Class 10: Quadratic Equations

Solve each of the following equations:

$\displaystyle \textbf{Question 1: } \text{Solve }\frac{2x}{x-3}+\frac{1}{2x+3}+\frac{3x+9}{(x-3)(2x+3)}=0,\ x\neq3,\ x\neq-\frac{3}{2}.$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{2x}{x-3}+\frac{1}{2x+3}+\frac{3x+9}{(x-3)(2x+3)}=0$
$\displaystyle \frac{2x(2x+3)+(x-3)+(3x+9)}{(x-3)(2x+3)}=0$
$\displaystyle 4x^2+6x+x-3+3x+9=0$
$\displaystyle 4x^2+10x+6=0$
$\displaystyle 2x^2+5x+3=0$
$\displaystyle 2x^2+2x+3x+3=0$
$\displaystyle 2x(x+1)+3(x+1)=0$
$\displaystyle (2x+3)(x+1)=0$
$\displaystyle \Rightarrow x=-\frac{3}{2}\text{ or }-1$
$\displaystyle \text{But }x\neq-\frac{3}{2}$
$\displaystyle \therefore x=-1$
$\\$

$\displaystyle \textbf{Question 2: } \text{Solve }(2x+3)^2=81.$
$\displaystyle \text{Answer:}$
$\displaystyle (2x+3)^2=81$
$\displaystyle 4x^2+12x+9=81$
$\displaystyle 4x^2+12x-72=0$
$\displaystyle x^2+3x-18=0$
$\displaystyle x^2+6x-3x-18=0$
$\displaystyle x(x+6)-3(x+6)=0$
$\displaystyle (x-3)(x+6)=0$
$\displaystyle \Rightarrow x=3\text{ or }-6$
$\\$

$\displaystyle \textbf{Question 3: } \text{Solve }a^2x^2-b^2=0.$
$\displaystyle \text{Answer:}$
$\displaystyle a^2x^2-b^2=0$
$\displaystyle (ax-b)(ax+b)=0$
$\displaystyle \Rightarrow ax=b\text{ or }ax=-b$
$\displaystyle \Rightarrow x=\frac{b}{a}\text{ or }-\frac{b}{a}$
$\\$

$\displaystyle \textbf{Question 4: } \text{Solve }x^2-\frac{11}{4}x+\frac{15}{8}=0.$
$\displaystyle \text{Answer:}$
$\displaystyle x^2-\frac{11}{4}x+\frac{15}{8}=0$
$\displaystyle \text{Multiplying the equation by }8,$
$\displaystyle 8x^2-22x+15=0$
$\displaystyle 8x^2-10x-12x+15=0$
$\displaystyle 4x(2x-3)-5(2x-3)=0$
$\displaystyle (2x-3)(4x-5)=0$
$\displaystyle \Rightarrow x=\frac{3}{2}\text{ or }\frac{5}{4}$
$\\$

$\displaystyle \textbf{Question 5: } \text{Solve }x+\frac{4}{x}=-4,\ x\neq0.$
$\displaystyle \text{Answer:}$
$\displaystyle x+\frac{4}{x}=-4$
$\displaystyle x^2+4=-4x$
$\displaystyle x^2+4x+4=0$
$\displaystyle x^2+2x+2x+4=0$
$\displaystyle (x+2)(x+2)=0$
$\displaystyle \Rightarrow x=-2$
$\\$

$\displaystyle \textbf{Question 6: } \text{Solve }2x^4-5x^2+3=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }x^2=y$
$\displaystyle 2y^2-5y+3=0$
$\displaystyle 2y^2-3y-2y+3=0$
$\displaystyle 2y(y-1)-3(y-1)=0$
$\displaystyle (2y-3)(y-1)=0$
$\displaystyle \Rightarrow y=\frac{3}{2}\text{ or }1$
$\displaystyle \text{When }y=1,\ x^2=1$
$\displaystyle \Rightarrow x=\pm1$
$\displaystyle \text{When }y=\frac{3}{2},\ x^2=\frac{3}{2}$
$\displaystyle \Rightarrow x=\pm\sqrt{\frac{3}{2}}$
$\\$

