Class 8: Quadratic Equations

$\displaystyle \text{Solve each of the following quadratic equations:}$

$\displaystyle \textbf{Question 1: }\quad x^2-11x+28=0$
$\displaystyle \text{Answer:}$
$\displaystyle x^2-11x+28=0$
$\displaystyle \Rightarrow x^2-7x-4x+28=0$
$\displaystyle \Rightarrow x(x-7)-4(x-7)=0$
$\displaystyle \Rightarrow (x-4)(x-7)=0$
$\displaystyle \Rightarrow x-4=0\quad\text{or}\quad x-7=0$
$\displaystyle \therefore x=4\quad\text{or}\quad x=7.$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\quad x^2-13x=68$
$\displaystyle \text{Answer:}$
$\displaystyle x^2-13x=68$
$\displaystyle \Rightarrow x^2-13x-68=0$
$\displaystyle \Rightarrow x^2-17x+4x-68=0$
$\displaystyle \Rightarrow x(x-17)+4(x-17)=0$
$\displaystyle \Rightarrow (x+4)(x-17)=0$
$\displaystyle \Rightarrow x+4=0\quad\text{or}\quad x-17=0$
$\displaystyle \therefore x=-4\quad\text{or}\quad x=17.$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\quad x^2+2x=48$
$\displaystyle \text{Answer:}$
$\displaystyle x^2+2x=48$
$\displaystyle \Rightarrow x^2+2x-48=0$
$\displaystyle \Rightarrow x^2+8x-6x-48=0$
$\displaystyle \Rightarrow x(x+8)-6(x+8)=0$
$\displaystyle \Rightarrow (x-6)(x+8)=0$
$\displaystyle \Rightarrow x-6=0\quad\text{or}\quad x+8=0$
$\displaystyle \therefore x=6\quad\text{or}\quad x=-8.$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\quad x^2-2x=3$
$\displaystyle \text{Answer:}$
$\displaystyle x^2-2x=3$
$\displaystyle \Rightarrow x^2-2x-3=0$
$\displaystyle \Rightarrow x^2-3x+x-3=0$
$\displaystyle \Rightarrow x(x-3)+1(x-3)=0$
$\displaystyle \Rightarrow (x+1)(x-3)=0$
$\displaystyle \Rightarrow x+1=0\quad\text{or}\quad x-3=0$
$\displaystyle \therefore x=-1\quad\text{or}\quad x=3.$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\quad x^2-49=0$
$\displaystyle \text{Answer:}$
$\displaystyle x^2-49=0$
$\displaystyle \Rightarrow x^2-7^2=0$
$\displaystyle \Rightarrow (x-7)(x+7)=0$
$\displaystyle \Rightarrow x-7=0\quad\text{or}\quad x+7=0$
$\displaystyle \therefore x=7\quad\text{or}\quad x=-7.$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\quad 3x^2-243=0$
$\displaystyle \text{Answer:}$
$\displaystyle 3x^2-243=0$
$\displaystyle \Rightarrow 3(x^2-81)=0$
$\displaystyle \Rightarrow 3(x^2-9^2)=0$
$\displaystyle \Rightarrow 3(x-9)(x+9)=0$
$\displaystyle \Rightarrow (x-9)(x+9)=0$
$\displaystyle \Rightarrow x-9=0\quad\text{or}\quad x+9=0$
$\displaystyle \therefore x=9\quad\text{or}\quad x=-9.$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\quad 16x^2=9$
$\displaystyle \text{Answer:}$
$\displaystyle 16x^2=9$
$\displaystyle \Rightarrow 16x^2-9=0$
$\displaystyle \Rightarrow (4x)^2-3^2=0$
$\displaystyle \Rightarrow (4x-3)(4x+3)=0$
$\displaystyle \Rightarrow 4x-3=0\quad\text{or}\quad 4x+3=0$
$\displaystyle \therefore x=\frac{3}{4}\quad\text{or}\quad x=-\frac{3}{4}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\quad 3x^2+13x=10$
$\displaystyle \text{Answer:}$
$\displaystyle 3x^2+13x=10$
$\displaystyle \Rightarrow 3x^2+13x-10=0$
$\displaystyle \Rightarrow 3x^2-2x+15x-10=0$
$\displaystyle \Rightarrow x(3x-2)+5(3x-2)=0$
$\displaystyle \Rightarrow (3x-2)(x+5)=0$
$\displaystyle \Rightarrow 3x-2=0\quad\text{or}\quad x+5=0$
$\displaystyle \therefore x=\frac{2}{3}\quad\text{or}\quad x=-5.$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\quad 2x^2-7x+3=0$
