Class 10: Quadratic Equations

Solve each of the following equations for $\displaystyle x$

$\displaystyle \textbf{Question 1: } \text{Solve for }x:\ x^2-8x+5=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Comparing with }ax^2+bx+c=0,\ a=1,\ b=-8,\ c=5$
$\displaystyle \text{Since }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle x=\frac{-(-8)\pm\sqrt{(-8)^2-4(1)(5)}}{2(1)}$
$\displaystyle =\frac{8\pm\sqrt{64-20}}{2}$
$\displaystyle =\frac{8\pm\sqrt{44}}{2}$
$\displaystyle =4\pm\sqrt{11}$
$\displaystyle \Rightarrow x=4+\sqrt{11}\approx7.32$
$\displaystyle \text{or }x=4-\sqrt{11}\approx0.68$
$\\$

$\displaystyle \textbf{Question 2: } \text{Solve for }x:\ 5x^2+10x-3=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Comparing with }ax^2+bx+c=0,\ a=5,\ b=10,\ c=-3$
$\displaystyle \text{Since }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle x=\frac{-10\pm\sqrt{10^2-4(5)(-3)}}{2(5)}$
$\displaystyle =\frac{-10\pm\sqrt{100+60}}{10}$
$\displaystyle =\frac{-10\pm\sqrt{160}}{10}$
$\displaystyle =\frac{-10\pm4\sqrt{10}}{10}$
$\displaystyle =-1\pm\frac{2\sqrt{10}}{5}$
$\displaystyle \Rightarrow x=-1+\frac{2\sqrt{10}}{5}\approx0.26$
$\displaystyle \text{or }x=-1-\frac{2\sqrt{10}}{5}\approx-2.26$
$\\$

$\displaystyle \textbf{Question 3: } \text{Solve for }x:\ 2x^2-10x+5=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Comparing with }ax^2+bx+c=0,\ a=2,\ b=-10,\ c=5$
$\displaystyle \text{Since }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle x=\frac{-(-10)\pm\sqrt{(-10)^2-4(2)(5)}}{2(2)}$
$\displaystyle =\frac{10\pm\sqrt{100-40}}{4}$
$\displaystyle =\frac{10\pm\sqrt{60}}{4}$
$\displaystyle =\frac{5\pm\sqrt{15}}{2}$
$\displaystyle \Rightarrow x=\frac{5+\sqrt{15}}{2}\approx4.44$
$\displaystyle \text{or }x=\frac{5-\sqrt{15}}{2}\approx0.56$
$\\$

$\displaystyle \textbf{Question 4: } \text{Solve for }x:\ 4x+\frac{6}{x}+13=0.$
$\displaystyle \text{Answer:}$
$\displaystyle 4x+\frac{6}{x}+13=0$
$\displaystyle \Rightarrow 4x^2+13x+6=0$
$\displaystyle \text{Comparing with }ax^2+bx+c=0,\ a=4,\ b=13,\ c=6$
$\displaystyle \text{Since }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle x=\frac{-13\pm\sqrt{13^2-4(4)(6)}}{2(4)}$
$\displaystyle =\frac{-13\pm\sqrt{169-96}}{8}$
$\displaystyle =\frac{-13\pm\sqrt{73}}{8}$
$\displaystyle \Rightarrow x=\frac{-13+\sqrt{73}}{8}\approx-0.56$
$\displaystyle \text{or }x=\frac{-13-\sqrt{73}}{8}\approx-2.69$
$\\$

$\displaystyle \textbf{Question 5: } \text{Solve for }x:\ x^2-3x-9=0.\hfill\text{[ICSE 2007]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Comparing with }ax^2+bx+c=0,\ a=1,\ b=-3,\ c=-9$
$\displaystyle \text{Since }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle x=\frac{-(-3)\pm\sqrt{(-3)^2-4(1)(-9)}}{2(1)}$
$\displaystyle =\frac{3\pm\sqrt{9+36}}{2}$
$\displaystyle =\frac{3\pm\sqrt{45}}{2}$
$\displaystyle =\frac{3\pm3\sqrt{5}}{2}$
$\displaystyle \Rightarrow x=\frac{3+3\sqrt{5}}{2}\approx4.85$
$\displaystyle \text{or }x=\frac{3-3\sqrt{5}}{2}\approx-1.85$
$\\$

