\displaystyle \textbf{Question 1: } \text{Solve:}
\displaystyle \text{(i) }2x^2-7x=39 \qquad \text{(ii) }x^2=5x \qquad \text{(iii) }x^2=16
\displaystyle \text{Answer:}
\displaystyle \text{(i) }2x^2-7x=39
\displaystyle \Rightarrow 2x^2-7x-39=0
\displaystyle \Rightarrow 2x^2-13x+6x-39=0
\displaystyle \Rightarrow x(2x-13)+3(2x-13)=0
\displaystyle \Rightarrow (2x-13)(x+3)=0
\displaystyle \Rightarrow 2x-13=0 \text{ or } x+3=0
\displaystyle \Rightarrow x=\frac{13}{2} \text{ or } x=-3
\displaystyle \text{(ii) }x^2=5x
\displaystyle \Rightarrow x^2-5x=0
\displaystyle \Rightarrow x(x-5)=0
\displaystyle \Rightarrow x=0 \text{ or } x-5=0
\displaystyle \Rightarrow x=0 \text{ or } x=5
\displaystyle \text{(iii) }x^2=16
\displaystyle \Rightarrow x^2-16=0
\displaystyle \Rightarrow (x+4)(x-4)=0
\displaystyle \Rightarrow x+4=0 \text{ or } x-4=0
\displaystyle \Rightarrow x=-4 \text{ or } x=4
\\

\displaystyle \textbf{Question 2: } \text{Solve: }\frac{x}{x-1}+\frac{x-1}{x}=2\frac{1}{2}
\displaystyle \text{Answer:}
\displaystyle \frac{x}{x-1}+\frac{x-1}{x}=2\frac{1}{2}
\displaystyle \Rightarrow \frac{x^2+(x-1)^2}{x(x-1)}=\frac{5}{2}
\displaystyle \Rightarrow 2(x^2+x^2-2x+1)=5(x^2-x)
\displaystyle \Rightarrow 4x^2-4x+2=5x^2-5x
\displaystyle \Rightarrow -x^2+x+2=0
\displaystyle \Rightarrow x^2-x-2=0
\displaystyle \Rightarrow (x-2)(x+1)=0
\displaystyle \Rightarrow x-2=0 \text{ or } x+1=0
\displaystyle \Rightarrow x=2 \text{ or } x=-1
\\

\displaystyle \textbf{Question 3: } \text{Find the quadratic equation whose solution set is }\{-2,3\}.
\displaystyle \text{Answer:}
\displaystyle \text{Since, solution set }=\{-2,3\}
\displaystyle \Rightarrow x=-2 \text{ or } x=3
\displaystyle \Rightarrow x+2=0 \text{ or } x-3=0
\displaystyle \Rightarrow (x+2)(x-3)=0
\displaystyle \Rightarrow x^2-3x+2x-6=0
\displaystyle \Rightarrow x^2-x-6=0
\displaystyle \therefore \text{Required quadratic equation is }x^2-x-6=0.
\\

\displaystyle \textbf{Question 4: } \text{Use the substitution }x=3y+1\text{ to solve for }y,\text{ if }5(3y+1)^2+6(3y+1)-8=0.
\displaystyle \text{Answer:}
\displaystyle 5(3y+1)^2+6(3y+1)-8=0
\displaystyle \text{Put }x=3y+1
\displaystyle \Rightarrow 5x^2+6x-8=0
\displaystyle \Rightarrow (x+2)(5x-4)=0
\displaystyle \Rightarrow x=-2 \text{ or } x=\frac{4}{5}
\displaystyle \Rightarrow \text{When }x=-2,\ 3y+1=-2 \Rightarrow y=-1
\displaystyle \Rightarrow \text{When }x=\frac{4}{5},\ 3y+1=\frac{4}{5} \Rightarrow y=-\frac{1}{15}
\displaystyle \therefore y=-1 \text{ or } y=-\frac{1}{15}
\\

\displaystyle \textbf{Question 5: } \text{Without solving the quadratic equation }3x^2-2x-1=0,\text{ find whether }x=1\text{ is a root.}
\displaystyle \text{Answer:}
\displaystyle \text{Substituting }x=1\text{ in }3x^2-2x-1=0,\text{ we get}
\displaystyle 3(1)^2-2(1)-1=0
\displaystyle \Rightarrow 3-2-1=0
\displaystyle \text{This is true.}
\displaystyle \therefore x=1\text{ is a root of the given quadratic equation.}
\\

\displaystyle \textbf{Question 6: } \text{Without solving the quadratic equation }x^2-x+1=0,\text{ find whether }x=-1\text{ is a root.}
\displaystyle \text{Answer:}
\displaystyle \text{Substituting }x=-1\text{ in }x^2-x+1=0,\text{ we get}
\displaystyle (-1)^2-(-1)+1=0
\displaystyle \Rightarrow 1+1+1=3\neq 0
\displaystyle \text{This is not true.}
\displaystyle \therefore x=-1\text{ is not a root of the given quadratic equation.}
\\

\displaystyle \textbf{Question 7: } \text{Find the value of }k\text{ for which }x=2\text{ is a root of }kx^2+2x-3=0.
\displaystyle \text{Answer:}
\displaystyle \text{Substituting }x=2\text{ in }kx^2+2x-3=0,\text{ we get}
\displaystyle k(2)^2+2(2)-3=0
\displaystyle \Rightarrow 4k+4-3=0
\displaystyle \Rightarrow 4k=-1
\displaystyle \Rightarrow k=-\frac{1}{4}
\\

\displaystyle \textbf{Question 8: } \text{If }x=2\text{ and }x=3\text{ are roots of }3x^2-2mx+2n=0,\text{ find }m\text{ and }n.
\displaystyle \text{Answer:}
\displaystyle \text{Since }x=2\text{ is a root,}
\displaystyle 3(2)^2-2m(2)+2n=0
\displaystyle \Rightarrow 12-4m+2n=0
\displaystyle \Rightarrow 2m-n=6 \qquad \text{... (i)}
\displaystyle \text{Since }x=3\text{ is a root,}
\displaystyle 3(3)^2-2m(3)+2n=0
\displaystyle \Rightarrow 27-6m+2n=0
\displaystyle \Rightarrow 6m-2n=27 \qquad \text{... (ii)}
\displaystyle \text{Multiplying (i) by }2,\text{ we get }4m-2n=12
\displaystyle \text{Subtracting from (ii), }2m=15
\displaystyle \Rightarrow m=\frac{15}{2}
\displaystyle \text{Substituting in (i), }2\left(\frac{15}{2}\right)-n=6
\displaystyle \Rightarrow 15-n=6
\displaystyle \Rightarrow n=9
\displaystyle \therefore m=\frac{15}{2},\ n=9
\\

\displaystyle \textbf{Question 9: } \text{If one root of }2x^2+ax-6=0\text{ is }2,\text{ find }a\text{ and the other root.}
\displaystyle \text{Answer:}
\displaystyle \text{Since }x=2\text{ is a root of }2x^2+ax-6=0,
\displaystyle 2(2)^2+2a-6=0
\displaystyle \Rightarrow 8+2a-6=0
\displaystyle \Rightarrow a=-1
\displaystyle \text{Substituting }a=-1,\text{ the equation becomes}
\displaystyle 2x^2-x-6=0
\displaystyle \Rightarrow 2x^2-4x+3x-6=0
\displaystyle \Rightarrow 2x(x-2)+3(x-2)=0
\displaystyle \Rightarrow (x-2)(2x+3)=0
\displaystyle \Rightarrow x=2 \text{ or } x=-\frac{3}{2}
\displaystyle \therefore \text{The other root is }-\frac{3}{2}.
\\

\displaystyle \textbf{Question 10: } \text{Solve each of the following equations by using the formula:}
\displaystyle \text{(i) }5x^2-2x-3=0 \qquad \text{(ii) }x^2=18x-77 \qquad \text{(iii) }\sqrt{3}x^2+11x+6\sqrt{3}=0
\displaystyle \text{Answer:}
\displaystyle \text{(i) Comparing }5x^2-2x-3=0\text{ with }ax^2+bx+c=0,
\displaystyle \text{we get }a=5,\ b=-2\text{ and }c=-3
\displaystyle \text{We know that }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
\displaystyle x=\frac{2\pm\sqrt{(-2)^2-4(5)(-3)}}{2(5)}
\displaystyle =\frac{2\pm\sqrt{64}}{10}=\frac{2\pm8}{10}
\displaystyle \Rightarrow x=1\text{ or }x=-\frac{3}{5}
\displaystyle \text{(ii) }x^2=18x-77
\displaystyle \Rightarrow x^2-18x+77=0
\displaystyle \text{Comparing with }ax^2+bx+c=0,\text{ we get }a=1,\ b=-18,\ c=77
\displaystyle x=\frac{18\pm\sqrt{(-18)^2-4(1)(77)}}{2(1)}
\displaystyle =\frac{18\pm\sqrt{16}}{2}=\frac{18\pm4}{2}
\displaystyle \Rightarrow x=11\text{ or }x=7
\displaystyle \text{(iii) Comparing }\sqrt{3}x^2+11x+6\sqrt{3}=0\text{ with }ax^2+bx+c=0,
\displaystyle \text{we get }a=\sqrt{3},\ b=11\text{ and }c=6\sqrt{3}
\displaystyle x=\frac{-11\pm\sqrt{(11)^2-4(\sqrt{3})(6\sqrt{3})}}{2\sqrt{3}}
\displaystyle =\frac{-11\pm\sqrt{49}}{2\sqrt{3}}=\frac{-11\pm7}{2\sqrt{3}}
\displaystyle \Rightarrow x=-\frac{2\sqrt{3}}{3}\text{ or }x=-3\sqrt{3}
\\

\displaystyle \textbf{Question 11: } \text{Without solving, examine the nature of the roots of the equations:}
\displaystyle \text{(i) }5x^2-6x+7=0 \qquad \text{(ii) }x^2+6x+9=0 \qquad \text{(iii) }2x^2+6x+3=0
\displaystyle \text{Answer:}
\displaystyle \text{(i) Comparing }5x^2-6x+7=0\text{ with }ax^2+bx+c=0,
\displaystyle \text{we get }a=5,\ b=-6\text{ and }c=7
\displaystyle \text{Discriminant }D=b^2-4ac=(-6)^2-4(5)(7)
\displaystyle =36-140=-104
\displaystyle \text{Since }D<0,\text{ the roots are not real.}
\displaystyle \therefore \text{The roots are imaginary.}
\displaystyle \text{(ii) Comparing }x^2+6x+9=0\text{ with }ax^2+bx+c=0,
\displaystyle \text{we get }a=1,\ b=6\text{ and }c=9
\displaystyle \text{Discriminant }D=b^2-4ac=(6)^2-4(1)(9)
\displaystyle =36-36=0
\displaystyle \text{Since }D=0,\text{ the roots are real and equal.}
\displaystyle \text{(iii) Comparing }2x^2+6x+3=0\text{ with }ax^2+bx+c=0,
\displaystyle \text{we get }a=2,\ b=6\text{ and }c=3
\displaystyle \text{Discriminant }D=b^2-4ac=(6)^2-4(2)(3)
\displaystyle =36-24=12
\displaystyle \text{Since }D>0,\text{ the roots are real and unequal.}
\\

\displaystyle \textbf{Question 12: } \text{Find the value of }p\text{ if the roots of }(p-3)x^2+6x+9=0\text{ are equal.}
\displaystyle \text{Answer:}
\displaystyle \text{Comparing }(p-3)x^2+6x+9=0\text{ with }ax^2+bx+c=0,
\displaystyle \text{we get }a=p-3,\ b=6\text{ and }c=9
\displaystyle \text{Since the roots are equal, }b^2-4ac=0
\displaystyle \Rightarrow 6^2-4(p-3)(9)=0
\displaystyle \Rightarrow 36-36p+108=0
\displaystyle \Rightarrow 144-36p=0
\displaystyle \Rightarrow p=4
\\

\displaystyle \textbf{Question 13: } \text{Find the value of }m\text{ if the roots of }(4+m)x^2+(m+1)x+1=0\text{ are equal.}
\displaystyle \text{Answer:}
\displaystyle \text{Comparing }(4+m)x^2+(m+1)x+1=0\text{ with }ax^2+bx+c=0,
\displaystyle \text{we get }a=4+m,\ b=m+1\text{ and }c=1
\displaystyle \text{Since the roots are equal, }b^2-4ac=0
\displaystyle \Rightarrow (m+1)^2-4(4+m)=0
\displaystyle \Rightarrow m^2+2m+1-16-4m=0
\displaystyle \Rightarrow m^2-2m-15=0
\displaystyle \Rightarrow (m-5)(m+3)=0
\displaystyle \Rightarrow m=5 \text{ or } m=-3
\\

\displaystyle \textbf{Question 14: } \text{Solve each equation for }x\text{ correct to }2\text{ decimal places.}
\displaystyle \text{(i) }x^2-10x+6=0 \qquad \text{(ii) }3x^2+5x-9=0
\displaystyle \text{Answer:}
\displaystyle \text{(i) Comparing }x^2-10x+6=0\text{ with }ax^2+bx+c=0,
\displaystyle \text{we get }a=1,\ b=-10\text{ and }c=6
\displaystyle b^2-4ac=(-10)^2-4(1)(6)=100-24=76
\displaystyle \sqrt{76}\approx8.718
\displaystyle x=\frac{10\pm8.718}{2}
\displaystyle \Rightarrow x\approx9.359\text{ or }x\approx0.641
\displaystyle \therefore x\approx9.36\text{ or }x\approx0.64
\displaystyle \text{(ii) Comparing }3x^2+5x-9=0\text{ with }ax^2+bx+c=0,
\displaystyle \text{we get }a=3,\ b=5\text{ and }c=-9
\displaystyle b^2-4ac=(5)^2-4(3)(-9)=25+108=133
\displaystyle \sqrt{133}\approx11.533
\displaystyle x=\frac{-5\pm11.533}{2(3)}
\displaystyle \Rightarrow x\approx1.089\text{ or }x\approx-2.756
\displaystyle \therefore x\approx1.09\text{ or }x\approx-2.76
\\

\displaystyle \textbf{Question 15: } \text{Solve the following equation: }x-\frac{18}{x}=6.\hfill\text{[ICSE 2011]}
\displaystyle \text{Give your answer correct to two significant figures.}
\displaystyle \text{Answer:}
\displaystyle x-\frac{18}{x}=6
\displaystyle \Rightarrow x^2-18=6x
\displaystyle \Rightarrow x^2-6x-18=0
\displaystyle \text{Comparing with }ax^2+bx+c=0,\text{ we get }a=1,\ b=-6,\ c=-18
\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
\displaystyle =\frac{6\pm\sqrt{(-6)^2-4(1)(-18)}}{2}
\displaystyle =\frac{6\pm\sqrt{108}}{2}=\frac{6\pm10.392}{2}
\displaystyle \Rightarrow x\approx8.196\text{ or }x\approx-2.196
\displaystyle \therefore \text{Correct to two significant figures, }x\approx8.2\text{ or }x\approx-2.2
\\

\displaystyle \textbf{Question 16: } \text{Solve:}
\displaystyle \text{(i) }2x^4-5x^2+3=0 \qquad \text{(ii) }(x^2+3x)^2-(x^2+3x)-6=0,\ x\in R
\displaystyle \text{Answer:}
\displaystyle \text{(i) }2x^4-5x^2+3=0
\displaystyle \text{Let }x^2=y
\displaystyle \Rightarrow 2y^2-5y+3=0
\displaystyle \Rightarrow (y-1)(2y-3)=0
\displaystyle \Rightarrow y=1\text{ or }y=\frac{3}{2}
\displaystyle \text{When }y=1,\ x=\pm1
\displaystyle \text{When }y=\frac{3}{2},\ x=\pm\sqrt{\frac{3}{2}}
\displaystyle \therefore x=\pm1,\ \pm\sqrt{\frac{3}{2}}
\displaystyle \text{(ii) }(x^2+3x)^2-(x^2+3x)-6=0
\displaystyle \text{Let }x^2+3x=y
\displaystyle \Rightarrow y^2-y-6=0
\displaystyle \Rightarrow (y-3)(y+2)=0
\displaystyle \Rightarrow y=3\text{ or }y=-2
\displaystyle \text{When }y=3,\ x^2+3x-3=0
\displaystyle \Rightarrow x=\frac{-3\pm\sqrt{9+12}}{2}=\frac{-3\pm\sqrt{21}}{2}
\displaystyle \text{When }y=-2,\ x^2+3x+2=0
\displaystyle \Rightarrow (x+1)(x+2)=0
\displaystyle \Rightarrow x=-1\text{ or }x=-2
\displaystyle \therefore x=\frac{-3+\sqrt{21}}{2},\ \frac{-3-\sqrt{21}}{2},\ -1,\ -2
\\

\displaystyle \textbf{Question 17: } \text{Solve: }\sqrt{\frac{x}{1-x}}+\sqrt{\frac{1-x}{x}}=2\frac{1}{6},\ x\neq0,\ x\neq1
\displaystyle \text{Answer:}
\displaystyle \text{Let }\sqrt{\frac{x}{1-x}}=y
\displaystyle \Rightarrow \sqrt{\frac{1-x}{x}}=\frac{1}{y}
\displaystyle \therefore y+\frac{1}{y}=\frac{13}{6}
\displaystyle \Rightarrow 6y^2+6=13y
\displaystyle \Rightarrow 6y^2-13y+6=0
\displaystyle \Rightarrow (2y-3)(3y-2)=0
\displaystyle \Rightarrow y=\frac{3}{2}\text{ or }y=\frac{2}{3}
\displaystyle \text{When }y=\frac{3}{2},\ \sqrt{\frac{x}{1-x}}=\frac{3}{2}
\displaystyle \Rightarrow \frac{x}{1-x}=\frac{9}{4}
\displaystyle \Rightarrow 4x=9-9x
\displaystyle \Rightarrow x=\frac{9}{13}
\displaystyle \text{When }y=\frac{2}{3},\ \sqrt{\frac{x}{1-x}}=\frac{2}{3}
\displaystyle \Rightarrow \frac{x}{1-x}=\frac{4}{9}
\displaystyle \Rightarrow 9x=4-4x
\displaystyle \Rightarrow x=\frac{4}{13}
\displaystyle \therefore x=\frac{9}{13}\text{ or }x=\frac{4}{13}
\\

\displaystyle \textbf{Question 18: } \text{Find the solution set of }3x^2-8x-3=0,\text{ when}
\displaystyle \text{(i) }x\in Z\text{ (integers)} \qquad \text{(ii) }x\in Q\text{ (rational numbers)}
\displaystyle \text{Answer:}
\displaystyle 3x^2-8x-3=0
\displaystyle \Rightarrow 3x^2-9x+x-3=0
\displaystyle \Rightarrow 3x(x-3)+1(x-3)=0
\displaystyle \Rightarrow (x-3)(3x+1)=0
\displaystyle \Rightarrow x=3\text{ or }x=-\frac{1}{3}
\displaystyle \text{(i) When }x\in Z,\text{ solution set }=\{3\}
\displaystyle \text{(ii) When }x\in Q,\text{ solution set }=\left\{3,-\frac{1}{3}\right\}
\\

\displaystyle \textbf{Question 19: } \text{Solve: }(2x-3)^2=25.
\displaystyle \text{Answer:}
\displaystyle (2x-3)^2=25
\displaystyle \Rightarrow 4x^2-12x+9-25=0
\displaystyle \Rightarrow 4x^2-12x-16=0
\displaystyle \Rightarrow x^2-3x-4=0
\displaystyle \Rightarrow (x-4)(x+1)=0
\displaystyle \Rightarrow x=4\text{ or }x=-1
\\

\displaystyle \textbf{Question 20: } \text{Solve for }x:4\left(x-\frac{1}{x}\right)^2+8\left(x+\frac{1}{x}\right)=29,\ x\neq0.
\displaystyle \text{Answer:}
\displaystyle \text{Let }x+\frac{1}{x}=y
\displaystyle \left(x-\frac{1}{x}\right)^2=\left(x+\frac{1}{x}\right)^2-4=y^2-4
\displaystyle \Rightarrow 4(y^2-4)+8y=29
\displaystyle \Rightarrow 4y^2+8y-45=0
\displaystyle \Rightarrow (2y+9)(2y-5)=0
\displaystyle \Rightarrow y=-\frac{9}{2}\text{ or }y=\frac{5}{2}
\displaystyle \text{When }y=-\frac{9}{2},\ x+\frac{1}{x}=-\frac{9}{2}
\displaystyle \Rightarrow 2x^2+9x+2=0
\displaystyle \Rightarrow x=\frac{-9\pm\sqrt{65}}{4}
\displaystyle \text{When }y=\frac{5}{2},\ x+\frac{1}{x}=\frac{5}{2}
\displaystyle \Rightarrow 2x^2-5x+2=0
\displaystyle \Rightarrow (x-2)(2x-1)=0
\displaystyle \Rightarrow x=2\text{ or }x=\frac{1}{2}
\\

\displaystyle \textbf{Question 21: } \text{Solve: }\frac{a}{ax-1}+\frac{b}{bx-1}=a+b,\text{ where }a+b\neq0,\ ab\neq0.
\displaystyle \text{Answer:}
\displaystyle \frac{a}{ax-1}+\frac{b}{bx-1}=a+b
\displaystyle \Rightarrow a(bx-1)+b(ax-1)=(a+b)(ax-1)(bx-1)
\displaystyle \Rightarrow 2abx-a-b=(a+b)(abx^2-ax-bx+1)
\displaystyle \Rightarrow ab(a+b)x^2-\left((a+b)^2+2ab\right)x+2(a+b)=0
\displaystyle \Rightarrow (ax+bx-2)(abx-a-b)=0
\displaystyle \Rightarrow (a+b)x=2\text{ or }abx=a+b
\displaystyle \Rightarrow x=\frac{2}{a+b}\text{ or }x=\frac{a+b}{ab}
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