$\displaystyle \text{Answer:}$
$\displaystyle 2x^2-7x+3=0$
$\displaystyle \Rightarrow 2x^2-6x-x+3=0$
$\displaystyle \Rightarrow 2x(x-3)-1(x-3)=0$
$\displaystyle \Rightarrow (2x-1)(x-3)=0$
$\displaystyle \Rightarrow 2x-1=0\quad\text{or}\quad x-3=0$
$\displaystyle \therefore x=\frac{1}{2}\quad\text{or}\quad x=3.$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\quad 6x^2+5x=6$
$\displaystyle \text{Answer:}$
$\displaystyle 6x^2+5x=6$
$\displaystyle \Rightarrow 6x^2+5x-6=0$
$\displaystyle \Rightarrow 6x^2+9x-4x-6=0$
$\displaystyle \Rightarrow 3x(2x+3)-2(2x+3)=0$
$\displaystyle \Rightarrow (3x-2)(2x+3)=0$
$\displaystyle \Rightarrow 3x-2=0\quad\text{or}\quad 2x+3=0$
$\displaystyle \therefore x=\frac{2}{3}\quad\text{or}\quad x=-\frac{3}{2}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\quad 10x^2+11x+3=0$
$\displaystyle \text{Answer:}$
$\displaystyle 10x^2+11x+3=0$
$\displaystyle \Rightarrow 10x^2+6x+5x+3=0$
$\displaystyle \Rightarrow 2x(5x+3)+1(5x+3)=0$
$\displaystyle \Rightarrow (2x+1)(5x+3)=0$
$\displaystyle \Rightarrow 2x+1=0\quad\text{or}\quad 5x+3=0$
$\displaystyle \therefore x=-\frac{1}{2}\quad\text{or}\quad x=-\frac{3}{5}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: }\quad 9x^2+8=22x$
$\displaystyle \text{Answer:}$
$\displaystyle 9x^2+8=22x$
$\displaystyle \Rightarrow 9x^2-22x+8=0$
$\displaystyle \Rightarrow 9x^2-18x-4x+8=0$
$\displaystyle \Rightarrow 9x(x-2)-4(x-2)=0$
$\displaystyle \Rightarrow (x-2)(9x-4)=0$
$\displaystyle \Rightarrow x-2=0\quad\text{or}\quad 9x-4=0$
$\displaystyle \therefore x=2\quad\text{or}\quad x=\frac{4}{9}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: }\quad 22x^2-x=6$
$\displaystyle \text{Answer:}$
$\displaystyle 22x^2-x=6$
$\displaystyle \Rightarrow 22x^2-x-6=0$
$\displaystyle \Rightarrow 22x^2+11x-12x-6=0$
$\displaystyle \Rightarrow 11x(2x+1)-6(2x+1)=0$
$\displaystyle \Rightarrow (2x+1)(11x-6)=0$
$\displaystyle \Rightarrow 2x+1=0\quad\text{or}\quad 11x-6=0$
$\displaystyle \therefore x=-\frac{1}{2}\quad\text{or}\quad x=\frac{6}{11}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 14: }\quad 2x^2+5x-7=0$
$\displaystyle \text{Answer:}$
$\displaystyle 2x^2+5x-7=0$
$\displaystyle \Rightarrow 2x^2+7x-2x-7=0$
$\displaystyle \Rightarrow 2x(x-1)+7(x-1)=0$
$\displaystyle \Rightarrow (x-1)(2x+7)=0$
$\displaystyle \Rightarrow x-1=0\quad\text{or}\quad 2x+7=0$
$\displaystyle \therefore x=1\quad\text{or}\quad x=-\frac{7}{2}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 15: }\quad 6y^2=7y+20$
$\displaystyle \text{Answer:}$
$\displaystyle 6y^2=7y+20$
$\displaystyle \Rightarrow 6y^2-7y-20=0$
$\displaystyle \Rightarrow 6y^2-15y+8y-20=0$
$\displaystyle \Rightarrow 3y(2y-5)+4(2y-5)=0$
$\displaystyle \Rightarrow (2y-5)(3y+4)=0$
$\displaystyle \Rightarrow 2y-5=0\quad\text{or}\quad 3y+4=0$
$\displaystyle \therefore y=\frac{5}{2}\quad\text{or}\quad y=-\frac{4}{3}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 16: }\quad 4z^2=11z+3$
$\displaystyle \text{Answer:}$
$\displaystyle 4z^2=11z+3$
$\displaystyle \Rightarrow 4z^2-11z-3=0$
$\displaystyle \Rightarrow 4z^2+z-12z-3=0$
$\displaystyle \Rightarrow z(4z+1)-3(4z+1)=0$
$\displaystyle \Rightarrow (4z+1)(z-3)=0$
$\displaystyle \Rightarrow 4z+1=0\quad\text{or}\quad z-3=0$
$\displaystyle \therefore z=-\frac{1}{4}\quad\text{or}\quad z=3.$
$\displaystyle \\$