$\displaystyle \textbf{Question 6: } \text{Solve for }x:\ x^2-5x-10=0.\hfill\text{[ICSE 2013]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Comparing with }ax^2+bx+c=0,\ a=1,\ b=-5,\ c=-10$
$\displaystyle \text{Since }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle x=\frac{-(-5)\pm\sqrt{(-5)^2-4(1)(-10)}}{2(1)}$
$\displaystyle =\frac{5\pm\sqrt{25+40}}{2}$
$\displaystyle =\frac{5\pm\sqrt{65}}{2}$
$\displaystyle \Rightarrow x=\frac{5+\sqrt{65}}{2}\approx6.53$
$\displaystyle \text{or }x=\frac{5-\sqrt{65}}{2}\approx-1.53$
$\\$

$\displaystyle \textbf{Question 7: } \text{Solve for }x:\ 3x^2-12x-1=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Comparing with }ax^2+bx+c=0,\ a=3,\ b=-12,\ c=-1$
$\displaystyle \text{Since }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle x=\frac{-(-12)\pm\sqrt{(-12)^2-4(3)(-1)}}{2(3)}$
$\displaystyle =\frac{12\pm\sqrt{144+12}}{6}$
$\displaystyle =\frac{12\pm\sqrt{156}}{6}$
$\displaystyle =\frac{6\pm\sqrt{39}}{3}$
$\displaystyle \Rightarrow x=\frac{6+\sqrt{39}}{3}\approx4.082$
$\displaystyle \text{or }x=\frac{6-\sqrt{39}}{3}\approx-0.082$
$\\$

$\displaystyle \textbf{Question 8: } \text{Solve for }x:\ x^2-16x+6=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Comparing with }ax^2+bx+c=0,\ a=1,\ b=-16,\ c=6$
$\displaystyle \text{Since }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle x=\frac{-(-16)\pm\sqrt{(-16)^2-4(1)(6)}}{2(1)}$
$\displaystyle =\frac{16\pm\sqrt{256-24}}{2}$
$\displaystyle =\frac{16\pm\sqrt{232}}{2}$
$\displaystyle =8\pm\sqrt{58}$
$\displaystyle \Rightarrow x=8+\sqrt{58}\approx15.616$
$\displaystyle \text{or }x=8-\sqrt{58}\approx0.384$
$\\$

$\displaystyle \textbf{Question 9: } \text{Solve for }x:\ 2x^2+11x+4=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Comparing with }ax^2+bx+c=0,\ a=2,\ b=11,\ c=4$
$\displaystyle \text{Since }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle x=\frac{-11\pm\sqrt{11^2-4(2)(4)}}{2(2)}$
$\displaystyle =\frac{-11\pm\sqrt{121-32}}{4}$
$\displaystyle =\frac{-11\pm\sqrt{89}}{4}$
$\displaystyle \Rightarrow x=\frac{-11+\sqrt{89}}{4}\approx-0.392$
$\displaystyle \text{or }x=\frac{-11-\sqrt{89}}{4}\approx-5.108$
$\\$

$\displaystyle \textbf{Question 10: } \text{Solve for }x:\ x^4-2x^2-3=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }y=x^2$
$\displaystyle y^2-2y-3=0$
$\displaystyle \text{Comparing with }ay^2+by+c=0,\ a=1,\ b=-2,\ c=-3$
$\displaystyle \text{Since }y=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle y=\frac{-(-2)\pm\sqrt{(-2)^2-4(1)(-3)}}{2(1)}$
$\displaystyle =\frac{2\pm\sqrt{4+12}}{2}$
$\displaystyle =\frac{2\pm4}{2}$
$\displaystyle \Rightarrow y=3\text{ or }-1$
$\displaystyle \therefore x^2=3\text{ or }x^2=-1$
$\displaystyle \Rightarrow x=\pm\sqrt{3}\text{ or }\pm i$
$\displaystyle \text{Real roots are }x=\pm\sqrt{3}\approx\pm1.732$
$\\$

$\displaystyle \textbf{Question 11: } \text{Solve for }x:\ x^4-10x^2+9=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }y=x^2$
$\displaystyle y^2-10y+9=0$
$\displaystyle \text{Comparing with }ay^2+by+c=0,\ a=1,\ b=-10,\ c=9$
$\displaystyle \text{Since }y=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle y=\frac{-(-10)\pm\sqrt{(-10)^2-4(1)(9)}}{2(1)}$
$\displaystyle =\frac{10\pm\sqrt{100-36}}{2}$
$\displaystyle =\frac{10\pm8}{2}$
$\displaystyle \Rightarrow y=9\text{ or }1$
$\displaystyle \therefore x^2=9\text{ or }x^2=1$
$\displaystyle \Rightarrow x=\pm3\text{ or }\pm1$
$\\$

