Solving a quadratic equation (factorization method)

In polynomials, we studied that a polynomial of degree 1 is called a linear polynomials.

For example: x - 5, 7x, 3 - 2x are linear polynomials which may be monomials or binomials.

A polynomial of degree 2 (two) is called a quadratic polynomial.

For example: 3x², x² + 7 , x² – 3x + 4 are quadratic polynomials which may be monomials, binomials or trinomials.



What is known as quadratic equation?

When these quadratic polynomials are equated to zero, equation is formed and is known as a quadratic equation.

The standard form of quadratic equation is ax² + bx + c = 0. Here a, b, c are real numbers and a ≠ 0. The power of x in the equation must be a non-negative integer.

(i) 3x² - 6x + 1 = 0 is a quadratic equation.

(ii) x + (1/x) = 5 is a quadratic equation.

On solving, we get x × x + (1/x) × x = 5 × x

⇒ x² + 1 = 5x

⇒ x² - 5x + 1 = 0

(iii) √2x² - x - 7 = 0 is a quadratic equation.

(iv) 3x² - √x + 1 = 0 is not a quadratic equation, since the power of x must be a positive integer.

(v) x² - (1/x) + 7 = 0 is not a quadratic equation, since on solving it becomes an equation of degree 3.

(vi) x² - 4 = 0 is a quadratic equation.

(vii) x² = 0 is a quadratic equation.

Write the quadratic equation in the standard form, i.e.,

ax² + bx + c = 0.

Express it as the product of two linear factors, say (px + q) and (rx + s), where p, q, r, S are real numbers and p, r are not equal to zero.

Then, ax² + bx + c = 0

(px + q) (rx + s) = 0

Put each of the linear factors equal to zero

i.e., px + q = 0     and     rx + s = 0

⇒ px = - q           ⇒ rx = - s

⇒ x = -q/p           ⇒ x = -s/r

Thus, the two values of x are called the roots of the quadratic equation.

Therefore, the solution set = {-q/p, -s/r}

Worked-out problems on solving quadratic equation will help the students to understand the detailed explanation showing the step-by-step quadratic equation solution.

1. Solve: x² + 6x + 5 = 0

Solution:

x² + 6x + 5 = 0

⇒ x² + 5x + x + 5 = 0

⇒ x(x + 5) + 1(x + 5) = 0

⇒ (x + 1) (x + 5) = 0

⇒ x + 1 = 0 and x + 5 = 0

⇒ x = -1   and   x = -5

Therefore, solution set = {-1, -5}

2. Solve: 8x² = 21 + 22x

Solution:

8x² = 21 + 22x

⇒ 8x² - 21 - 22x = 0

⇒ 8x² - 22x - 21 = 0

⇒ 8x² - 28x + 6x - 21 = 0

⇒ 4x (2x - 7) + 3(2x - 7) = 0

⇒ (4x + 3) (2x - 7) = 0

⇒ 4x + 3 = 0 and 2x - 7 = 0

⇒ 4x = -3 and 2x = 7

⇒ x = -3/₄ and x = ⁷/₂

Therefore, solution set = {-3/₄, ⁷/₂}

3. 1/(x + 4) - 1/(x - 7) = 11/30

Solution:

1/(x + 4) - 1/(x - 7) = 11/30

⇒ [(x - 7) - (x + 4)]/(x + 4) (x - 7) = ¹¹/₃₀

⇒ [x - 7 - x - 4]/(x² - 3x - 28) = ¹¹/₃₀

⇒ - 11/(x² - 3x - 28) = ¹¹/₃₀

⇒ -1/(x² - 3x - 28) = ¹/₃₀

⇒ -30 = x² - 3x - 28

⇒ x² - 3x + 2 = 0

⇒ x² - 2x - x + 2 = 0

⇒ x(x - 2) - 1(x - 2) = 0

⇒ (x - 1) (x - 2) = 0

⇒ x - 1 = 0 and x - 2 = 0

⇒ x = 1 and x = 2

Therefore, Solution set = {1, 2}

4. Solve (2x - 3)/(x + 2) = (3x - 7)/(x + 3)

Solution:

(2x - 3)/(x + 2) = (3x - 7)/(x + 3)

⇒ (2x - 3) (x + 3) = (x + 2) (3x - 7)

⇒ 2x² + 6x - 3x - 9 = 3x² - 7x + 6x - 14

⇒ 2x² + 6x - 3x - 9 - 3x² + 7x - 6x + 14 = 0

⇒ 2x² - 3x² + 6x - 3x + 7x - 6x - 9 + 14 = 0

⇒ -x² - 4x + 5 = 0

⇒ x² + 4x - 5 = 0

⇒ x² + 5x - x - 5 = 0

⇒ x (x + 5) -1 (x + 5) = 0

⇒ (x - 1) (x + 5) = 0

⇒ x - 1 = 0 and x + 5 = 0

⇒ x = 1 and x = -5

Therefore, solution set = {1, -5}

5. Solve x² - 9/5 + x² = -5/₉

Solution:

x² - 9/5 + x² = -5/₉

⇒ 9(x² - 9) = -5 (5 + x²)

⇒ 9x² - 81 = -25 - 5x²

⇒ 9x² + 5 x² = -25 + 81

⇒ 14x² = 56

⇒ x² = 56/14

⇒ x² = 4

⇒ x² - 4 = 0

⇒ x² - 2² = 0

⇒ (x - 2) (x + 2) = 0

⇒ x - 2 = 0 and x + 2 = 0

⇒ x = 2 and x = -2

Therefore, solution set = {2, -2}

These are the above examples on quadratic equations which are explained to show the exact way to solve.

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