$\displaystyle \textbf{Question 17: }\quad (t-2)^2=36$
$\displaystyle \text{Answer:}$
$\displaystyle (t-2)^2=36$
$\displaystyle \Rightarrow t^2-4t+4=36$
$\displaystyle \Rightarrow t^2-4t-32=0$
$\displaystyle \Rightarrow t^2-8t+4t-32=0$
$\displaystyle \Rightarrow t(t-8)+4(t-8)=0$
$\displaystyle \Rightarrow (t+4)(t-8)=0$
$\displaystyle \Rightarrow t+4=0\quad\text{or}\quad t-8=0$
$\displaystyle \therefore t=-4\quad\text{or}\quad t=8.$
$\displaystyle \\$

$\displaystyle \textbf{Question 18: }\quad (2x+3)(x-4)=6$
$\displaystyle \text{Answer:}$
$\displaystyle (2x+3)(x-4)=6$
$\displaystyle \Rightarrow 2x^2-8x+3x-12=6$
$\displaystyle \Rightarrow 2x^2-5x-18=0$
$\displaystyle \Rightarrow 2x^2+4x-9x-18=0$
$\displaystyle \Rightarrow 2x(x+2)-9(x+2)=0$
$\displaystyle \Rightarrow (x+2)(2x-9)=0$
$\displaystyle \Rightarrow x+2=0\quad\text{or}\quad 2x-9=0$
$\displaystyle \therefore x=-2\quad\text{or}\quad x=\frac{9}{2}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 19: }\quad 12x+7=\frac{10}{x}$
$\displaystyle \text{Answer:}$
$\displaystyle 12x+7=\frac{10}{x},\qquad x\ne0$
$\displaystyle \Rightarrow 12x^2+7x=10$
$\displaystyle \Rightarrow 12x^2+7x-10=0$
$\displaystyle \Rightarrow 12x^2+15x-8x-10=0$
$\displaystyle \Rightarrow 3x(4x+5)-2(4x+5)=0$
$\displaystyle \Rightarrow (4x+5)(3x-2)=0$
$\displaystyle \Rightarrow 4x+5=0\quad\text{or}\quad 3x-2=0$
$\displaystyle \therefore x=-\frac{5}{4}\quad\text{or}\quad x=\frac{2}{3}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 20: }\quad x^2-10x+21=0$
$\displaystyle \text{Answer:}$
$\displaystyle x^2-10x+21=0$
$\displaystyle \Rightarrow x^2-7x-3x+21=0$
$\displaystyle \Rightarrow x(x-7)-3(x-7)=0$
$\displaystyle \Rightarrow (x-7)(x-3)=0$
$\displaystyle \Rightarrow x-7=0\quad\text{or}\quad x-3=0$
$\displaystyle \therefore x=7\quad\text{or}\quad x=3.$
$\displaystyle \\$

$\displaystyle \textbf{Question 21: }\quad \frac{x}{3}-\frac{6}{x}=1$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{x}{3}-\frac{6}{x}=1,\qquad x\ne0$
$\displaystyle \Rightarrow x^2-18=3x$
$\displaystyle \Rightarrow x^2-3x-18=0$
$\displaystyle \Rightarrow x^2-6x+3x-18=0$
$\displaystyle \Rightarrow x(x-6)+3(x-6)=0$
$\displaystyle \Rightarrow (x+3)(x-6)=0$
$\displaystyle \Rightarrow x+3=0\quad\text{or}\quad x-6=0$
$\displaystyle \therefore x=-3\quad\text{or}\quad x=6.$
$\displaystyle \\$

$\displaystyle \textbf{Question 22: }\quad \frac{x+1}{x+5}=\frac{2x-3}{x+3}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{x+1}{x+5}=\frac{2x-3}{x+3}, \qquad x\ne-5,-3$
$\displaystyle \Rightarrow (x+1)(x+3)=(2x-3)(x+5)$
$\displaystyle \Rightarrow x^2+4x+3=2x^2+7x-15$
$\displaystyle \Rightarrow x^2+3x-18=0$
$\displaystyle \Rightarrow x^2+6x-3x-18=0$
$\displaystyle \Rightarrow x(x+6)-3(x+6)=0$
$\displaystyle \Rightarrow (x+6)(x-3)=0$
$\displaystyle \Rightarrow x+6=0\quad\text{or}\quad x-3=0$
$\displaystyle \therefore x=-6\quad\text{or}\quad x=3.$
$\displaystyle \\$

