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
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