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

Solve by factorization:

Question 1: $x^2 -10x -24=0$

$x^2 -10x -24=0$

$x^2-6x-4x-24=0$

$x(x-6)-4(x-6)=0$

$(x-4)(x-6)=0$

$\Rightarrow x = 4 \ or \ 6$

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Question 2: $x^2-16=0$

$x^2-16=0$

$(x-4)(x+4)=0$

$\Rightarrow x = 4 \ or \ -4$

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Question 3: $2x^2-\frac{1}{2}x = 0$

$2x^2-\frac{1}{2}x = 0$

$4x^2-x=0$

$x(2x-1)=0$

$\Rightarrow x = 0 \ or \ \frac{1}{2}$

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Question 4: $x(x-5)=24$

$x(x-5)=24$

$x^2-5x-24=0$

$x^2-8x+3x-24=0$

$x(x-8)+3(x-8)=0$

$(x+3)(x-8)=0$

$\Rightarrow x = -3 \ or \ 8$

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Question 5: $\frac{9}{2}x = 5+x^2$

$\frac{9}{2}x = 5+x^2$

$9x=10+2x^2$

$2x^2-9x+10=0$

$2x^2-5x-4x+10=0$

$2x(x-2)-5(x-2)=0$

$(2x-5)(x-2)=0$

$\Rightarrow x = 2 \ or \ \frac{5}{2}$

$\\$

Question 6: $\frac{6}{x}=1+x$

$\frac{6}{x}=1+x$

$6=x+x^2$

$x^2+x-6=0$

$x^2+3x-2x-6=0$

$x(x+3)-2(x+3)=0$

$(x-2)(x+3)=0$

$\Rightarrow x = 2 \ or \ -3$

$\\$

Question 7: $x = \frac{3x+1}{4x}$

$x = \frac{3x+1}{4x}$

$4x^2=3x+1$

$4x^2-3x-1=0$

$4x^2-4x+x-1=0$

$4x(x-1)+(x-1)=0$

$(4x+1)(x-1)=0$

$\Rightarrow x = 1 \ or -\frac{1}{4}$

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Question 8: $x+\frac{1}{x} = 2.5$

$x+\frac{1}{x} = 2.5$

$\frac{x^2+1}{x}=\frac{5}{2}$

$2x^2+2=5x$

$2x^2-5x+2=0$

$2x^2-4x-x+2=0$

$2x(x-2)-1(x-2)=0$

$(2x-1)(x-2)=0$

$\Rightarrow x = 2 \ or \ \frac{1}{2}$

$\\$

Question 9: $(2x-3)^2=49$

$(2x-3)^2=49$

$4x^2+9-12x=49$

$4x^2-12x-40=0$

$x^2-3x-10=0$

$x^2-5x+2x-10=0$

$x(x-5)+2(x-5)=0$

$(x+2)(x-5)=0$

$\Rightarrow x = -2 \ or 5$

$\\$

Question 10: $2(x^2-6)=3(x-4)$

$2(x^2-6)=3(x-4)$

$2x^2-12=3x-12$

$2x^2-3x=0$

$x(2x-3)=0$

$\Rightarrow x = 0 \ or \frac{3}{2}$

$\\$

Question 11: $(x+1)(2x+8)=(x+7)(x+3)$

$(x+1)(2x+8)=(x+7)(x+3)$

$2x^2+2x+8x+8=x^2+7x+3x+21$

$2x^2+10x+8=x^2+10x+21$

$x^2=13$

$\Rightarrow x = \pm \sqrt{13}$

$\\$

Question 12: $x^2-(a+b)x+ab=0$

$x^2-(a+b)x+ab=0$

$x^2-ax-bx+ab=0$

$x(x-b)-a(x-b)=0$

$(x-a)(x-b) = 0$

$\Rightarrow x = a \ or \ b$

$\\$

Question 13: $(x+3)^2-4(x+3)-5=0$

$(x+3)^2-4(x+3)-5=0$

Let $x+3 = y$

$y^2-4y-5=0$

$y^2-5y+y-5=0$

$y(y-5)+ (1(y-5)=0$

$(y+1)(y-5)=0$

$\Rightarrow y = -1 \ or \ 5$

Hence when $y = -1$,

$x+3 = 1 \Rightarrow x = -4$ and

when $y = 5$,

$x+3 = 5 \Rightarrow x = 2$

Hence $\Rightarrow y = -4 \ or 2$

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Question 14: $4(2x-3)^2-(2x-3)-14=0$

$4(2x-3)^2-(2x-3)-14=0$

Let $2x-3 = y$

$4y^2-y-14=0$

$4y^2-8y+7y-14=0$

$4y(y-2)+7(y-2)=0$

$(4y+7)(y-2)=0$

$\Rightarrow y = 2 \ or \frac{-7}{4}$

Hence when $y = 2$,

$2x-3 = 2 \Rightarrow x = \frac{5}{2}$ and

when $y = \frac{-7}{4}$,

$x+3 = \frac{-7}{4} \Rightarrow x = \frac{-19}{4}$

Hence $\Rightarrow y = \frac{5}{2} \ or \frac{-19}{4}$

$\\$

Question 15: $\frac{3x-2}{2x-3}=\frac{3x-8}{x+4}$

$\frac{3x-2}{2x-3}=\frac{3x-8}{x+4}$

$(3x-2)(x+4)=(2x-3)(3x-8)$

$3x^2-2x+12x-8=6x^2-9x-16x+24$

$3x^2+10x-8=6x^2-25x+24$

$3x^2-35x+32=0$

$3x^2-3x-32x+32=0$

$3x(x-3)-32(x-3)=0$

$(x-3)(3x-32)=0$

$\Rightarrow x = 3 \ or \frac{32}{3}$

$\\$

Question 16: $\frac{100}{x}-\frac{100}{x+5}=+ 1$

$\frac{100}{x}-\frac{100}{x+5}=+ 1$

$100x+500-100x=x(x+5)$

$x^2+5x-500=0$

$x^2+25x-20x-500=0$

$x(x+25)-20(x+25)=0$

$(x+25)(x-20)=0$

$\Rightarrow x = 20 \ or \ -25$

$\\$

Question 17: $\frac{x-3}{x+3}+\frac{x+3}{x-3}=2\frac{1}{2}$

$\frac{x-3}{x+3}+\frac{x+3}{x-3}=2\frac{1}{2}$

$\frac{x^2+9-6x+x^2+9+6x}{x^2-9} = \frac{5}{2}$

$2(x^2+18)=5(x^2-9)$

$4x^2+36=5x^2-45$

$x^2=81$

$\Rightarrow x = 9 \ or \ -9$

$\\$

Question 18: $\frac{4}{x+2}-\frac{1}{x+3}=\frac{4}{2x+1}$

$\frac{4}{x+2}-\frac{1}{x+3}=\frac{4}{2x+1}$

$(2x+1)[4(x+3)-(x+2)]=4(x+2)(x+3)$

$(2x+1)(4x+12-x-2)=4(x+2)(x+3)$

$(2x+1)(3x+10)=4(x+2)(x+3)$

$6x^2+3x+20x+10=4(x^2+2x+3x+6)$

$6x^2+23x+10=4x^2+20x+24$

$2x^2+3x-14=0$

$2x^2+7x-4x-14=0$

$2x(x-2)+7(x-2)=0$

$(x-2)(2x+7)=0$

$\Rightarrow x = 2 \ or \ \frac{7}{2}$

$\\$

Question 19: $\frac{5}{x-2}-\frac{3}{x+6}=\frac{4}{x}$

$\frac{5}{x-2}-\frac{3}{x+6}=\frac{4}{x}$

$x[5(x+6)-3(x-2)]=4(x-2)(x+6)$

$x(2x+36)=4(x62-2x+6x-12)$

$2x^2+36x=4x^2+16x-48$

$2x^2-20x-48=0$

$x^2-10x-24=0$

$x^2-6x-4x-24=0$

$x(x-6)-4(x-6)=0$

$(x-4)(x-6)=0$

$\Rightarrow x = 4 \ or \ 6$

$\\$

Question 20: $(1+\frac{1}{x+1})(1-\frac{1}{x-1})=\frac{7}{8}$

$(1+\frac{1}{x+1})(1-\frac{1}{x-1})=\frac{7}{8}$

$(\frac{x+2}{x+1})(\frac{x-2}{x-1})=\frac{7}{8}$

$8(x^2-4)=7(x^2-1)$

$8x^2-32=7x^2-7$

$x^2=25$

$\Rightarrow x = 5 \ or \ -5$

$\\$

Question 21: Find the quadratic equation whose solution set is i) $\{3, 5 \}$ ii) $\{-2, 3 \}$ iii)   $\{5, -4 \}$ iv)   $\{-3, -\frac{2}{5} \}$

i) $\{3, 5 \}$

$(x-3)(x-5)=0$

$x^2-8x+15=0$

ii) $\{-2, 3 \}$

$(x+2)(x-3)=0$

$x^2-x-6=0$

iii)   $\{5, -4 \}$

$(x-5)(x+4)=0$

$x^2-x-20=0$

iv)   $\{-3, -\frac{2}{5} \}$

$(x+3)(x+\frac{2}{5})=0$

$(x+3)(5x+2)=0$

$5x^2+17x+6=0$

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Question 22: Find the value of $x$ , if $a+1 = 0$ and $x^2+ax-6=0$.

$x^2+ax-6=0$

$a+1 =0 \Rightarrow a = -1$

Substituting

$x^2-x-6=0$

$x^2-3x+2x-6=0$

$x(x+2)-3(x+2)=0$

$(x+2)(x-3)=0$

$\Rightarrow x = -2 \ or \ 3$

$\\$

Question 23: Find the value of $x$ , if $a+7=0; b+10=0$ and $12x^2=ax-b$

$a+7=0 \Rightarrow a = -7$

$b+10=0 \Rightarrow b = -10$

Substituting

$12x^2=ax-b$

$12x^2=-7x+10$

$12x^2+7x-10=0$

$12x^2+15x-8x-10=0$

$4x(3x-2)+5(3x-2)=0$

$(4x+5)(3x-2)=0$

$\Rightarrow x = -\frac{5}{4} \ or \ x = \frac{2}{3}$

$\\$

Question 24:  Use the substitution $y = 2x+3$ to solve for $x$ , if $4(2x+3)^2-(2x+3)-14=0$.

$y = 2x+3$

$4(2x+3)^2-(2x+3)-14=0$

Therefore

$4y^2-y-14=0$

$4y^2-8y+7y-14=0$

$4y(y-2)+7(y-2)=0$

$(4y+7)(y-2)=0$

$\Rightarrow y = -\frac{7}{4} \ or \ 2$

Hence, if $y = -\frac{7}{4}$

$2x+3 = -\frac{7}{4}$

$2x = -\frac{19}{4}$

$x = -\frac{19}{8}$

Hence, if $y = 2$

$2x+3 =2$

$2x = -1$

$x = -\frac{1}{2}$

$\\$

Question 25: Without solving for the quadratic equation $6x^2-x-2=0$ , find whether $x=\frac{2}{3}$ is a solution of this equation or not.

$x=\frac{2}{3}$

Substituting

LHS $= 6x^2-x-2$

$= 6(\frac{2}{3})^2-(\frac{2}{3})-2$

$= \frac{8-2-6}{3} =0 =$ RHS

Hence $x=\frac{2}{3}$ is a root of the equation.

$\\$

Question 26: Determine whether $x=-1$ is a root of the equation $x^2-3x+2 = 0$ or not.

$x= -1$

$LHS = x^2-3x +2$

$= (-1)^2-3(-1)+2$

$= 1+3+2= 6 \neq RHS$

Hence $x=-1$ is a not root of the equation.

$\\$

Question 27: If $x = \frac{2}{3}$ is a solution of the quadratic equation $7x^2+mx-3=0$ ; find the value of $m$.

$x = \frac{2}{3}$

$7x^2+mx-3=0$

Substituting

$7(\frac{2}{3})^2+m(\frac{2}{3})-3=0$

$\frac{28}{9}+\frac{2}{3}m-3=0$

$\frac{2}{3}m= -\frac{1}{9}$

$m =-\frac{3}{18} = - \frac{1}{6}$

$\\$

Question 28: If $x = -3$ and $x = \frac{2}{3}$ are solutions of quadratic equation $mx^2+7x+n=0$ , find the value of $m \ and \ n$.

$x = -3$ and $x = \frac{2}{3}$

Substituting

$m(-3)^2+7(-3)+n=0$

$9m+n=21$ … … … … i)

Similarly,

$m(\frac{2}{3})^2+7(\frac{2}{3})+n=0$

$4m+9n=-42$  … … … … ii)

Solving i) and ii)

We get $m = 3 \ and \ n = -6$

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Question 29: In quadratic equation $x^2-(m+1)x+6 =0$ has one root as $x=3$ ; find the value of m and the other root of the equation.

$x^2-(m+1)x+6 =0$

$x = 3$

Substituting

$(3)^2-(m+1)(3)+6=0$

$12=3m \Rightarrow m = 4$

Sustituting

$x^2-5x+6=0$

$x^2-3x-2x+6=0$

$x(x-3)-2(x-3)=0$

$(x-2)(x-3)=0$

$\Rightarrow x = 2 \ or \ 3$

Hence the other root is $2$

$\\$