Solve each of the following quadratic equations:

Question 1: x^2-11x+28 = 0

Answer:

x^2-11x+28 = 0

\Rightarrow x^2 - 7x - 4x + 28 = 0

\Rightarrow x (x-4)-7 (x-4) = 0

\Rightarrow (x-4)(x-7) = 0

\therefore \ x = 4 \ or\ x = 7

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Question 2: x^2-13x = 68

Answer:

x^2-13x = 68

\Rightarrow x^2-13x-68x = 0

\Rightarrow x^2-17x+4x-68x = 0

\Rightarrow x(x-17)+4(x-17) = 0

\therefore \ x = 17, \ or\ x = -4

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Question 3: x^2+2x = 48

Answer:

x^2+2x = 48

\Rightarrow x^2+2x-48 = 0

\Rightarrow x^2+8x-6x-48 = 0

\Rightarrow x(x+8)-6(x+8) = 0

\Rightarrow (x-6)(x+8) = 0

\therefore \ x = 6 \ or\ x = -8

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Question 4: x^2-2x = 3

Answer:

x^2-2x = 3

\Rightarrow x^2-2x-3 = 0

\Rightarrow x^2-3x+1x-3 = 0

\Rightarrow (x+1)(x-3) = 0

\Rightarrow x(x-3)+1(x-3) = 0

\Rightarrow x = -1 \ or\ x = 3

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Question 5: x^2-49 = 0

Answer:

x^2-49 = 0

\Rightarrow (x-7)(x+7) = 0

\Rightarrow x = 7 \ or\ x = -7

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Question 6: 3x^2-243 = 0

Answer:

3x^2-243 = 0

\Rightarrow 3 (x^2-81) = 0

\Rightarrow (x-9)(x+9) = 0

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

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Question 7: 16x^2 = 9

Answer:

16x^2 = 9

\Rightarrow 16 x^2-9 = 0

\Rightarrow (4x-3)(4x+3) = 0

\Rightarrow x = \frac{3}{4} \ or\ x = \frac{(-3)}{4}

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Question 8: 3x^2+13x = 10

Answer:

3x^2+13x = 10

\Rightarrow 3x^2+13x-10 = 0

\Rightarrow 3x^2-2x+15x-10 = 0

\Rightarrow x(3x-2)+5(3x-2) = 0

\Rightarrow (3x-2)(x+5) = 0

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

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Question 9: 2x^2-7x+3 = 0

Answer:

2x^2-7x+3 = 0

\Rightarrow 2 x^2-6x-x+3 = 0

\Rightarrow 2x(x-3)-1(x-3) = 0

\Rightarrow (2x-1)(x-3) = 0

\Rightarrow x = \frac{1}{2} \ or\ x = 3

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Question 10: 6x^2+5x = 6

Answer:

6x^2+5x = 6

\Rightarrow 6x^2+5x-6 = 0

\Rightarrow 6x^2+9x-4x-6 = 0

\Rightarrow 3x(2x+3)-2(2x+3) = 0

\Rightarrow (3x-2)(2x+3) = 0

\Rightarrow x = \frac{2}{3} \ or\ x = \frac{(-3)}{2}

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Question 11: 10x^2+11x+3 = 0

Answer:

10x^2+11x+3 = 0

\Rightarrow 10x^2+6x+5x+3 = 0

\Rightarrow 5x(2x+1)+3(2x+1) = 0

\Rightarrow (2x+1)(5x+3) = 0

\Rightarrow x = \frac{(-1)}{2} \ or\ x = \frac{(-3)}{5}

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Question 12: 9x^2+8 = 22x

Answer:

9x^2+8 = 22x

\Rightarrow 9x^2-22x+8 = 0

\Rightarrow 9x^2-18x-4x+8 = 0

\Rightarrow 9x(x-2)-4(x-2) = 0

\Rightarrow (x-2)(9x-4) = 0

\Rightarrow x = 2 \ or\ x = \frac{4}{9}

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Question 13: 22x^2+x = 6

Answer:

22x^2+x = 6

\Rightarrow 22x^2+x-6 = 0

\Rightarrow 22x^2+11x-12x-6 = 0

\Rightarrow 11x(2x+1)-6(2x+1) = 0

\Rightarrow (2x+1)(11x-6) = 0

\Rightarrow x = \frac{(-1)}{2} \ or\ x = \frac{6}{11}

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Question 14: 2x^2+5x-7 = 0

Answer:

2x^2+5x-7 = 0

\Rightarrow 2x^2+7x-2x-7 = 0

\Rightarrow 2x(x-1))+7(x-1) = 0

\Rightarrow (x-1)(2x+7) = 0

\Rightarrow x = 1 \ or\ x = \frac{(-7)}{2}

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Question 15: 6y^2 = 7y+20

Answer:

6y^2 = 7y+20

\Rightarrow 6y^2-7y-20 = 0

\Rightarrow 6y^2-15y+8y-20 = 0

\Rightarrow 3y(2y-5)+4(2y-5) = 0

\Rightarrow (2y-5)(3y+4) = 0

\Rightarrow y = \frac{5}{2} \ or \ y = \frac{(-4)}{3}

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Question 16: 4z^2 = 11z+3

Answer:

4z^2 = 11z+3

\Rightarrow 4z^2-11z-3 = 0

\Rightarrow 4z^2+z-12x-3 = 0

\Rightarrow 4z(z-3)+(z-3) = 0

\Rightarrow (z-3)(4z+1) = 0

\Rightarrow z = 3 \ or\ z = \frac{(-1)}{4}

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Question 17: (t-2)^2 = 36

Answer:

(t-2)^2 = 36

\Rightarrow t^2+4-4t = 36

\Rightarrow t^2-4t-32 = 0

\Rightarrow t^2-8t+4t-32 = 0

\Rightarrow t(t+4)-8(t+4) = 0

\Rightarrow (t+4)(t-8) = 0

\Rightarrow t = -4 \ or\ t = 8

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Question 18: (2x+3)(x-4) = 6

Answer:

(2x+3)(x-4) = 6

\Rightarrow 2x^2+3x-8x-12 = 6

\Rightarrow 2x^2-5x-18 = 0

\Rightarrow 2x^2-9x+4x-18 = 0

\Rightarrow 2x(x+2)-9(x+2) = 0

\Rightarrow 2x (x+2)(2x-9) = 0

\Rightarrow x = -2 \ or\ x = \frac{9}{2}

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Question 19: 12x+7 = \frac{10}{x}

Answer:

12x+7 = \frac{10}{x}

\Rightarrow 12x^2+7x-10 = 0

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

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

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

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

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Question 20: x^2-10x+21 = 0

Answer:

x^2-10x+21 = 0

\Rightarrow x^2-7x-3x+21 = 0

\Rightarrow x(x-7)-3(x-7) = 0

\Rightarrow (x-7)(x-3) = 0

\Rightarrow x = 7 \ or \ x = 3

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Question 21: \frac{x}{3}  - \frac{6}{x}  = 1

Answer:

\frac{x}{3} - \frac{6}{x}  = 1

\Rightarrow x^2-18 = 3x

\Rightarrow x^2-3x-18 = 0

\Rightarrow x^2-6x+3x-18 = 0

\Rightarrow x(x+3)-6(x+3)

\Rightarrow (x+3)(x-6) = 0

\Rightarrow x = -3 \ or\ x = 6

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Question 22: \frac{(x+1)}{(x+5)} = \frac{(2x-3)}{(x+3)}

Answer:

\frac{(x+1)}{(x+5)} = \frac{(2x-3)}{(x+3)}

\Rightarrow x^2+x+3x+3 = 2x^2-3x+10x-15

\Rightarrow x^2+3x-18 = 0

\Rightarrow x^2+6x-3x-18 = 0

\Rightarrow x(x-3)+6(x-3) = 0

\Rightarrow (x-3)(x+6)

\Rightarrow x = 3 \ or\ x = -6

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Question 23: \frac{(x+1)}{(3x-7)} = \frac{(x-1)}{(2x-5)}

Answer:

\frac{(x+1)}{(3x-7)} = \frac{(x-1)}{(2x-5)}

\Rightarrow 2x^2+2x-5x-5 = 3x^2-3x-7x+7

\Rightarrow 2x^2-3x-5 = 3x^2-10x+7

\Rightarrow x^2-7x+12 = 0

\Rightarrow x^2-3x-4x+12 = 0

\Rightarrow x(x-3)-4(x-3) = 0

\Rightarrow (x-3)(x-4) = 0

\Rightarrow x = 3 \ or\ x = 4

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Question 24: \frac{1}{(x-1)} - \frac{1}{(x+2)} = \frac{3}{4}

Answer:

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

\Rightarrow (x+2)-(x-1) = \frac{3}{4} (x-1)(x+2)

\Rightarrow 3 = \frac{3}{4} (x^2-x+2x-2)

\Rightarrow 4 = x^2+x-2

\Rightarrow x^2+x-6 = 0

\Rightarrow x^2+3x-2x-6 = 0

\Rightarrow x(x-2)+3(x-2) = 0

\Rightarrow (x-2)(x+3) = 0

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

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Question 25: \frac{4}{(x-1)} - \frac{3}{x} = \frac{5}{x+2}

Answer:

\frac{4}{(x-1)} - \frac{3}{x} = \frac{5}{x+2}

Multiply LHS and RHS by (x-1)(x)(x+2)

\Rightarrow 4(x)(x+2)-3(x-1)(x+2) = 5(x-1)(x)

\Rightarrow 4x^2+8x-3x^2-3x+6 = 5x^2-5x

\Rightarrow 4x^2+8x-3x^2-3x+6 = 5x^2-5x

\Rightarrow x^2+5x+6 = 5x^2-5x

\Rightarrow 4x^2-10x-6 = 0

\Rightarrow 2x^2-5x-3 = 0

\Rightarrow 2x^2-6x+x-3 = 0

\Rightarrow 2x (x-3)+(x-3) = 1

\Rightarrow (x-3)(2x+1) = 0

\Rightarrow x = 3 \ or\ - \frac{1}{2}

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Question 26: \frac{(x+2)}{(x-1)} - \frac{4-x}{2x}  = \frac{7}{2}

Answer:

\frac{(x+2)}{(x-1)} - \frac{4-x}{2x}  = \frac{7}{2}

\Rightarrow 2x(x+2)-(x-1)(4-x) = \frac{7}{2} (x-1)(2x)

\Rightarrow 2x^2+4x-4x+4+x^2-x = 7(x^2-x)

\Rightarrow 3x^2-x+4 = 7x^2-7x

\Rightarrow 4x^2-6x-4 = 0

\Rightarrow 2x^2-3x-2 = 0

\Rightarrow 2x^2-4x+x-2 = 0

\Rightarrow 2x (x-2)+1(x-2) = 0

\Rightarrow (x-2)(2x+1) = 0

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