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  

Answer:

\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  

Answer:

\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  

Answer:

\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  

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 \\

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

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 \\

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

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 \\

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

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 \\

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

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 \\

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

Answer:

\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)  

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 \\

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

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  

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

\displaystyle \\

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

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 \\

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

Answer:

\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  

Answer:

\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}  

Answer:

\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  

Answer:

\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}  

Answer:

\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}  

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)(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}  

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(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}  

Answer:

\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} \}  

Answer:

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 .

Answer:

\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  

Answer:

\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 .

Answer:

\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.

Answer:

\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.

Answer:

\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 .

Answer:

\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 .

Answer:

\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.

Answer:

\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