Class 10: Quadratic Equations

Notes: Factorization of Trinomials of the form $\displaystyle Ax^2+Bx+C=0$ . To factorize this $\displaystyle a+b = B$ and $\displaystyle ab=AC$ . We will use this all across the solution.

Solve by factorization:

$\displaystyle \textbf{Question 1: } \text{Solve by factorization: }x^2-10x-24=0.$
$\displaystyle \text{Answer:}$
$\displaystyle x^2-10x-24=0$
$\displaystyle x^2-12x+2x-24=0$
$\displaystyle x(x-12)+2(x-12)=0$
$\displaystyle (x+2)(x-12)=0$
$\displaystyle \Rightarrow x=-2\text{ or }12$
$\\$

$\displaystyle \textbf{Question 2: } \text{Solve by factorization: }x^2-16=0.$
$\displaystyle \text{Answer:}$
$\displaystyle x^2-16=0$
$\displaystyle (x-4)(x+4)=0$
$\displaystyle \Rightarrow x=4\text{ or }-4$
$\\$

$\displaystyle \textbf{Question 3: } \text{Solve by factorization: }2x^2-\frac{1}{2}x=0.$
$\displaystyle \text{Answer:}$
$\displaystyle 2x^2-\frac{1}{2}x=0$
$\displaystyle 4x^2-x=0$
$\displaystyle x(4x-1)=0$
$\displaystyle \Rightarrow x=0\text{ or }\frac{1}{4}$
$\\$

$\displaystyle \textbf{Question 4: } \text{Solve by factorization: }x(x-5)=24.$
$\displaystyle \text{Answer:}$
$\displaystyle x(x-5)=24$
$\displaystyle x^2-5x-24=0$
$\displaystyle x^2-8x+3x-24=0$
$\displaystyle x(x-8)+3(x-8)=0$
$\displaystyle (x+3)(x-8)=0$
$\displaystyle \Rightarrow x=-3\text{ or }8$
$\\$

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

$\displaystyle \textbf{Question 6: } \text{Solve by factorization: }\frac{6}{x}=1+x.$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{6}{x}=1+x$
$\displaystyle 6=x+x^2$
$\displaystyle x^2+x-6=0$
$\displaystyle x^2+3x-2x-6=0$
$\displaystyle x(x+3)-2(x+3)=0$
$\displaystyle (x-2)(x+3)=0$
$\displaystyle \Rightarrow x=2\text{ or }-3$
$\\$

$\displaystyle \textbf{Question 7: } \text{Solve by factorization: }x=\frac{3x+1}{4x}.$
$\displaystyle \text{Answer:}$
$\displaystyle x=\frac{3x+1}{4x}$
$\displaystyle 4x^2=3x+1$
$\displaystyle 4x^2-3x-1=0$
$\displaystyle 4x^2-4x+x-1=0$
$\displaystyle 4x(x-1)+(x-1)=0$
$\displaystyle (4x+1)(x-1)=0$
$\displaystyle \Rightarrow x=1\text{ or }-\frac{1}{4}$
$\\$

$\displaystyle \textbf{Question 8: } \text{Solve by factorization: }x+\frac{1}{x}=2.5.$
$\displaystyle \text{Answer:}$
$\displaystyle x+\frac{1}{x}=2.5$
$\displaystyle \frac{x^2+1}{x}=\frac{5}{2}$
$\displaystyle 2x^2+2=5x$
$\displaystyle 2x^2-5x+2=0$
$\displaystyle 2x^2-4x-x+2=0$
$\displaystyle 2x(x-2)-1(x-2)=0$
$\displaystyle (2x-1)(x-2)=0$
$\displaystyle \Rightarrow x=2\text{ or }\frac{1}{2}$
$\\$

$\displaystyle \textbf{Question 9: } \text{Solve by factorization: }(2x-3)^2=49.$
$\displaystyle \text{Answer:}$
$\displaystyle (2x-3)^2=49$
$\displaystyle 4x^2-12x+9=49$
$\displaystyle 4x^2-12x-40=0$
$\displaystyle x^2-3x-10=0$
$\displaystyle x^2-5x+2x-10=0$
$\displaystyle x(x-5)+2(x-5)=0$
$\displaystyle (x+2)(x-5)=0$
$\displaystyle \Rightarrow x=-2\text{ or }5$
$\\$

$\displaystyle \textbf{Question 10: } \text{Solve by factorization: }2(x^2-6)=3(x-4).$
$\displaystyle \text{Answer:}$
$\displaystyle 2(x^2-6)=3(x-4)$
$\displaystyle 2x^2-12=3x-12$
$\displaystyle 2x^2-3x=0$
$\displaystyle x(2x-3)=0$
$\displaystyle \Rightarrow x=0\text{ or }\frac{3}{2}$
$\\$

$\displaystyle \textbf{Question 11: } \text{Solve by factorization: }(x+1)(2x+8)=(x+7)(x+3).$
$\displaystyle \text{Answer:}$
$\displaystyle (x+1)(2x+8)=(x+7)(x+3)$
$\displaystyle 2x^2+2x+8x+8=x^2+7x+3x+21$
$\displaystyle 2x^2+10x+8=x^2+10x+21$
$\displaystyle x^2-13=0$
$\displaystyle (x-\sqrt{13})(x+\sqrt{13})=0$
$\displaystyle \Rightarrow x=\pm\sqrt{13}$
$\\$

$\displaystyle \textbf{Question 12: } \text{Solve by factorization: }x^2-(a+b)x+ab=0.$
$\displaystyle \text{Answer:}$
$\displaystyle x^2-(a+b)x+ab=0$
$\displaystyle x^2-ax-bx+ab=0$
$\displaystyle x(x-b)-a(x-b)=0$
$\displaystyle (x-a)(x-b)=0$
$\displaystyle \Rightarrow x=a\text{ or }b$
$\\$

$\displaystyle \textbf{Question 13: } \text{Solve by factorization: }(x+3)^2-4(x+3)-5=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }x+3=y$
$\displaystyle y^2-4y-5=0$
$\displaystyle y^2-5y+y-5=0$
$\displaystyle y(y-5)+1(y-5)=0$
$\displaystyle (y+1)(y-5)=0$
$\displaystyle \Rightarrow y=-1\text{ or }5$
$\displaystyle \text{When }y=-1,\ x+3=-1\Rightarrow x=-4$
$\displaystyle \text{When }y=5,\ x+3=5\Rightarrow x=2$
$\displaystyle \therefore x=-4\text{ or }2$
$\\$

$\displaystyle \textbf{Question 14: } \text{Solve by factorization: }4(2x-3)^2-(2x-3)-14=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }2x-3=y$
$\displaystyle 4y^2-y-14=0$
$\displaystyle 4y^2-8y+7y-14=0$
$\displaystyle 4y(y-2)+7(y-2)=0$
$\displaystyle (4y+7)(y-2)=0$
$\displaystyle \Rightarrow y=2\text{ or }-\frac{7}{4}$
$\displaystyle \text{When }y=2,\ 2x-3=2\Rightarrow x=\frac{5}{2}$
$\displaystyle \text{When }y=-\frac{7}{4},\ 2x-3=-\frac{7}{4}$
$\displaystyle \Rightarrow 2x=\frac{5}{4}\Rightarrow x=\frac{5}{8}$
$\displaystyle \therefore x=\frac{5}{2}\text{ or }\frac{5}{8}$
$\\$

$\displaystyle \textbf{Question 15: } \text{Solve by factorization: }\frac{3x-2}{2x-3}=\frac{3x-8}{x+4}.$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{3x-2}{2x-3}=\frac{3x-8}{x+4}$
$\displaystyle (3x-2)(x+4)=(2x-3)(3x-8)$
$\displaystyle 3x^2+12x-2x-8=6x^2-16x-9x+24$
$\displaystyle 3x^2+10x-8=6x^2-25x+24$
$\displaystyle 3x^2-35x+32=0$
$\displaystyle 3x^2-3x-32x+32=0$
$\displaystyle 3x(x-1)-32(x-1)=0$
$\displaystyle (x-1)(3x-32)=0$
$\displaystyle \Rightarrow x=1\text{ or }\frac{32}{3}$
$\\$

$\displaystyle \textbf{Question 16: } \text{Solve by factorization: }\frac{100}{x}-\frac{100}{x+5}=1.$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{100}{x}-\frac{100}{x+5}=1$
$\displaystyle 100(x+5)-100x=x(x+5)$
$\displaystyle 500=x^2+5x$
$\displaystyle x^2+5x-500=0$
$\displaystyle x^2+25x-20x-500=0$
$\displaystyle x(x+25)-20(x+25)=0$
$\displaystyle (x+25)(x-20)=0$
$\displaystyle \Rightarrow x=20\text{ or }-25$
$\\$

$\displaystyle \textbf{Question 17: } \text{Solve by factorization: }\frac{x-3}{x+3}+\frac{x+3}{x-3}=2\frac{1}{2}.$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{x-3}{x+3}+\frac{x+3}{x-3}=2\frac{1}{2}$
$\displaystyle \frac{(x-3)^2+(x+3)^2}{(x+3)(x-3)}=\frac{5}{2}$
$\displaystyle \frac{x^2-6x+9+x^2+6x+9}{x^2-9}=\frac{5}{2}$
$\displaystyle \frac{2x^2+18}{x^2-9}=\frac{5}{2}$
$\displaystyle 2(2x^2+18)=5(x^2-9)$
$\displaystyle 4x^2+36=5x^2-45$
$\displaystyle x^2-81=0$
$\displaystyle (x-9)(x+9)=0$
$\displaystyle \Rightarrow x=9\text{ or }-9$
$\\$

$\displaystyle \textbf{Question 18: } \text{Solve by factorization: }\frac{4}{x+2}-\frac{1}{x+3}=\frac{4}{2x+1}.$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{4}{x+2}-\frac{1}{x+3}=\frac{4}{2x+1}$
$\displaystyle (2x+1)[4(x+3)-(x+2)]=4(x+2)(x+3)$
$\displaystyle (2x+1)(4x+12-x-2)=4(x+2)(x+3)$
$\displaystyle (2x+1)(3x+10)=4(x^2+5x+6)$
$\displaystyle 6x^2+23x+10=4x^2+20x+24$
$\displaystyle 2x^2+3x-14=0$
$\displaystyle 2x^2+7x-4x-14=0$
$\displaystyle x(2x+7)-2(2x+7)=0$
$\displaystyle (x-2)(2x+7)=0$
$\displaystyle \Rightarrow x=2\text{ or }-\frac{7}{2}$
$\\$

$\displaystyle \textbf{Question 19: } \text{Solve by factorization: }\frac{5}{x-2}-\frac{3}{x+6}=\frac{4}{x}.$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{5}{x-2}-\frac{3}{x+6}=\frac{4}{x}$
$\displaystyle x[5(x+6)-3(x-2)]=4(x-2)(x+6)$
$\displaystyle x(5x+30-3x+6)=4(x^2+4x-12)$
$\displaystyle x(2x+36)=4x^2+16x-48$
$\displaystyle 2x^2+36x=4x^2+16x-48$
$\displaystyle x^2-10x-24=0$
$\displaystyle x^2-12x+2x-24=0$
$\displaystyle x(x-12)+2(x-12)=0$
$\displaystyle (x+2)(x-12)=0$
$\displaystyle \Rightarrow x=-2\text{ or }12$
$\\$

$\displaystyle \textbf{Question 20: } \text{Solve by factorization: }\left(1+\frac{1}{x+1}\right)\left(1-\frac{1}{x-1}\right)=\frac{7}{8}.$
$\displaystyle \text{Answer:}$
$\displaystyle \left(1+\frac{1}{x+1}\right)\left(1-\frac{1}{x-1}\right)=\frac{7}{8}$
$\displaystyle \left(\frac{x+2}{x+1}\right)\left(\frac{x-2}{x-1}\right)=\frac{7}{8}$
$\displaystyle 8(x^2-4)=7(x^2-1)$
$\displaystyle 8x^2-32=7x^2-7$
$\displaystyle x^2=25$
$\displaystyle (x-5)(x+5)=0$
$\displaystyle \Rightarrow x=5\text{ or }-5$
$\\$

$\displaystyle \textbf{Question 21: } \text{Find the quadratic equation whose solution set is:}$
$\displaystyle \text{(i) }\{3,5\}\qquad\text{(ii) }\{-2,3\}\qquad\text{(iii) }\{5,-4\}\qquad\text{(iv) }\left\{-3,-\frac{2}{5}\right\}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }\{3,5\}$
$\displaystyle (x-3)(x-5)=0$
$\displaystyle x^2-8x+15=0$
$\displaystyle \text{(ii) }\{-2,3\}$
$\displaystyle (x+2)(x-3)=0$
$\displaystyle x^2-x-6=0$
$\displaystyle \text{(iii) }\{5,-4\}$
$\displaystyle (x-5)(x+4)=0$
$\displaystyle x^2-x-20=0$
$\displaystyle \text{(iv) }\left\{-3,-\frac{2}{5}\right\}$
$\displaystyle (x+3)\left(x+\frac{2}{5}\right)=0$
$\displaystyle (x+3)(5x+2)=0$
$\displaystyle 5x^2+17x+6=0$
$\\$

$\displaystyle \textbf{Question 22: } \text{Find the value of }x,\text{ if }a+1=0\text{ and }x^2+ax-6=0.$
$\displaystyle \text{Answer:}$
$\displaystyle a+1=0\Rightarrow a=-1$
$\displaystyle \text{Substituting in }x^2+ax-6=0,$
$\displaystyle x^2-x-6=0$
$\displaystyle x^2-3x+2x-6=0$
$\displaystyle x(x-3)+2(x-3)=0$
$\displaystyle (x+2)(x-3)=0$
$\displaystyle \Rightarrow x=-2\text{ or }3$
$\\$

$\displaystyle \textbf{Question 23: } \text{Find the value of }x,\text{ if }a+7=0,\ b+10=0\text{ and }12x^2=ax-b.$
$\displaystyle \text{Answer:}$
$\displaystyle a+7=0\Rightarrow a=-7$
$\displaystyle b+10=0\Rightarrow b=-10$
$\displaystyle \text{Substituting in }12x^2=ax-b,$
$\displaystyle 12x^2=-7x+10$
$\displaystyle 12x^2+7x-10=0$
$\displaystyle 12x^2+15x-8x-10=0$
$\displaystyle 3x(4x+5)-2(4x+5)=0$
$\displaystyle (4x+5)(3x-2)=0$
$\displaystyle \Rightarrow x=-\frac{5}{4}\text{ or }\frac{2}{3}$
$\\$

$\displaystyle \textbf{Question 24: } \text{Use the substitution }y=2x+3\text{ to solve }4(2x+3)^2-(2x+3)-14=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }y=2x+3$
$\displaystyle 4y^2-y-14=0$
$\displaystyle 4y^2-8y+7y-14=0$
$\displaystyle 4y(y-2)+7(y-2)=0$
$\displaystyle (4y+7)(y-2)=0$
$\displaystyle \Rightarrow y=-\frac{7}{4}\text{ or }2$
$\displaystyle \text{When }y=-\frac{7}{4},\ 2x+3=-\frac{7}{4}$
$\displaystyle 2x=-\frac{19}{4}$
$\displaystyle x=-\frac{19}{8}$
$\displaystyle \text{When }y=2,\ 2x+3=2$
$\displaystyle 2x=-1$
$\displaystyle x=-\frac{1}{2}$
$\displaystyle \therefore x=-\frac{19}{8}\text{ or }-\frac{1}{2}$
$\\$

$\displaystyle \textbf{Question 25: } \text{Without solving the quadratic equation }6x^2-x-2=0,\text{ find whether }x=\frac{2}{3}\text{ is a solution.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Substituting }x=\frac{2}{3}\text{ in the LHS,}$
$\displaystyle \text{LHS}=6x^2-x-2$
$\displaystyle =6\left(\frac{2}{3}\right)^2-\frac{2}{3}-2$
$\displaystyle =\frac{8-2-6}{3}=0=\text{RHS}$
$\displaystyle \therefore x=\frac{2}{3}\text{ is a root of the equation.}$
$\\$

$\displaystyle \textbf{Question 26: } \text{Determine whether }x=-1\text{ is a root of the equation }x^2-3x+2=0\text{ or not.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Substituting }x=-1\text{ in the LHS,}$
$\displaystyle \text{LHS}=x^2-3x+2$
$\displaystyle =(-1)^2-3(-1)+2$
$\displaystyle =1+3+2=6\neq0=\text{RHS}$
$\displaystyle \therefore x=-1\text{ is not a root of the equation.}$
$\\$

$\displaystyle \textbf{Question 27: } \text{If }x=\frac{2}{3}\text{ is a solution of }7x^2+mx-3=0,\text{ find the value of }m.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Substituting }x=\frac{2}{3}\text{ in }7x^2+mx-3=0,$
$\displaystyle 7\left(\frac{2}{3}\right)^2+m\left(\frac{2}{3}\right)-3=0$
$\displaystyle \frac{28}{9}+\frac{2m}{3}-3=0$
$\displaystyle \frac{2m}{3}=-\frac{1}{9}$
$\displaystyle m=-\frac{1}{6}$
$\\$

$\displaystyle \textbf{Question 28: } \text{If }x=-3\text{ and }x=\frac{2}{3}\text{ are solutions of }mx^2+7x+n=0,\text{ find }m\text{ and }n.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Substituting }x=-3,$
$\displaystyle m(-3)^2+7(-3)+n=0$
$\displaystyle 9m+n=21\qquad\ldots\text{(i)}$
$\displaystyle \text{Substituting }x=\frac{2}{3},$
$\displaystyle m\left(\frac{2}{3}\right)^2+7\left(\frac{2}{3}\right)+n=0$
$\displaystyle \frac{4m}{9}+\frac{14}{3}+n=0$
$\displaystyle 4m+9n=-42\qquad\ldots\text{(ii)}$
$\displaystyle \text{Solving (i) and (ii),}$
$\displaystyle m=3\text{ and }n=-6$
$\\$

$\displaystyle \textbf{Question 29: } \text{If one root of }x^2-(m+1)x+6=0\text{ is }3,\text{ find }m\text{ and the other root.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Substituting }x=3,$
$\displaystyle 3^2-(m+1)(3)+6=0$
$\displaystyle 9-3m-3+6=0$
$\displaystyle 12=3m\Rightarrow m=4$
$\displaystyle \text{Substituting }m=4,$
$\displaystyle x^2-5x+6=0$
$\displaystyle x^2-3x-2x+6=0$
$\displaystyle x(x-3)-2(x-3)=0$
$\displaystyle (x-2)(x-3)=0$
$\displaystyle \Rightarrow x=2\text{ or }3$
$\displaystyle \therefore \text{The other root is }2.$
$\\$

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Class 10: Quadratic Equations – Exercise 7(a)

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