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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