$\displaystyle \textbf{Question 12: } \text{Solve for }x:\ (x^2-x)^2-5(x^2-x)+4=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }y=x^2-x$
$\displaystyle y^2-5y+4=0$
$\displaystyle \text{Comparing with }ay^2+by+c=0,\ a=1,\ b=-5,\ c=4$
$\displaystyle \text{Since }y=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle y=\frac{-(-5)\pm\sqrt{(-5)^2-4(1)(4)}}{2(1)}$
$\displaystyle =\frac{5\pm\sqrt{25-16}}{2}$
$\displaystyle =\frac{5\pm3}{2}$
$\displaystyle \Rightarrow y=4\text{ or }1$
$\displaystyle \text{When }y=4,\ x^2-x=4$
$\displaystyle x^2-x-4=0$
$\displaystyle x=\frac{1\pm\sqrt{1+16}}{2}=\frac{1\pm\sqrt{17}}{2}$
$\displaystyle \text{When }y=1,\ x^2-x=1$
$\displaystyle x^2-x-1=0$
$\displaystyle x=\frac{1\pm\sqrt{1+4}}{2}=\frac{1\pm\sqrt{5}}{2}$
$\displaystyle \therefore x=\frac{1\pm\sqrt{17}}{2},\frac{1\pm\sqrt{5}}{2}$
$\\$

$\displaystyle \textbf{Question 13: } \text{Solve for }x:\ (x^2-3x)^2-16(x^2-3x)-36=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }y=x^2-3x$
$\displaystyle y^2-16y-36=0$
$\displaystyle \text{Comparing with }ay^2+by+c=0,\ a=1,\ b=-16,\ c=-36$
$\displaystyle \text{Since }y=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle y=\frac{-(-16)\pm\sqrt{(-16)^2-4(1)(-36)}}{2(1)}$
$\displaystyle =\frac{16\pm\sqrt{256+144}}{2}$
$\displaystyle =\frac{16\pm20}{2}$
$\displaystyle \Rightarrow y=18\text{ or }-2$
$\displaystyle \text{When }y=18,\ x^2-3x=18$
$\displaystyle x^2-3x-18=0$
$\displaystyle (x-6)(x+3)=0$
$\displaystyle \Rightarrow x=6\text{ or }-3$
$\displaystyle \text{When }y=-2,\ x^2-3x=-2$
$\displaystyle x^2-3x+2=0$
$\displaystyle (x-1)(x-2)=0$
$\displaystyle \Rightarrow x=1\text{ or }2$
$\displaystyle \therefore x=1,2,6,-3$
$\\$

$\displaystyle \textbf{Question 14: } \text{Solve for }x:\ \sqrt{\frac{x}{x-3}}+\sqrt{\frac{x-3}{x}}=\frac{5}{2}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }y=\sqrt{\frac{x}{x-3}}$
$\displaystyle \therefore \sqrt{\frac{x-3}{x}}=\frac{1}{y}$
$\displaystyle y+\frac{1}{y}=\frac{5}{2}$
$\displaystyle 2y^2-5y+2=0$
$\displaystyle \text{Comparing with }ay^2+by+c=0,\ a=2,\ b=-5,\ c=2$
$\displaystyle \text{Since }y=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle y=\frac{-(-5)\pm\sqrt{(-5)^2-4(2)(2)}}{2(2)}$
$\displaystyle =\frac{5\pm\sqrt{25-16}}{4}$
$\displaystyle =\frac{5\pm3}{4}$
$\displaystyle \Rightarrow y=2\text{ or }\frac{1}{2}$
$\displaystyle \text{When }y=2,\ \sqrt{\frac{x}{x-3}}=2$
$\displaystyle \frac{x}{x-3}=4$
$\displaystyle x=4x-12$
$\displaystyle x=4$
$\displaystyle \text{When }y=\frac{1}{2},\ \sqrt{\frac{x}{x-3}}=\frac{1}{2}$
$\displaystyle \frac{x}{x-3}=\frac{1}{4}$
$\displaystyle 4x=x-3$
$\displaystyle x=-1$
$\displaystyle \therefore x=-1\text{ or }4$
$\\$

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

$\displaystyle \textbf{Question 16: } \text{Solve for }x:\ \frac{3x+1}{x+1}+\frac{x+1}{3x+1}=\frac{5}{2}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\frac{3x+1}{x+1}=y$
$\displaystyle \therefore y+\frac{1}{y}=\frac{5}{2}$
$\displaystyle 2(y^2+1)=5y$
$\displaystyle 2y^2-5y+2=0$
$\displaystyle 2y^2-4y-y+2=0$
$\displaystyle 2y(y-2)-1(y-2)=0$
$\displaystyle (2y-1)(y-2)=0$
$\displaystyle \Rightarrow y=2\text{ or }\frac{1}{2}$
$\displaystyle \text{When }y=2,\ \frac{3x+1}{x+1}=2$
$\displaystyle 3x+1=2x+2$
$\displaystyle x=1$
$\displaystyle \text{When }y=\frac{1}{2},\ \frac{3x+1}{x+1}=\frac{1}{2}$
$\displaystyle 6x+2=x+1$
$\displaystyle x=-\frac{1}{5}$
$\displaystyle \therefore x=1\text{ or }-\frac{1}{5}$
$\\$

