SOLVING QUADRATIC EQUATIONS - BY FACTORING, EXAMPLES, EXERCISE, LINKS TO QUADRATIC FORMULA

Please study
about Quadratic Equation before Solving Quadratic Equations
if you have not already done so.

There, we gave introduction to Quadratic polynomial,
Quadratic Equation, Methods to solve the Quadratic
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.

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

In Factoring of Trinomials (Quadratics) , we follow the five steps.

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, x = 2, 1⁄2 are the two roots of the given 6th Degree equation. The other four roots are not real. Ans.

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

In Factoring of Trinomials (Quadratics), we follow the five steps.

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, x = -4, 3, -3, 2 are the four roots of the given 4th Degree equation. Ans.

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.

Answers to Exercise : Solving Quadratic Equations

1. 1⁄2, 1, four imaginary roots
2. -5, 2, -4, 1