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
Quadratic Equation
Introduction to Quadratic Equation
Formation of Quadratic Equation in One Variable
General Properties of Quadratic Equation
Methods of Solving Quadratic Equations
Examine the Roots of a Quadratic Equation
Problems on Quadratic Equations
Quadratic Equations by Factoring
Word Problems Using Quadratic Formula
Examples on Quadratic Equations
Word Problems on Quadratic Equations by Factoring
Worksheet on Formation of Quadratic Equation in One Variable
Worksheet on Quadratic Formula
Worksheet on Nature of the Roots of a Quadratic Equation
Worksheet on Word Problems on Quadratic Equations by Factoring
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