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.
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 nonnegative 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}
How to solve quadratic equations?
Workedout problems on solving quadratic equation will help the students to understand the detailed explanation showing the stepbystep 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.
8th Grade Math Practice
From Quadratic Equations to HOME PAGE
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
