# Worksheet on Quadratic Equations by Factoring

Practice the questions given in the worksheet on quadratic equations by factoring.

We know to factorize the quadratic equation we first need to convert the given expression in the standard form of a quadratic equation i.e., ax^2 + bx + c = 0, where a, b and c are all real numbers and a ≠ 0.

Solve each of the following equations by factorization:

(i) 3x$$^{2}$$ - 20 = 160 - 2x$$^{2}$$

(ii) (2x - 3)$$^{2}$$ = 49

(iii) 16x$$^{2}$$ = 25

(iv) (2x + 1)$$^{2}$$ + (x + 1)$$^{2}$$ = 6x + 47

(v) 2x$$^{2}$$ + x - 6 = 0

(vi) 3x$$^{2}$$ = x + 4

(vii) (x - 7)(x - 9) = 195

(viii) x$$^{2}$$ - (a + b)x + ab = 0

(ix) $$\frac{4}{3}$$x$$^{2}$$- 2x + $$\frac{3}{4}$$ = 0

(x) $$\frac{x}{3}$$ + $$\frac{3}{x}$$ = $$\frac{15}{x}$$

(xi) $$\frac{100}{x}$$- $$\frac{100}{x + 5}$$= 1

(xii) $$\frac{4}{x + 2}$$ - $$\frac{1}{x + 3}$$ = $$\frac{4}{2x + 1}$$

(xiii) $$\frac{1}{a + b + x}$$ = $$\frac{1}{a}$$+ $$\frac{1}{b}$$ + $$\frac{1}{x}$$

(xiv) $$\frac{a}{x - a}$$ + $$\frac{b}{x - b}$$ = $$\frac{2c}{x - c}$$

(xv) (1 + $$\frac{1}{x + 1}$$)(1 - $$\frac{1}{x - 1}$$) = $$\frac{7}{8}$$

(xvi) $$\frac{5}{x - 2}$$ - $$\frac{3}{x + 6}$$ = $$\frac{4}{x}$$

(xvii) ($$\frac{x + a}{x - a}$$)$$^{2}$$ - 5($$\frac{x + a}{x - a}$$) + 6 = 0

(xviii) $$\frac{x + 3}{x - 3}$$ + 6($$\frac{x - 3}{x + 3}$$) = 5

(xix) $$\frac{3x - 2}{2x - 3}$$ = $$\frac{3x - 8}{x + 4}$$

(xx) $$\frac{x - 3}{x + 3}$$ + $$\frac{x + 3}{x - 3}$$ = 2$$\frac{1}{2}$$

Answers for the worksheet on quadratic equations by factoring are given below.

(i) 6, -6

(ii) -2, 5

(iii) - $$\frac{5}{4}$$, $$\frac{5}{4}$$

(iv) -3, 3

(v) -2, $$\frac{3}{2}$$

(vi) -1, $$\frac{4}{3}$$

(vii) -6, 22

(viii) a, b

(ix) $$\frac{3}{4}$$ only

(x) -6, 6

(xi) -25, 20

(xii) - 3$$\frac{1}{2}$$, 2

(xiii) - a, - b

(xiv) 0, $$\frac{2ab - bc - ac}{a + b - 2c}$$

(xv) -5 , 5

(xvi) -2, 12

(xvii) 2a, 3a

(xviii) 6, 9

(xix) 1, 10$$\frac{2}{3}$$

(xx) -9, 9

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