$\displaystyle \textbf{Question 23: }\quad \frac{x+1}{3x-7}=\frac{x-1}{2x-5}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{x+1}{3x-7}=\frac{x-1}{2x-5}, \qquad x\ne\frac{7}{3},\frac{5}{2}$
$\displaystyle \Rightarrow (x+1)(2x-5)=(x-1)(3x-7)$
$\displaystyle \Rightarrow 2x^2-3x-5=3x^2-10x+7$
$\displaystyle \Rightarrow x^2-7x+12=0$
$\displaystyle \Rightarrow x^2-3x-4x+12=0$
$\displaystyle \Rightarrow x(x-3)-4(x-3)=0$
$\displaystyle \Rightarrow (x-3)(x-4)=0$
$\displaystyle \Rightarrow x-3=0\quad\text{or}\quad x-4=0$
$\displaystyle \therefore x=3\quad\text{or}\quad x=4.$
$\displaystyle \\$

$\displaystyle \textbf{Question 24: }\quad \frac{1}{x-1}-\frac{1}{x+2}=\frac{3}{4}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{1}{x-1}-\frac{1}{x+2}=\frac{3}{4}, \qquad x\ne1,-2$
$\displaystyle \Rightarrow \frac{(x+2)-(x-1)}{(x-1)(x+2)}=\frac{3}{4}$
$\displaystyle \Rightarrow 4\{(x+2)-(x-1)\}=3(x-1)(x+2)$
$\displaystyle \Rightarrow 12=3(x^2+x-2)$
$\displaystyle \Rightarrow 4=x^2+x-2$
$\displaystyle \Rightarrow x^2+x-6=0$
$\displaystyle \Rightarrow x^2+3x-2x-6=0$
$\displaystyle \Rightarrow x(x+3)-2(x+3)=0$
$\displaystyle \Rightarrow (x+3)(x-2)=0$
$\displaystyle \Rightarrow x+3=0\quad\text{or}\quad x-2=0$
$\displaystyle \therefore x=-3\quad\text{or}\quad x=2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 25: }\quad \frac{4}{x-1}-\frac{3}{x}=\frac{5}{x+2}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{4}{x-1}-\frac{3}{x}=\frac{5}{x+2}, \qquad x\ne1,0,-2$
$\displaystyle \text{Multiplying throughout by }x(x-1)(x+2),$
$\displaystyle \Rightarrow 4x(x+2)-3(x-1)(x+2)=5x(x-1)$
$\displaystyle \Rightarrow 4x^2+8x-3(x^2+x-2)=5x^2-5x$
$\displaystyle \Rightarrow 4x^2+8x-3x^2-3x+6=5x^2-5x$
$\displaystyle \Rightarrow x^2+5x+6=5x^2-5x$
$\displaystyle \Rightarrow 4x^2-10x-6=0$
$\displaystyle \Rightarrow 2x^2-5x-3=0$
$\displaystyle \Rightarrow 2x^2-6x+x-3=0$
$\displaystyle \Rightarrow 2x(x-3)+1(x-3)=0$
$\displaystyle \Rightarrow (x-3)(2x+1)=0$
$\displaystyle \Rightarrow x-3=0\quad\text{or}\quad 2x+1=0$
$\displaystyle \therefore x=3\quad\text{or}\quad x=-\frac{1}{2}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 26: }\quad \frac{x+2}{x-1}-\frac{4-x}{2x}=\frac{7}{2}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{x+2}{x-1}-\frac{4-x}{2x}=\frac{7}{2}, \qquad x\ne1,0$
$\displaystyle \text{Multiplying throughout by }2x(x-1),$
$\displaystyle \Rightarrow 2x(x+2)-(x-1)(4-x)=7x(x-1)$
$\displaystyle \Rightarrow 2x^2+4x-\left(-x^2+5x-4\right)=7x^2-7x$
$\displaystyle \Rightarrow 3x^2-x+4=7x^2-7x$
$\displaystyle \Rightarrow 4x^2-6x-4=0$
$\displaystyle \Rightarrow 2x^2-3x-2=0$
$\displaystyle \Rightarrow 2x^2-4x+x-2=0$
$\displaystyle \Rightarrow 2x(x-2)+1(x-2)=0$
$\displaystyle \Rightarrow (x-2)(2x+1)=0$
$\displaystyle \Rightarrow x-2=0\quad\text{or}\quad 2x+1=0$
$\displaystyle \therefore x=2\quad\text{or}\quad x=-\frac{1}{2}.$
$\displaystyle \\$