$\displaystyle \textbf{Question 7: } \text{Solve }x^4-2x^2-3=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }x^2=y$
$\displaystyle y^2-2y-3=0$
$\displaystyle y^2-3y+y-3=0$
$\displaystyle y(y-3)+1(y-3)=0$
$\displaystyle (y+1)(y-3)=0$
$\displaystyle \Rightarrow y=-1\text{ or }3$
$\displaystyle \text{When }y=-1,\ x^2=-1$
$\displaystyle \Rightarrow x\text{ is imaginary}$
$\displaystyle \text{When }y=3,\ x^2=3$
$\displaystyle \Rightarrow x=\pm\sqrt{3}$
$\\$

$\displaystyle \textbf{Question 8: } \text{Solve }9\left(x^2+\frac{1}{x^2}\right)-9\left(x+\frac{1}{x}\right)-52=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }x+\frac{1}{x}=y$
$\displaystyle x^2+\frac{1}{x^2}+2=y^2$
$\displaystyle x^2+\frac{1}{x^2}=y^2-2$
$\displaystyle \therefore 9(y^2-2)-9y-52=0$
$\displaystyle 9y^2-9y-70=0$
$\displaystyle 9y^2-30y+21y-70=0$
$\displaystyle 3y(3y-10)+7(3y-10)=0$
$\displaystyle (3y+7)(3y-10)=0$
$\displaystyle \Rightarrow y=\frac{10}{3}\text{ or }-\frac{7}{3}$
$\displaystyle \text{When }y=\frac{10}{3},\ x+\frac{1}{x}=\frac{10}{3}$
$\displaystyle 3x^2-10x+3=0$
$\displaystyle 3x^2-9x-x+3=0$
$\displaystyle 3x(x-3)-1(x-3)=0$
$\displaystyle (3x-1)(x-3)=0$
$\displaystyle \Rightarrow x=\frac{1}{3}\text{ or }3$
$\displaystyle \text{When }y=-\frac{7}{3},\ x+\frac{1}{x}=-\frac{7}{3}$
$\displaystyle 3x^2+7x+3=0$
$\displaystyle x=\frac{-7\pm\sqrt{7^2-4(3)(3)}}{2(3)}$
$\displaystyle =\frac{-7\pm\sqrt{13}}{6}$
$\displaystyle \therefore x=\frac{1}{3},3,\frac{-7+\sqrt{13}}{6},\frac{-7-\sqrt{13}}{6}$
$\\$

$\displaystyle \textbf{Question 9: } \text{Solve }2\left(x^2+\frac{1}{x^2}\right)-\left(x+\frac{1}{x}\right)=11.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }x+\frac{1}{x}=y$
$\displaystyle \therefore x^2+\frac{1}{x^2}=y^2-2$
$\displaystyle 2(y^2-2)-y=11$
$\displaystyle 2y^2-y-15=0$
$\displaystyle 2y^2-6y+5y-15=0$
$\displaystyle 2y(y-3)+5(y-3)=0$
$\displaystyle (2y+5)(y-3)=0$
$\displaystyle \Rightarrow y=3\text{ or }-\frac{5}{2}$
$\displaystyle \text{When }y=3,$
$\displaystyle x+\frac{1}{x}=3$
$\displaystyle x^2-3x+1=0$
$\displaystyle x=\frac{3\pm\sqrt{9-4}}{2}=\frac{3\pm\sqrt{5}}{2}$
$\displaystyle \text{When }y=-\frac{5}{2},$
$\displaystyle x+\frac{1}{x}=-\frac{5}{2}$
$\displaystyle 2x^2+5x+2=0$
$\displaystyle 2x^2+4x+x+2=0$
$\displaystyle 2x(x+2)+1(x+2)=0$
$\displaystyle (2x+1)(x+2)=0$
$\displaystyle \Rightarrow x=-2\text{ or }-\frac{1}{2}$
$\displaystyle \therefore x=\frac{3+\sqrt{5}}{2},\ \frac{3-\sqrt{5}}{2},\ -2,\ -\frac{1}{2}$
$\\$

