# STEPS OF QUADRATIC FORMULA - SOLVED EXAMPLES AND PRACTICE PROBLEMS

if you have not already done so.

There, We presented the Derivation
Of Quadratic Formula in a lucid way.
An Example in applying the formula
is given. Another problem for practice
is also given.

Here we present some more Solved Examples.

We also give problems for practice in Exercise. Great Deals on School & Homeschool Curriculum Books

## Example 1 : Steps of Quadratic Formula

Solve the following equation using Quadratic Formula

2x2 + 3x - 3 = 0

Solution of Example 1 of Steps of Quadratic Formula :

To solve 2x2 + 3x - 3 = 0 using Quadratic Formula
Comparing this equation with ax2 + bx + c = 0, we get
a = 2, b = 3 and c = -3
We know by Quadratic Formula, x = {(-b) ± √(b2 - 4ac)}⁄2a
Applying this Quadratic Formula here, we get
x = {(-b) ± √(b2 - 4ac)}⁄2a
= [ (-3) ± √{(3)2 - 4(2)(-3)}]⁄2(2)
= [ (-3) ± √{9 + 24}]⁄4 = [-3 ± √(33)]⁄4 Ans.

Example 1 of Steps of Quadratic Formula is thus solved.

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## Example 2 : Steps of Quadratic Formula

Solve the following equation using Quadratic Formula
2x4 - 9x3 + 14x2 - 9x + 2 = 0

Solution of Example 2 of Steps of Quadratic Formula :

The given equation in descending powers of x is
2x4 - 9x3 + 14x2 - 9x + 2 = 0
This is an equation of the form ax4 + bx3 + cx2 + bx + a = 0
in which the coefficients of terms equidistant from first and last are equal, is called a RECIPROCAL EQUATION. By dividing this reciprocal equationwith x2 and with a proper substitution, the reciprocal equationcan be reduced to a Quadratic Equation.
Dividing the equation by x2, we get
2x2 - 9x + 14 - 9x-1 + 2x-2 = 0
⇒ 2x2 + 2⁄x2 - 9x - 9⁄x + 14 = 0
⇒ 2(x2 + 1⁄x2) -9(x + 1⁄x) + 14 = 0
Putting x + 1⁄x = t, x2 + 1⁄x2 = (x + 1⁄x)2 - 2 = t2 - 2
∴ the given equation becomes 2(t2 - 2) -9t + 14 = 0
⇒ 2t2 - 4 - 9t + 14 = 0 ⇒ 2t2 - 9t + 10 = 0
Comparing this equation with at2 + bt + c = 0, we get
a = 2, b = -9 and c = 10
We know by Quadratic Formula, t = {-b ± √(b2 - 4ac)}⁄2a
Applying this Quadratic Formula here, we get
t = {-b ± √(b2 - 4ac)}⁄2a
= [-(-9) ± √{(-9)2 - 4(2)(10)}]⁄2(2)
= [ (+9) ± √{81 - 80}]⁄4 = [ (9) ± √{1}]⁄4 = [ (9) ± 1]⁄4
= (9 + 1)⁄4, (9 - 1)⁄4 = 10⁄4, 8⁄4 = 5⁄2, 2

But t = x + 1⁄x

Taking t = 5⁄2,
t = 5⁄2 ⇒ x + 1⁄x = 5⁄2
Multiplying both sides with 2x, we get
2x2 + 2 = 5x ⇒ 2x2 - 5x + 2 = 0Comparing this equation with ax2 + bx + c = 0, we get
a = 2, b = -5 and c = 2
We know by Quadratic Formula, x = {-b ± √(b2 - 4ac)}⁄2a
Applying this Quadratic Formula here, we get
x = {-b ± √(b2 - 4ac)}⁄2a
= [ -(-5) ± √{(-5)2 - 4(2)(2)}]⁄2(2)
= [ (+5) ± √{25 - 16}]⁄4 = [ (5) ± √{9}]⁄4 = [ (5) ± 3]⁄4
= (5 + 3)⁄4, (5 - 3)⁄4 = 8⁄4, 2⁄4 = 2, 1⁄2

Taking t = 2,
t = 2 ⇒ x + 1⁄x = 2
Multiplying both sides with x, we get
x2 + 1 = 2xx2 - 2x + 1 = 0
Comparing this equation with ax2 + bx + c = 0, we get
a = 1, b = -2 and c = 1
We know by Quadratic Formula, x = {-b ± √(b2 - 4ac)}⁄2a
Applying this Quadratic Formula here, we get
x = {-b ± √(b2 - 4ac)}⁄2a
= [ -(-2) ± √{(-2)2 - 4(1)(1)}]⁄2(1)
= [ (+2) ± √{4 - 4}]⁄2 = [ (2) ± √{0}]⁄2 = [ (2) ± 0]⁄2
= (2 + 0)⁄2, (2 - 0)⁄2 = 2⁄2, 2⁄2 = 1, 1

Thus the solution of the given equation is x = { 1, 1, 2, 1⁄2 } Ans.

Example 2 of Steps of Quadratic Formula is thus solved.

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## Exercise : Steps of Quadratic Formula

1. Solve the following equations using Steps of Quadratic Formula
12x2 + 3x - 99 = 0
2. Solve the following equation using Quadratic Formula
6x4 - 35x3 + 62x2 - 35x + 6 = 0

For Answers See at the bottom of the Page.

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