Quadratic Equations


In quadratic equations we will learn about ………

Solving a quadratic equation (factorization method)

Roots of the quadratic equation.


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. 

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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.



Examples of quadratic equation

(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.

Solving a quadratic equation and finding the roots of quadratic equation

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

        ax² + bx + c = 0.

Factorize the quadratic equation.


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}


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How to solve quadratic equations?

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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