Quadratic Equations by Factoring

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

Solving Quadratic Equations

General Properties of Quadratic Equation

Methods of Solving Quadratic Equations

Roots of a Quadratic Equation

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

9th Grade Math

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