Class 8, Exercise 26, Quadratic Equations

Class 8: Quadratic Equations – Exercise 26

Date: November 28, 2016Author: ICSE CBSE ISC Board Mathematics Portal for Students 0 Comments

Solve each of the following quadratic equations:

Question 1: $\displaystyle x^2-11x+28 = 0$

Answer:

$\displaystyle x^2-11x+28 = 0$

$\displaystyle \Rightarrow x^2 - 7x - 4x + 28 = 0$

$\displaystyle \Rightarrow x (x-4)-7 (x-4) = 0$

$\displaystyle \Rightarrow (x-4)(x-7) = 0$

$\displaystyle \therefore \ x = 4 \ or\ x = 7$

$\displaystyle \\$

Question 2: $\displaystyle x^2-13x = 68$

Answer:

$\displaystyle x^2-13x = 68$

$\displaystyle \Rightarrow x^2-13x-68x = 0$

$\displaystyle \Rightarrow x^2-17x+4x-68x = 0$

$\displaystyle \Rightarrow x(x-17)+4(x-17) = 0$

$\displaystyle \therefore \ x = 17, \ or\ x = -4$

$\displaystyle \\$

Question 3: $\displaystyle x^2+2x = 48$

Answer:

$\displaystyle x^2+2x = 48$

$\displaystyle \Rightarrow x^2+2x-48 = 0$

$\displaystyle \Rightarrow x^2+8x-6x-48 = 0$

$\displaystyle \Rightarrow x(x+8)-6(x+8) = 0$

$\displaystyle \Rightarrow (x-6)(x+8) = 0$

$\displaystyle \therefore \ x = 6 \ or\ x = -8$

$\displaystyle \\$

Question 4: $\displaystyle x^2-2x = 3$

Answer:

$\displaystyle x^2-2x = 3$

$\displaystyle \Rightarrow x^2-2x-3 = 0$

$\displaystyle \Rightarrow x^2-3x+1x-3 = 0$

$\displaystyle \Rightarrow (x+1)(x-3) = 0$

$\displaystyle \Rightarrow x(x-3)+1(x-3) = 0$

$\displaystyle \Rightarrow x = -1 \ or\ x = 3$

$\displaystyle \\$

Question 5: $\displaystyle x^2-49 = 0$

Answer:

$\displaystyle x^2-49 = 0$

$\displaystyle \Rightarrow (x-7)(x+7) = 0$

$\displaystyle \Rightarrow x = 7 \ or\ x = -7$

$\displaystyle \\$

Question 6: $\displaystyle 3x^2-243 = 0$

Answer:

$\displaystyle 3x^2-243 = 0$

$\displaystyle \Rightarrow 3 (x^2-81) = 0$

$\displaystyle \Rightarrow (x-9)(x+9) = 0$

$\displaystyle \Rightarrow x = 9 \ or\ x = -9$

$\displaystyle \\$

Question 7: $\displaystyle 16x^2 = 9$

Answer:

$\displaystyle 16x^2 = 9$

$\displaystyle \Rightarrow 16 x^2-9 = 0$

$\displaystyle \Rightarrow (4x-3)(4x+3) = 0$

$\displaystyle \Rightarrow x = \frac{3}{4} \ or\ x = \frac{(-3)}{4}$

$\displaystyle \\$

Question 8: $\displaystyle 3x^2+13x = 10$

Answer:

$\displaystyle 3x^2+13x = 10$

$\displaystyle \Rightarrow 3x^2+13x-10 = 0$

$\displaystyle \Rightarrow 3x^2-2x+15x-10 = 0$

$\displaystyle \Rightarrow x(3x-2)+5(3x-2) = 0$

$\displaystyle \Rightarrow (3x-2)(x+5) = 0$

$\displaystyle \Rightarrow x = \frac{2}{3} \ or\ x = -5$

$\displaystyle \\$

Question 9: $\displaystyle 2x^2-7x+3 = 0$

Answer:

$\displaystyle 2x^2-7x+3 = 0$

$\displaystyle \Rightarrow 2 x^2-6x-x+3 = 0$

$\displaystyle \Rightarrow 2x(x-3)-1(x-3) = 0$

$\displaystyle \Rightarrow (2x-1)(x-3) = 0$

$\displaystyle \Rightarrow x = \frac{1}{2} \ or\ x = 3$

$\displaystyle \\$

Question 10: $\displaystyle 6x^2+5x = 6$

Answer:

$\displaystyle 6x^2+5x = 6$

$\displaystyle \Rightarrow 6x^2+5x-6 = 0$

$\displaystyle \Rightarrow 6x^2+9x-4x-6 = 0$

$\displaystyle \Rightarrow 3x(2x+3)-2(2x+3) = 0$

$\displaystyle \Rightarrow (3x-2)(2x+3) = 0$

$\displaystyle \Rightarrow x = \frac{2}{3} \ or\ x = \frac{(-3)}{2}$

$\displaystyle \\$

Question 11: $\displaystyle 10x^2+11x+3 = 0$

Answer:

$\displaystyle 10x^2+11x+3 = 0$

$\displaystyle \Rightarrow 10x^2+6x+5x+3 = 0$

$\displaystyle \Rightarrow 5x(2x+1)+3(2x+1) = 0$

$\displaystyle \Rightarrow (2x+1)(5x+3) = 0$

$\displaystyle \Rightarrow x = \frac{(-1)}{2} \ or\ x = \frac{(-3)}{5}$

$\displaystyle \\$

Question 12: $\displaystyle 9x^2+8 = 22x$

Answer:

$\displaystyle 9x^2+8 = 22x$

$\displaystyle \Rightarrow 9x^2-22x+8 = 0$

$\displaystyle \Rightarrow 9x^2-18x-4x+8 = 0$

$\displaystyle \Rightarrow 9x(x-2)-4(x-2) = 0$

$\displaystyle \Rightarrow (x-2)(9x-4) = 0$

$\displaystyle \Rightarrow x = 2 \ or\ x = \frac{4}{9}$

$\displaystyle \\$

Question 13: $\displaystyle 22x^2+x = 6$

Answer:

$\displaystyle 22x^2+x = 6$

$\displaystyle \Rightarrow 22x^2+x-6 = 0$

$\displaystyle \Rightarrow 22x^2+11x-12x-6 = 0$

$\displaystyle \Rightarrow 11x(2x+1)-6(2x+1) = 0$

$\displaystyle \Rightarrow (2x+1)(11x-6) = 0$

$\displaystyle \Rightarrow x = \frac{(-1)}{2} \ or\ x = \frac{6}{11}$

$\displaystyle \\$

Question 14: $\displaystyle 2x^2+5x-7 = 0$

Answer:

$\displaystyle 2x^2+5x-7 = 0$

$\displaystyle \Rightarrow 2x^2+7x-2x-7 = 0$

$\displaystyle \Rightarrow 2x(x-1))+7(x-1) = 0$

$\displaystyle \Rightarrow (x-1)(2x+7) = 0$

$\displaystyle \Rightarrow x = 1 \ or\ x = \frac{(-7)}{2}$

$\displaystyle \\$

Question 15: $\displaystyle 6y^2 = 7y+20$

Answer:

$\displaystyle 6y^2 = 7y+20$

$\displaystyle \Rightarrow 6y^2-7y-20 = 0$

$\displaystyle \Rightarrow 6y^2-15y+8y-20 = 0$

$\displaystyle \Rightarrow 3y(2y-5)+4(2y-5) = 0$

$\displaystyle \Rightarrow (2y-5)(3y+4) = 0$

$\displaystyle \Rightarrow y = \frac{5}{2} \ or \ y = \frac{(-4)}{3}$

$\displaystyle \\$

Question 16: $\displaystyle 4z^2 = 11z+3$

Answer:

