Question

Solve the quadratic equations in questions 1 – 5 by factoring.

1. x2 – 49 = 0

2. 3x3 – 12x = 0

3. 12x2 + 14x + 12 = 18

4. –x3 + 22x2 – 121x = 0

5. x2 – 4x = 5

Answer 1

The solutions for the given equations are:

1. x² - 49 = 0; x = {-7, 7}
2. 3x³ - 12x = 0; x = {-2, 0, 2}
3. 12x² + 14x + 12 = 18; x = {-3/2, 1/3}
4. -x³ + 22x² - 121x = 0; x = {0, 11, 11}
5. x² - 4x = 5; x = {-1, 5}

What is factorization?

Writing a number or an equation as a product of its factors is said to be the factorization.

A linear equation has only one factor, a quadratic equation has 2 factors and a cubic equation has 3 factors.

Calculation:

1. Solving x² - 49 = 0; (quadratic equation)

⇒ x² - 7² = 0

This is in the form of a² - b². So, a² - b² = (a + b)(a - b)

⇒ (x + 7)(x - 7) =0

By the zero-product rule,

x = -7 and 7.

2. Solving 3x³ - 12x = 0

⇒ 3x(x² - 4) = 0

⇒ 3x(x² - 2²) = 0

⇒ 3x(x + 2)(x - 2) = 0

So, by the zero product rule, x = -2, 0, 2

3. Solving 12x² + 14x + 12 = 18; (quadratic equation)

⇒ 12x² + 14x + 12 - 18 = 0

⇒ 12x² + 14x - 6 = 0

⇒ 2(6x² + 7x - 3) = 0

⇒ 6x² + 9x - 2x - 3 = 0

⇒ 3x(2x + 3) - (2x + 3) = 0

⇒ (3x - 1)(2x + 3) = 0

∴ x = 1/3, -3/2

4. Solving -x³ + 22x² - 121x = 0

⇒ -x³ + 22x² - 121x = 0

⇒ -x(x² - 22x + 121) = 0

⇒ -x(x² - 11x - 11x + 121) = 0

⇒ -x(x(x - 11) - 11(x - 11)) = 0

⇒ -x(x - 11)² = 0

∴ x = 0, 11, 11

5. Solving x² - 4x = 5; (quadratic equation)

⇒ x² - 4x - 5 = 0

⇒ x² -5x + x - 5 = 0

⇒ x(x - 5) + (x - 5) = 0

⇒ (x + 1)(x - 5) =0

∴ x = -1, 5

Hence all the given equations are solved.

Learn more about the factorization here:

https://brainly.com/question/25829061

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