if you have not already done so.

There, we gave introduction to Quadratic polynomial,
Equations and about Method of Solving by Factoring.

That knowledge is a prerequisite here.

Here, we apply the Method to solve problems.
Solved Examples and Exercise problems are given.

## Example 1 : Solving Quadratic Equations

Solve 8x6 - 65x3 + 8 = 0
with the knowledge of Solving Quadratic Equations.

Solution of Example 1 of Solving Quadratic Equations :

Factoring of LHS of the given Equation is given below.

-*-*-*-*-*-*-*
Let P = 8x6 - 65x3 + 8
Put x3 = t; Then P = 8t2 - 65t + 8

Step 1: Coefficient of t2 x constant term
= 8 x 8 = 64

Step 2: We have to express 64 as two factors whose sum
= coefficient of t = -65 ;
64 = 64 x 1 = -64 x -1; {(-64) + (-1) = -65}

Step 3: P = 8t2 - 65t + 8 = 8t2 - 64t - t + 8

Step 4: P = 8t(t - 8) - 1(t - 8)

Step 5: P = (t - 8)(8t - 1)

But t = x3; ∴ P = (t - 8)(8t - 1) = (x3 - 8)(8x3 - 1)
⇒ P = (x3 - 23){(2x)3 - 13}

In each of the two brackets, there is difference
of cubes of two terms for which we have a formula
a3 - b3 = (a - b)(a2 + ab + b2) [See Formula 5]

Applying this Formula here, we get
P = (x - 2){x2 + x(2) + 22}(2x - 1){(2x)2 + (2x)(1) + 12}
= (x - 2)(x2 + 2x + 4)(2x - 1)(4x2 + 2x + 1)

Thus, Factoring of Quadratic Polynomial, 8x6 - 65x3 + 8
and then applying Algebra formulas,
we get the Factors as (x - 2)(x2 + 2x + 4)(2x - 1)(4x2 + 2x + 1)
-*-*-*-*-*-*-*

8x6 - 65x3 + 8 = 0 ⇒ (x - 2)(x2 + 2x + 4)(2x - 1)(4x2 + 2x + 1) = 0
⇒ (x - 2) = 0 or (2x - 1) = 0 or (x2 + 2x + 4) = 0 or (4x2 + 2x + 1) = 0
(x - 2) = 0 ⇒ x = 2
(2x - 1) = 0 ⇒ 2x = 1 ⇒ x = 1⁄2
(x2 + 2x + 4) = 0 has no real roots as (x2 + 2x + 4) has no real factors.
(4x2 + 2x + 1) = 0 has no real roots as (4x2 + 2x + 1) has no real factors.

Thus, Knowledge of Solving Quadratic Equations
helped in presenting the solution as
x = 2, 1⁄2 are the two roots of the given 6th Degree equation. The other four roots are not real. Ans.

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## Example 2 : Solving Quadratic Equations

Solve (x2 + x)2 -18(x2 + x) + 721 = 0
with the knowledge of Solving Quadratic Equations.

Solution of Example 2 of Solving Quadratic Equations :

The Factoring of LHS of the Equation is given below.

-*-*-*-*-*-*
Let P = (x2 + x)2 -18(x2 + x) + 72
Put (x2 + x) = t; Then P = t2 -18t + 72

Step 1: Coefficient of t2 x constant term
= 1 x 72 = 72

Step 2: We have to express 72 as two factors whose sum
= coefficient of t = -18 ;
72 = 12 x 6 = -12 x -6; [(-12) + (-6) = -18]

Step 3: P = t2 -18t + 72 = t2 - 12t - 6t + 72

Step 4: P = t(t - 12) - 6(t - 12)

Step 5: P = (t - 12)(t - 6)

But t = (x2 + x); ∴ P = (t - 12)(t - 6) = (x2 + x - 12)(x2 + x - 6)

In each of these two brackets, there is a Quadratic Polynomial
which can be factorised using the five steps above.

x2 + x - 12 = x2 + 4x - 3x - 12 = x(x + 4) - 3(x + 4) = (x + 4)(x - 3)

x2 + x - 6 = x2 + 3x - 2x - 6 = x(x + 3) - 2(x + 3) = (x + 3)(x - 2)

See how these two Quadratic Polynomials are factorised with the knowledge of the 5 steps. You might have mastered the 5 stepsof factorisation by this time, to write directly like this.

Thus P = (x2 + x)2 -18(x2 + x) + 72
= (x2 + x - 12)(x2 + x - 6)
= (x + 4)(x - 3)(x + 3)(x - 2)

Thus, Factoring of Quadratic Polynomial,(x2 + x)2 -18(x2 + x) + 72 twice,
we get the Factors as (x + 4)(x - 3)(x + 3)(x - 2)
-*-*-*-*-*-*

(x2 + x)2 -18(x2 + x) + 72 = 0 ⇒ (x + 4)(x - 3)(x + 3)(x - 2) = 0
⇒ (x + 4) = 0 or (x - 3) = 0 or (x + 3) = 0 or (x - 2) = 0
(x + 4) = 0 ⇒ x = -4
(x - 3) = 0 ⇒ x = 3
(x + 3) = 0 ⇒ x = -3
(x - 2) = 0 ⇒ x = 2

Thus, Knowledge of Solving Quadratic Equations
helped in presenting the solution as
x = -4, 3, -3, 2 are the four roots of the given 4th Degree equation. Ans. Great Deals on School & Homeschool Curriculum Books

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## Exercise : Solving Quadratic Equations

With the knowledge of Solving Quadratic Equations,

1. Solve 8x6 - 9x3 + 1 = 0.
2. Solve (x2 + 3x)2 - 14(x2 + 3x) + 40 = 0.

For Answers See at the bottom of the Page.

NOTE: You may solve these problems of Exercise
using Quadratic Formula, after learning it.

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