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

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

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

Solve 3x² - 17x - 20 = 0.

Solution of Example 1 of Quadratic Equations :

The Factoring of LHS of the above Equation is given below.

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Let P = 3x² - 17x - 20

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

Step 1: Coefficient of x² x constant term
= 3 x -20 = -60

Step 2: We have to express -60 as two factors whose sum
= coefficient of x = -17 ;
-60 = -20 x 3; (-20 + 3 = -17)

Step 3: P = 3x² - 17x - 20 = 3x² - 20x + 3x - 20

Step 4: P = x(3x - 20) + 1(3x - 20)

Step 5: P = (3x - 20)(x + 1)

Thus, Factoring of Quadratic Polynomial, 3x² - 17x - 20, we get the Factors as (3x - 20)(x + 1)

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3x² - 17x - 20 = 0 ⇒ (3x - 20)(x + 1) = 0
⇒ (3x - 20) = 0 or (x + 1) = 0
(3x - 20) = 0 ⇒ 3x = 20 ⇒ x = 20⁄3
(x + 1) = 0 ⇒ x = -1
Thus, x = 20⁄3, -1 are the two roots of the given Quadratic equation. Ans.

Example 2 : Quadratic Equations

Solve 4x⁴ - 5x² + 1 = 0.

Solution of Example 2 of Quadratic Equations :

The Factoring of LHS of the above Equation is given below.

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Let P = 4x⁴ - 5x² + 1
Put x² = t; then P = 4t² - 5t + 1
In Factoring of Trinomials (Quadratics), we follow the five steps.

Step 1: Coefficient of t² x constant term
= 4 x 1 = 4

Step 2: We have to express 4 as two factors whose sum
= coefficient of t = -5 ;
4 = 4 x 1 = -4 x -1; [(-4) + (-1) = -5]

Step 3: P = 4t² - 5t + 1 = 4t² - 4t - t + 1

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

Step 5: P = (t - 1)(4t - 1)

But t = x²; ∴ P = (t - 1)(4t - 1) = (x² - 1)(4x² - 1)
P = (x² - 1²){(2x)² - 1²}

In each of the two brackets, there is difference of
squares of two terms which is equal to the product of
the sum and difference of the terms.[See Formula 3]

∴ P = (x + 1)(x - 1)(2x + 1)(2x - 1)

Thus, Factoring of Quadratic Polynomial, 4x⁴ - 5x² + 1
and then applying Algebra formulas,
we get the Factors as (x + 1)(x - 1)(2x + 1)(2x - 1)

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4x⁴ - 5x² + 1 = 0 ⇒ (x + 1)(x - 1)(2x + 1)(2x - 1) = 0
⇒ (x + 1) = 0 or (x - 1) = 0 or (2x + 1) = 0 or (2x - 1) = 0
(x + 1) = 0 ⇒ x = -1
(x - 1) = 0 ⇒ x = 1
(2x + 1) = 0 ⇒ 2x = -1 ⇒ x = -1⁄2
(2x - 1) = 0 ⇒ 2x = 1 ⇒ x = 1⁄2

Thus, x = -1, 1, -1⁄2, 1⁄2 are the four roots of the given 4th Degree equation. Ans.

Exercise : Quadratic Equations

1. Solve 7x² - 8x - 12 = 0.
2. Solve 4x⁴ - 25x² + 36 = 0.

For Answers See at the bottom of the Page.

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

Answers to Exercise : Quadratic Equations

1. -6⁄7, 2
2. -3⁄2, 3⁄2, -2, 2