$\displaystyle \textbf{Question 10: } \text{Solve }\left(x^2+\frac{1}{x^2}\right)-3\left(x+\frac{1}{x}\right)-2=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }x+\frac{1}{x}=y$
$\displaystyle \therefore x^2+\frac{1}{x^2}=y^2-2$
$\displaystyle (y^2-2)-3y-2=0$
$\displaystyle y^2-3y-4=0$
$\displaystyle y^2-4y+y-4=0$
$\displaystyle y(y-4)+1(y-4)=0$
$\displaystyle (y+1)(y-4)=0$
$\displaystyle \Rightarrow y=4\text{ or }-1$
$\displaystyle \text{When }y=4,$
$\displaystyle x+\frac{1}{x}=4$
$\displaystyle x^2-4x+1=0$
$\displaystyle x=\frac{4\pm\sqrt{16-4}}{2}=2\pm\sqrt{3}$
$\displaystyle \text{When }y=-1,$
$\displaystyle x+\frac{1}{x}=-1$
$\displaystyle x^2+x+1=0$
$\displaystyle x=\frac{-1\pm\sqrt{1-4}}{2}=\frac{-1\pm i\sqrt{3}}{2}$
$\displaystyle \therefore x=2\pm\sqrt{3},\ \frac{-1\pm i\sqrt{3}}{2}$
$\\$

$\displaystyle \textbf{Question 11: } \text{Solve }(x^2+5x+4)(x^2+5x+6)=120.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }x^2+5x=y$
$\displaystyle (y+4)(y+6)=120$
$\displaystyle y^2+10y+24=120$
$\displaystyle y^2+10y-96=0$
$\displaystyle y^2-6y+16y-96=0$
$\displaystyle y(y-6)+16(y-6)=0$
$\displaystyle (y+16)(y-6)=0$
$\displaystyle \Rightarrow y=6\text{ or }-16$
$\displaystyle \text{When }y=6,$
$\displaystyle x^2+5x=6$
$\displaystyle x^2+5x-6=0$
$\displaystyle x^2+6x-x-6=0$
$\displaystyle x(x+6)-1(x+6)=0$
$\displaystyle (x-1)(x+6)=0$
$\displaystyle \Rightarrow x=1\text{ or }-6$
$\displaystyle \text{When }y=-16,$
$\displaystyle x^2+5x=-16$
$\displaystyle x^2+5x+16=0$
$\displaystyle \text{This gives imaginary roots.}$
$\displaystyle \therefore \text{Real roots are }x=1\text{ or }-6$
$\\$

$\displaystyle \textbf{Question 12: } \text{Solve }x^2-5x-10=0.\hfill\text{[ICSE 2005]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Comparing with }ax^2+bx+c=0,\ a=1,\ b=-5,\ c=-10$
$\displaystyle \text{Since }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle x=\frac{-(-5)\pm\sqrt{(-5)^2-4(1)(-10)}}{2(1)}$
$\displaystyle =\frac{5\pm\sqrt{25+40}}{2}$
$\displaystyle =\frac{5\pm\sqrt{65}}{2}$
$\displaystyle \Rightarrow x=\frac{5+\sqrt{65}}{2}\approx6.53$
$\displaystyle \text{or }x=\frac{5-\sqrt{65}}{2}\approx-1.53$
$\\$

$\displaystyle \textbf{Question 13: } \text{Solve }3x^2-x-7=0.\hfill\text{[ICSE 2004]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Comparing with }ax^2+bx+c=0,\ a=3,\ b=-1,\ c=-7$
$\displaystyle \text{Since }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle x=\frac{-(-1)\pm\sqrt{(-1)^2-4(3)(-7)}}{2(3)}$
$\displaystyle =\frac{1\pm\sqrt{1+84}}{6}$
$\displaystyle =\frac{1\pm\sqrt{85}}{6}$
$\displaystyle \Rightarrow x=\frac{1+\sqrt{85}}{6}\approx1.703$
$\displaystyle \text{or }x=\frac{1-\sqrt{85}}{6}\approx-1.370$
$\\$

$\displaystyle \textbf{Question 14: } \text{Solve }\left(\frac{x}{x+2}\right)^2-7\left(\frac{x}{x+2}\right)+12=0,\ x\neq-2.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\frac{x}{x+2}=y$
$\displaystyle y^2-7y+12=0$
$\displaystyle y^2-3y-4y+12=0$
$\displaystyle y(y-3)-4(y-3)=0$
$\displaystyle (y-3)(y-4)=0$
$\displaystyle \Rightarrow y=3\text{ or }4$
$\displaystyle \text{When }y=3,\ \frac{x}{x+2}=3$
$\displaystyle x=3x+6$
$\displaystyle x=-3$
$\displaystyle \text{When }y=4,\ \frac{x}{x+2}=4$
$\displaystyle x=4x+8$
$\displaystyle x=-\frac{8}{3}$
$\displaystyle \therefore x=-3\text{ or }-\frac{8}{3}$
$\\$

