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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

\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}

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

Answer:

\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

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

Answer:

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

Answer:

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}

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

Answer:

(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

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

Answer:

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}

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

Answer:

(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}

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

Answer:

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

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

Answer:

(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

Answer:

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}

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

Answer:

\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}

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

Answer:

\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

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

Answer:

\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

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

Answer:

\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}

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

Answer:

\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

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

Answer:

(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

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

Answer:

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 .

Answer:

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

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

Answer:

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}

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

Answer:

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}

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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.

Answer:

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.

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

Answer:

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.

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

Answer:

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}

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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 .

Answer:

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.

Answer:

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

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