 + 27 = 28 = b.

By splitting the middle term, we get

x2 + 28x + 27 = 0

x2 + 1x + 27x + 27 = 0

x(x + 1) + 27(x + 1) = 0

(x + 1) (x + 27) = 0

x = 1, x = 27

The solution set = {1, 27}.

Few more Exercise Problems to Solve Quadratic Equations:

Exercise problem 3:

Solve the quadratic equation using quadratic formula: x2 - 28x + 27 = 0

Solution:

Coefficients are: a = 1, b = -28, c = 27

Quadratic Formula: x = [-b (b2-4ac)] / 2a

Put in a, b and c: x = [-(-28) [(-28)2-41 (27)] / (21)

Solve: x = [28 (784-108)]/2

x = [28 (676)]/2

x = (28 26)/2

x= (28+ 26)/2

x = (28-26)/2

x= (54)/2

x= (2)/2

x= 27

x= 1

So, the answer is x = 27, 1.

Exercise problem 4:

Solve the equation by completing the square x2 - 32x + 31 = 0.

Solution: x2 - 32x = -31. The term to be added =(-32/2)2= 256

Adding 256 on both sides we get

x2 - 32x + 256 = -31 + 256

(x 16)2 = 225

Taking square root on both sides x 16 = 15

x - 16 = 15 x 16 = 15

x=31

x=1

So, the answer is x=1 (or) 31.

by: mathqa

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