$\displaystyle \textbf{Question 15: } \text{Solve }x^2-11x-12=0,\text{ when }x\in N.$
$\displaystyle \text{Answer:}$
$\displaystyle x^2-11x-12=0$
$\displaystyle x^2-12x+x-12=0$
$\displaystyle x(x-12)+1(x-12)=0$
$\displaystyle (x+1)(x-12)=0$
$\displaystyle \Rightarrow x=12\text{ or }-1$
$\displaystyle \text{Since }x\in N,\ x=12$
$\\$

$\displaystyle \textbf{Question 16: } \text{Solve }x^2-4x-12=0,\text{ when }x\in I.$
$\displaystyle \text{Answer:}$
$\displaystyle x^2-4x-12=0$
$\displaystyle x^2-6x+2x-12=0$
$\displaystyle x(x-6)+2(x-6)=0$
$\displaystyle (x+2)(x-6)=0$
$\displaystyle \Rightarrow x=6\text{ or }-2$
$\\$

$\displaystyle \textbf{Question 17: } \text{Solve }2x^2-9x+10=0,\text{ when }x\in Q.$
$\displaystyle \text{Answer:}$
$\displaystyle 2x^2-9x+10=0$
$\displaystyle 2x^2-4x-5x+10=0$
$\displaystyle 2x(x-2)-5(x-2)=0$
$\displaystyle (2x-5)(x-2)=0$
$\displaystyle \Rightarrow x=2\text{ or }\frac{5}{2}$
$\\$

$\displaystyle \textbf{Question 18: } \text{Solve }(a+b)^2x^2-(a+b)x-6=0,\ a+b\neq0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }(a+b)x=y$
$\displaystyle y^2-y-6=0$
$\displaystyle y^2-3y+2y-6=0$
$\displaystyle y(y-3)+2(y-3)=0$
$\displaystyle (y+2)(y-3)=0$
$\displaystyle \Rightarrow y=3\text{ or }-2$
$\displaystyle \text{When }y=3,\ (a+b)x=3\Rightarrow x=\frac{3}{a+b}$
$\displaystyle \text{When }y=-2,\ (a+b)x=-2\Rightarrow x=-\frac{2}{a+b}$
$\\$

$\displaystyle \textbf{Question 19: } \text{Solve }\frac{1}{p}+\frac{1}{q}+\frac{1}{x}=\frac{1}{x+p+q}.$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{1}{p}+\frac{1}{q}+\frac{1}{x}-\frac{1}{x+p+q}=0$
$\displaystyle \frac{p+q}{pq}+\frac{p+q}{x(x+p+q)}=0$
$\displaystyle (p+q)\left[\frac{1}{pq}+\frac{1}{x(x+p+q)}\right]=0$
$\displaystyle x(x+p+q)+pq=0$
$\displaystyle x^2+(p+q)x+pq=0$
$\displaystyle (x+p)(x+q)=0$
$\displaystyle \Rightarrow x=-p\text{ or }-q$
$\\$

$\displaystyle \textbf{Question 20: } \text{Solve }x(x+1)+(x+2)(x+3)=42.$
$\displaystyle \text{Answer:}$
$\displaystyle x(x+1)+(x+2)(x+3)=42$
$\displaystyle x^2+x+x^2+5x+6=42$
$\displaystyle 2x^2+6x-36=0$
$\displaystyle x^2+3x-18=0$
$\displaystyle x^2-3x+6x-18=0$
$\displaystyle x(x-3)+6(x-3)=0$
$\displaystyle (x+6)(x-3)=0$
$\displaystyle \Rightarrow x=3\text{ or }-6$
$\\$

