Class 8: Chapter 26 – Quadratic Equations – Exercise 26 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Solve each of the following equation:

$1)\ \ x^2-11x+28=0$

$x^2-7x-4x+28=0$

$x (x-4)-7 (x-4)=0$

$(x-4)(x-7)=0$

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

$\\$

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

$x^2-13x-68x=0$

$x^2-17x+4x-68x=0$

$x(x-17)+4(x-17)=0$

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

$\\$

$3)\ \ x^2+2x=48$

$x^2+2x-48=0$

$x^2+8x-6x-48=0$

$x(x+8)-6(x+8)=0$

$(x-6)(x+8)=0$

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

$\\$

$4)\ \ x^2-2x=3$

$x^2-2x-3=0$

$x^2-3x+1x-3=0$

$(x+1)(x-3)=0$

$x(x-3)+1(x-3)=0$

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

$\\$

$5)\ \ x^2-49=0$

$(x-7)(x+7)=0$

$x=7 \ or\ x=-7$

$\\$

$6)\ \ 3x^2-243=0$

$3 (x^2-81)=0$

$(x-9)(x+9)=0$

$x=9 \ or\ x=-9$

$\\$

$7)\ \ 16x^2=9$

$16 x^2-9=0$

$(4x-3)(4x+3)=0$

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

$\\$

$8)\ \ 3x^2+13x=10$

$3x^2+13x-10=0$

$3x^2-2x+15x-10=0$

$x(3x-2)+5(3x-2)=0$

$(3x-2)(x+5)=0$

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

$\\$

$9)\ \ 2x^2-7x+3=0$

$2 x^2-6x-x+3=0$

$2x(x-3)-1(x-3)=0$

$(2x-1)(x-3)=0$

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

$\\$

$10)\ \ 6x^2+5x=6$

$6x^2+5x-6=0$

$6x^2+9x-4x-6=0$

$3x(2x+3)-2(2x+3)=0$

$(3x-2)(2x+3)=0$

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

$\\$

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

$10x^2+6x+5x+3=0$

$5x(2x+1)+3(2x+1)=0$

$(2x+1)(5x+3)=0$

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

$\\$

$12)\ \ 9x^2+8=22x$

$9x^2-22x+8=0$

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

$9x(x-2)-4(x-2)=0$

$(x-2)(9x-4)=0$

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

$\\$

$13)\ \ 22x^2+x=6$

$22x^2+x-6=0$

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

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

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

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

$\\$

$14)\ \ 2x^2+5x-7=0$

$2x^2+7x-2x-7=0$

$2x(x-1))+7(x-1)=0$

$(x-1)(2x+7)=0$

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

$\\$

$15)\ \ 6y^2=7y+20$

$6y^2-7y-20=0$

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

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

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

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

$\\$

$16)\ \ 4z^2=11z+3$

$4z^2-11z-3=0$

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

$4z(z-3)+(z-3)=0$

$(z-3)(4z+1)=0$

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

$\\$

$17)\ \ (t-2)^2=36$

$t^2+4-4t=36$

$t^2-4t-32=0$

$t^2-8t+4t-32=0$

$t(t+4)-8(t+4)=0$

$(t+4)(t-8)=0$

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

$\\$

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

$2x^2+3x-8x-12=6$

$2x^2-5x-18=0$

$2x^2-9x+4x-18=0$

$2x(x+2)-9(x+2)=0$

$2x (x+2)(2x-9)=0$

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

$\\$

$19)\ \ 12x+7=10/x$

$12x^2+7x-10=0$

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

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

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

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

$\\$

$20)\ \ x^2-10x+21=0$

$x^2-7x-3x+21=0$

$x(x-7)-3(x-7)=0$

$(x-7)(x-3)=0$

$x=7 \ or \ x=3$

$\\$

$21)$ $\ \ \frac{x}{3}-\frac{6}{x}$ $=1$

$x^2-18=3x$

$x^2-3x-18=0$

$x^2-6x+3x-18=0$

$x(x+3)-6(x+3)$

$(x+3)(x-6)=0$

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

$\\$

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

$x^2+x+3x+3=2x^2-3x+10x-15$

$x^2+3x-18=0$

$x^2+6x-3x-18=0$

$x(x-3)+6(x-3)=0$

$(x-3)(x+6)$

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

$\\$

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

$2x^2+2x-5x-5=3x^2-3x-7x+7$

$2x^2-3x-5=3x^2-10x+7$

$x^2-7x+12=0$

$x^2-3x-4x+12=0$

$x(x-3)-4(x-3)=0$

$(x-3)(x-4)=0$

$x=3 \ or\ x=4$

$\\$

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

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

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

$4=x^2+x-2$

$x^2+x-6=0$

$x^2+3x-2x-6=0$

$x(x-2)+3(x-2)=0$

$(x-2)(x+3)=0$

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

$\\$

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

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

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

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

$x^2+5x+6=5x^2-5x$

$4x^2-10x-6=0$

$2x^2-5x-3=0$

$2x^2-6x+x-3=0$

$2x (x-3)+(x-3)=1$

$(x-3)(2x+1)=0$

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

$\\$

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

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

$2x^2+4x-4x+4+x^2-x=7(x^2-x)$

$3x^2-x+4=7x^2-7x$

$4x^2-6x-4=0$

$2x^2-3x-2=0$

$2x^2-4x+x-2=0$

$2x (x-2)+1(x-2)=0$

$(x-2)(2x+1)=0$

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

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Class 8: Chapter 26 – Quadratic Equations – Exercise 26

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