The following steps will help us to solve quadratic equations by factoring:
Step I: Clear all the fractions and brackets, if necessary.
Step II: Transpose all the terms to the left hand side to get an equation in the form ax\(^{2}\) + bx + c = 0.
Step III: Factorize the expression on the left hand side.
Step IV: Put each factor equal to zero and solve.
1. Solve the quadratic equation 6m\(^{2}\) – 7m + 2 = 0 by factorization method.
Solution:
⟹ 6m\(^{2}\) – 4m – 3m + 2 = 0
⟹ 2m(3m – 2) – 1(3m – 2) = 0
⟹ (3m – 2) (2m – 1) = 0
⟹ 3m – 2 = 0 or 2m – 1 = 0
⟹ 3m = 2 or 2m = 1
⟹ m = \(\frac{2}{3}\) or m = \(\frac{1}{2}\)
Therefore, m = \(\frac{2}{3}\), \(\frac{1}{2}\)
2. Solve for x:
x\(^{2}\) + (4 – 3y)x – 12y = 0
Solution:
Here, x\(^{2}\) + 4x – 3xy – 12y = 0
⟹ x(x + 4)  3y(x + 4) = 0
or, (x + 4) (x – 3y) = 0
⟹ x + 4 = 0 or x – 3y = 0
⟹ x = 4 or x = 3y
Therefore, x = 4 or x = 3y
3. Find the integral values of x (i.e., x ∈ Z) which satisfy 3x\(^{2}\)  2x  8 = 0.
Solution:
Here the equation is 3x\(^{2}\) – 2x – 8 = 0
⟹ 3x\(^{2}\) – 6x + 4x – 8 = 0
⟹ 3x(x – 2) + 4(x – 2) = 0
⟹ (x – 2) (3x + 4) = 0
⟹ x – 2 = 0 or 3x + 4 = 0
⟹ x = 2 or x = \(\frac{4}{3}\)
Therefore, x = 2, \(\frac{4}{3}\)
But x is an integer (according to the question).
So, x ≠ \(\frac{4}{3}\)
Therefore, x = 2 is the only integral value of x.
4. Solve: 2(x\(^{2}\) + 1) = 5x
Solution:
Here the equation is 2x^2 + 2 = 5x
⟹ 2x\(^{2}\)  5x + 2 = 0
⟹ 2x\(^{2}\)  4x  x + 2 = 0
⟹ 2x(x  2)  1(x  2) = 0
⟹ (x – 2)(2x  1) = 0
⟹ x  2 = 0 or 2x  1 = 0 (by zero product rule)
⟹ x = 2 or x = \(\frac{1}{2}\)
Therefore, the solutions are x = 2, 1/2.
5. Find the solution set of the equation 3x\(^{2}\) – 8x – 3 = 0; when
(i) x ∈ Z (integers)
(ii) x ∈ Q (rational numbers)
Solution:
Here the equation is 3x\(^{2}\) – 8x – 3 = 0
⟹ 3x\(^{2}\) – 9x + x – 3 = 0
⟹ 3x(x – 3) + 1(x – 3) = 0
⟹ (x – 3) (3x + 1) = 0
⟹ x = 3 or x = \(\frac{1}{3}\)
(i) When x ∈ Z, the solution set = {3}
(ii) When x ∈ Q, the solution set = {3, \(\frac{1}{3}\)}
6. Solve: (2x  3)\(^{2}\) = 25
Solution:
Here the equation is (2x – 3)\(^{2}\) = 25
⟹ 4x\(^{2}\) – 12x + 9 – 25 = 0
⟹ 4x\(^{2}\) – 12x  16 = 0
⟹ x\(^{2}\) – 3x  4 = 0 (dividing each term by 4)
⟹ (x – 4) (x + 1) = 0
⟹ x = 4 or x = 1
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