$\displaystyle \textbf{Question 1: } \text{Solve }\frac{1}{x+1}-\frac{2}{x+2}=\frac{3}{x+3}-\frac{4}{x+4}.$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{1}{x+1}-\frac{2}{x+2}=\frac{3}{x+3}-\frac{4}{x+4}$
$\displaystyle \frac{(x+2)-2(x+1)}{(x+1)(x+2)}=\frac{3(x+4)-4(x+3)}{(x+3)(x+4)}$
$\displaystyle \frac{-x}{(x+1)(x+2)}=\frac{-x}{(x+3)(x+4)}$
$\displaystyle \text{For }x\neq0,\ (x+3)(x+4)=(x+1)(x+2)$
$\displaystyle x^2+7x+12=x^2+3x+2$
$\displaystyle 4x=-10$
$\displaystyle x=-\frac{5}{2}$
$\displaystyle \text{Also, }x=0\text{ satisfies the equation.}$
$\displaystyle \therefore x=0\text{ or }-\frac{5}{2}$
$\\$

$\displaystyle \textbf{Question 22: } \text{Find }m\text{ so that }(m-3)x^2-4x+1=0\text{ has equal roots.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Comparing with }ax^2+bx+c=0,\ a=m-3,\ b=-4,\ c=1$
$\displaystyle \text{For equal roots, }b^2-4ac=0$
$\displaystyle (-4)^2-4(m-3)(1)=0$
$\displaystyle 16-4m+12=0$
$\displaystyle 28-4m=0$
$\displaystyle m=7$
$\\$

$\displaystyle \textbf{Question 23: } \text{Find }m\text{ so that }3x^2+12x+(m+7)=0\text{ has equal roots.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Comparing with }ax^2+bx+c=0,\ a=3,\ b=12,\ c=m+7$
$\displaystyle \text{For equal roots, }b^2-4ac=0$
$\displaystyle 12^2-4(3)(m+7)=0$
$\displaystyle 144-12m-84=0$
$\displaystyle 12m=60$
$\displaystyle \Rightarrow m=5$
$\displaystyle \text{Substituting }m=5,$
$\displaystyle 3x^2+12x+12=0$
$\displaystyle x^2+4x+4=0$
$\displaystyle (x+2)^2=0$
$\displaystyle \Rightarrow x=-2$
$\\$

$\displaystyle \textbf{Question 24: } \text{Find }m\text{ so that }x^2-(m+2)x+(m+5)=0\text{ has equal roots.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Comparing with }ax^2+bx+c=0,\ a=1,\ b=-(m+2),\ c=m+5$
$\displaystyle \text{For equal roots, }b^2-4ac=0$
$\displaystyle (m+2)^2-4(m+5)=0$
$\displaystyle m^2+4m+4-4m-20=0$
$\displaystyle m^2-16=0$
$\displaystyle \Rightarrow m=\pm4$
$\displaystyle \text{When }m=4,$
$\displaystyle x^2-6x+9=0$
$\displaystyle (x-3)^2=0$
$\displaystyle \Rightarrow x=3$
$\displaystyle \text{When }m=-4,$
$\displaystyle x^2+2x+1=0$
$\displaystyle (x+1)^2=0$
$\displaystyle \Rightarrow x=-1$
$\\$

$\displaystyle \textbf{Question 25: } \text{Find }p\text{ so that }px^2-4x+3=0\text{ has equal roots.}\hfill\text{[ICSE 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Comparing with }ax^2+bx+c=0,\ a=p,\ b=-4,\ c=3$
$\displaystyle \text{For equal roots, }b^2-4ac=0$
$\displaystyle (-4)^2-4(p)(3)=0$
$\displaystyle 16-12p=0$
$\displaystyle \Rightarrow p=\frac{4}{3}$
$\\$

$\displaystyle \textbf{Question 26: } \text{Find }m\text{ so that }x^2+2(m-1)x+(m+5)=0\text{ has equal roots.}\hfill\text{[ICSE 2012]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Comparing with }ax^2+bx+c=0,\ a=1,\ b=2(m-1),\ c=m+5$
$\displaystyle \text{For equal roots, }b^2-4ac=0$
$\displaystyle [2(m-1)]^2-4(1)(m+5)=0$
$\displaystyle 4(m^2-2m+1)-4(m+5)=0$
$\displaystyle 4m^2-8m+4-4m-20=0$
$\displaystyle 4m^2-12m-16=0$
$\displaystyle m^2-3m-4=0$
$\displaystyle m^2-4m+m-4=0$
$\displaystyle m(m-4)+1(m-4)=0$
$\displaystyle (m+1)(m-4)=0$
$\displaystyle \Rightarrow m=4\text{ or }-1$
$\\$