$\displaystyle 4z^2 = 11z+3$

$\displaystyle \Rightarrow 4z^2-11z-3 = 0$

$\displaystyle \Rightarrow 4z^2+z-12x-3 = 0$

$\displaystyle \Rightarrow 4z(z-3)+(z-3) = 0$

$\displaystyle \Rightarrow (z-3)(4z+1) = 0$

$\displaystyle \Rightarrow z = 3 \ or\ z = \frac{(-1)}{4}$

$\displaystyle \\$

Question 17: $\displaystyle (t-2)^2 = 36$

Answer:

$\displaystyle (t-2)^2 = 36$

$\displaystyle \Rightarrow t^2+4-4t = 36$

$\displaystyle \Rightarrow t^2-4t-32 = 0$

$\displaystyle \Rightarrow t^2-8t+4t-32 = 0$

$\displaystyle \Rightarrow t(t+4)-8(t+4) = 0$

$\displaystyle \Rightarrow (t+4)(t-8) = 0$

$\displaystyle \Rightarrow t = -4 \ or\ t = 8$

$\displaystyle \\$

Question 18: $\displaystyle (2x+3)(x-4) = 6$

Answer:

$\displaystyle (2x+3)(x-4) = 6$

$\displaystyle \Rightarrow 2x^2+3x-8x-12 = 6$

$\displaystyle \Rightarrow 2x^2-5x-18 = 0$

$\displaystyle \Rightarrow 2x^2-9x+4x-18 = 0$

$\displaystyle \Rightarrow 2x(x+2)-9(x+2) = 0$

$\displaystyle \Rightarrow 2x (x+2)(2x-9) = 0$

$\displaystyle \Rightarrow x = -2 \ or\ x = \frac{9}{2}$

$\displaystyle \\$

$\displaystyle \text{Question 19: } 12x+7 = \frac{10}{x}$

Answer:

$\displaystyle 12x+7 = \frac{10}{x}$

$\displaystyle \Rightarrow 12x^2+7x-10 = 0$

$\displaystyle \Rightarrow 12x^2+15x-8x-10 = 0$

$\displaystyle \Rightarrow 4x(3x-2)+5(3x-2) = 0$

$\displaystyle \Rightarrow (3x-2)(4x+5) = 0$

$\displaystyle \Rightarrow x = \frac{2}{3} \ or \ x = \frac{(-5)}{4}$

$\displaystyle \\$

Question 20: $\displaystyle x^2-10x+21 = 0$

Answer:

$\displaystyle x^2-10x+21 = 0$

$\displaystyle \Rightarrow x^2-7x-3x+21 = 0$

$\displaystyle \Rightarrow x(x-7)-3(x-7) = 0$

$\displaystyle \Rightarrow (x-7)(x-3) = 0$

$\displaystyle \Rightarrow x = 7 \ or \ x = 3$

$\displaystyle \\$

$\displaystyle \text{Question 21: } \frac{x}{3} - \frac{6}{x} = 1$

Answer:

$\displaystyle \frac{x}{3} - \frac{6}{x} = 1$

$\displaystyle \Rightarrow x^2-18 = 3x$

$\displaystyle \Rightarrow x^2-3x-18 = 0$

$\displaystyle \Rightarrow x^2-6x+3x-18 = 0$

$\displaystyle \Rightarrow x(x+3)-6(x+3)$

$\displaystyle \Rightarrow (x+3)(x-6) = 0$

$\displaystyle \Rightarrow x = -3 \ or\ x = 6$

$\displaystyle \\$

$\displaystyle \text{Question 22: } \frac{(x+1)}{(x+5)} = \frac{(2x-3)}{(x+3)}$

Answer:

$\displaystyle \frac{(x+1)}{(x+5)} = \frac{(2x-3)}{(x+3)}$

$\displaystyle \Rightarrow x^2+x+3x+3 = 2x^2-3x+10x-15$

$\displaystyle \Rightarrow x^2+3x-18 = 0$

$\displaystyle \Rightarrow x^2+6x-3x-18 = 0$

$\displaystyle \Rightarrow x(x-3)+6(x-3) = 0$

$\displaystyle \Rightarrow (x-3)(x+6)$

$\displaystyle \Rightarrow x = 3 \ or\ x = -6$

$\displaystyle \\$

$\displaystyle \text{Question 23: } \frac{(x+1)}{(3x-7)} = \frac{(x-1)}{(2x-5)}$

Answer:

$\displaystyle \frac{(x+1)}{(3x-7)} = \frac{(x-1)}{(2x-5)}$

$\displaystyle \Rightarrow 2x^2+2x-5x-5 = 3x^2-3x-7x+7$

$\displaystyle \Rightarrow 2x^2-3x-5 = 3x^2-10x+7$

$\displaystyle \Rightarrow x^2-7x+12 = 0$

$\displaystyle \Rightarrow x^2-3x-4x+12 = 0$

$\displaystyle \Rightarrow x(x-3)-4(x-3) = 0$

$\displaystyle \Rightarrow (x-3)(x-4) = 0$

$\displaystyle \Rightarrow x = 3 \ or\ x = 4$

$\displaystyle \\$

$\displaystyle \text{Question 24: } \frac{1}{(x-1)} - \frac{1}{(x+2)} = \frac{3}{4}$

Answer:

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

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

$\displaystyle \Rightarrow 3 = \frac{3}{4} (x^2-x+2x-2)$

$\displaystyle \Rightarrow 4 = x^2+x-2$

$\displaystyle \Rightarrow x^2+x-6 = 0$

$\displaystyle \Rightarrow x^2+3x-2x-6 = 0$

$\displaystyle \Rightarrow x(x-2)+3(x-2) = 0$

$\displaystyle \Rightarrow (x-2)(x+3) = 0$

$\displaystyle \Rightarrow x = 2 \ or\ x = -3$

$\displaystyle \\$

$\displaystyle \text{Question 25: } \frac{4}{(x-1)} - \frac{3}{x} = \frac{5}{x+2}$

Answer:

$\displaystyle \frac{4}{(x-1)} - \frac{3}{x} = \frac{5}{x+2}$

Multiply LHS and RHS by $\displaystyle (x-1)(x)(x+2)$

$\displaystyle \Rightarrow 4(x)(x+2)-3(x-1)(x+2) = 5(x-1)(x)$

$\displaystyle \Rightarrow 4x^2+8x-3x^2-3x+6 = 5x^2-5x$

$\displaystyle \Rightarrow 4x^2+8x-3x^2-3x+6 = 5x^2-5x$

$\displaystyle \Rightarrow x^2+5x+6 = 5x^2-5x$

$\displaystyle \Rightarrow 4x^2-10x-6 = 0$

$\displaystyle \Rightarrow 2x^2-5x-3 = 0$

$\displaystyle \Rightarrow 2x^2-6x+x-3 = 0$

$\displaystyle \Rightarrow 2x (x-3)+(x-3) = 1$

$\displaystyle \Rightarrow (x-3)(2x+1) = 0$

$\displaystyle \Rightarrow x = 3 \ or\ - \frac{1}{2}$

$\displaystyle \\$

$\displaystyle \text{Question 26: } \frac{(x+2)}{(x-1)} - \frac{4-x}{2x} = \frac{7}{2}$

Answer:

$\displaystyle \frac{(x+2)}{(x-1)} - \frac{4-x}{2x} = \frac{7}{2}$

$\displaystyle \Rightarrow 2x(x+2)-(x-1)(4-x) = \frac{7}{2} (x-1)(2x)$

$\displaystyle \Rightarrow 2x^2+4x-4x+4+x^2-x = 7(x^2-x)$

$\displaystyle \Rightarrow 3x^2-x+4 = 7x^2-7x$

$\displaystyle \Rightarrow 4x^2-6x-4 = 0$

$\displaystyle \Rightarrow 2x^2-3x-2 = 0$

$\displaystyle \Rightarrow 2x^2-4x+x-2 = 0$

$\displaystyle \Rightarrow 2x (x-2)+1(x-2) = 0$

$\displaystyle \Rightarrow (x-2)(2x+1) = 0$

$\displaystyle \Rightarrow x = 2 \ or\ x = - \frac{1}{2}$

Related

Leave a Reply Cancel reply

%d bloggers like this: