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:

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

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

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

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

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

$\displaystyle \Rightarrow x = 4 \text{ or } 6$

$\displaystyle \\$

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

$\displaystyle x^2-16=0$

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

$\displaystyle \Rightarrow x = 4 \text{ or } -4$

$\displaystyle \\$

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

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

$\displaystyle 4x^2-x=0$

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

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

$\displaystyle \\$

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

$\displaystyle x(x-5)=24$

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

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

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

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

$\displaystyle \Rightarrow x = -3 \text{ or } 8$

$\displaystyle \\$

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

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

$\displaystyle 9x=10+2x^2$

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

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

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

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

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

$\displaystyle \\$

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

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

$\displaystyle 6=x+x^2$

$\displaystyle x^2+x-6=0$

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

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

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

$\displaystyle \Rightarrow x = 2 \text{ or } -3$

$\displaystyle \\$

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

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

$\displaystyle 4x^2=3x+1$

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

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

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

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

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

$\displaystyle \\$

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

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

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

$\displaystyle 2x^2+2=5x$

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

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

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

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

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

$\displaystyle \\$

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

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

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

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

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

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

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

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

$\displaystyle \Rightarrow x = -2 \text{ or } 5$

$\displaystyle \\$

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

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

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

$\displaystyle 2x^2-3x=0$

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

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

$\displaystyle \\$

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

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

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

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

$\displaystyle x^2=13$

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

$\displaystyle \\$

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

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

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

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

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

$\displaystyle \Rightarrow x = a \text{ or } b$

$\displaystyle \\$

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

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

Let $\displaystyle x+3 = y$

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

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

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

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

$\displaystyle \Rightarrow y = -1 \text{ or } 5$

Hence when $\displaystyle y = -1$ ,

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

when $\displaystyle y = 5$ ,

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

Hence $\displaystyle \Rightarrow y = -4 \text{ or } 2$

$\displaystyle \\$

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

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

Let $\displaystyle 2x-3 = y$

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

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

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

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

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

Hence when $\displaystyle y = 2$ ,

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

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

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

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

$\displaystyle \\$

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

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

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

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

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

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

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

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

$\displaystyle \Rightarrow x = 20 \text{ or } -25$

$\displaystyle \\$

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

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

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

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

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

$\displaystyle x^2=81$

$\displaystyle \Rightarrow x = 9 \text{ or } -9$

$\displaystyle \\$

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

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

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

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

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

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

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

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

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

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

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

$\displaystyle \Rightarrow x = 4 \text{ or } 6$

$\displaystyle \\$

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

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

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

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

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

$\displaystyle x^2=25$

$\displaystyle \Rightarrow x = 5 \text{ or } -5$

$\displaystyle \\$

Question 21: Find the quadratic equation whose solution set is

i) $\displaystyle \{3, 5 \}$      ii) $\displaystyle \{-2, 3 \}$      iii) $\displaystyle \{5, -4 \}$      iv) $\displaystyle \{-3, - \frac{2}{5} \}$

i) $\displaystyle \{3, 5 \}$

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

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

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

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

$\displaystyle x^2-x-6=0$

iii) $\displaystyle \{5, -4 \}$

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

$\displaystyle x^2-x-20=0$

iv) $\displaystyle \{-3, - \frac{2}{5} \}$

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

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

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

$\displaystyle \\$

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

$\displaystyle x^2+ax-6=0$

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

Substituting

$\displaystyle x^2-x-6=0$

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

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

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

$\displaystyle \Rightarrow x = -2 \text{ or } 3$

$\displaystyle \\$

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

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

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

Substituting

$\displaystyle 12x^2=ax-b$

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

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

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

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

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

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

$\displaystyle \\$

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

$\displaystyle y = 2x+3$

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

Therefore

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

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

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

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

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

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

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

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

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

Hence, if $\displaystyle y = 2$

$\displaystyle 2x+3 =2$

$\displaystyle 2x = -1$

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

$\displaystyle \\$

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

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

Substituting

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

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

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

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

$\displaystyle \\$

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

$\displaystyle x= -1$

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

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

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

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

$\displaystyle \\$

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

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

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

Substituting

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

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

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

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

$\displaystyle \\$

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

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

Substituting

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

$\displaystyle 9m+n=21$ … … … … i)

Similarly,

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

$\displaystyle 4m+9n=-42$ … … … … ii)

Solving i) and ii)

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

$\displaystyle \\$

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

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

$\displaystyle x = 3$

Substituting

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

$\displaystyle 12=3m \Rightarrow m = 4$

Sustituting

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

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

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

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

$\displaystyle \Rightarrow x = 2 \text{ or } 3$

Hence the other root is $\displaystyle 2$