$\displaystyle \textbf{Question 17: } \text{Solve for }x:\ 3\sqrt{\frac{x}{5}}+3\sqrt{\frac{5}{x}}=10.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\sqrt{\frac{x}{5}}=y$
$\displaystyle \therefore 3y+\frac{3}{y}=10$
$\displaystyle 3y^2+3=10y$
$\displaystyle 3y^2-10y+3=0$
$\displaystyle 3y^2-9y-y+3=0$
$\displaystyle 3y(y-3)-1(y-3)=0$
$\displaystyle (3y-1)(y-3)=0$
$\displaystyle \Rightarrow y=3\text{ or }\frac{1}{3}$
$\displaystyle \text{When }y=3,\ \sqrt{\frac{x}{5}}=3$
$\displaystyle \frac{x}{5}=9$
$\displaystyle x=45$
$\displaystyle \text{When }y=\frac{1}{3},\ \sqrt{\frac{x}{5}}=\frac{1}{3}$
$\displaystyle \frac{x}{5}=\frac{1}{9}$
$\displaystyle x=\frac{5}{9}$
$\displaystyle \therefore x=45\text{ or }\frac{5}{9}$
$\\$

$\displaystyle \textbf{Question 18: } \text{Solve for }x:\ 2x-\frac{1}{x}=7.\hfill\text{[ICSE 2006]}$
$\displaystyle \text{Answer:}$
$\displaystyle 2x-\frac{1}{x}=7$
$\displaystyle 2x^2-1=7x$
$\displaystyle 2x^2-7x-1=0$
$\displaystyle \text{Comparing with }ax^2+bx+c=0,\ a=2,\ b=-7,\ c=-1$
$\displaystyle \text{Since }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle x=\frac{-(-7)\pm\sqrt{(-7)^2-4(2)(-1)}}{2(2)}$
$\displaystyle =\frac{7\pm\sqrt{49+8}}{4}$
$\displaystyle =\frac{7\pm\sqrt{57}}{4}$
$\displaystyle \Rightarrow x=\frac{7+\sqrt{57}}{4}\approx3.64$
$\displaystyle \text{or }x=\frac{7-\sqrt{57}}{4}\approx-0.14$
$\\$

$\displaystyle \textbf{Question 19: } \text{Solve for }x:\ 5x^2-3x-4=0.\hfill\text{[ICSE 2012]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Comparing with }ax^2+bx+c=0,\ a=5,\ b=-3,\ c=-4$
$\displaystyle \text{Since }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle x=\frac{-(-3)\pm\sqrt{(-3)^2-4(5)(-4)}}{2(5)}$
$\displaystyle =\frac{3\pm\sqrt{9+80}}{10}$
$\displaystyle =\frac{3\pm\sqrt{89}}{10}$
$\displaystyle \Rightarrow x=\frac{3+\sqrt{89}}{10}\approx1.243$
$\displaystyle \text{or }x=\frac{3-\sqrt{89}}{10}\approx-0.643$
$\displaystyle \text{To three significant figures, }x=1.24\text{ or }-0.643$
$\\$

$\displaystyle \textbf{Question 20: } \text{Solve for }x:\ (x-1)^2-3x+4=0.\hfill\text{[ICSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle (x-1)^2-3x+4=0$
$\displaystyle x^2-2x+1-3x+4=0$
$\displaystyle x^2-5x+5=0$
$\displaystyle \text{Comparing with }ax^2+bx+c=0,\ a=1,\ b=-5,\ c=5$
$\displaystyle \text{Since }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle x=\frac{-(-5)\pm\sqrt{(-5)^2-4(1)(5)}}{2(1)}$
$\displaystyle =\frac{5\pm\sqrt{25-20}}{2}$
$\displaystyle =\frac{5\pm\sqrt{5}}{2}$
$\displaystyle \Rightarrow x=\frac{5+\sqrt{5}}{2}\approx3.618$
$\displaystyle \text{or }x=\frac{5-\sqrt{5}}{2}\approx1.382$
$\displaystyle \text{To two significant figures, }x=3.6\text{ or }1.4$
$\\$