Question 1: Solve: 

(i) 2x^2 - 7x = 39      (ii) x^2 = 5x       (iii) x^2 = 16

Answer:

(i)     \displaystyle 2x^2 - 7x = 39

\displaystyle  \Rightarrow 2x^2 - 7x - 39 = 0

\displaystyle  \Rightarrow 2x^2 - 13 x + 6 x - 39 = 0

\displaystyle  \Rightarrow x(2x-13) + 3( 2x - 13) = 0

\displaystyle  \Rightarrow (2x-13)(x+3) = 0

\displaystyle  \Rightarrow 2x-13 = 0  \text{   or   }  x + 3 = 0

\displaystyle  \Rightarrow x = \frac{13}{2}   \text{   or   }  x = - 3

(ii)    \displaystyle  x^2 = 5x

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

\displaystyle  \Rightarrow x(x-5) = 0

\displaystyle  \Rightarrow x = 0  \text{   or   }  x -5 = 0

\displaystyle  \Rightarrow x = 0  \text{   or   }  x = 5

(iii)    \displaystyle  x^2 = 16

\displaystyle  \Rightarrow x^2 - 16 = 0

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

\displaystyle  \Rightarrow x+ 4 = 0  \text{   or   }  x - 4 = 0

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

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\displaystyle \text{Question 2: Solve:  } \frac{x}{x-1} + \frac{x-1}{x} = 2\frac{1}{2}

Answer:

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

\displaystyle  \Rightarrow \frac{x^2 + ( x-1)^2}{x(x-1)} = \frac{5}{2}

\displaystyle  \Rightarrow 2 ( x^2 + x^2 - 2x + 1) = 5 ( x^2 - x)

\displaystyle  \Rightarrow 4x^2 - 4x + 2 = 5x^2 - 5x

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

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

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

\displaystyle  \Rightarrow x-2 = 0 \text{   or   }   x+1 = 0

\displaystyle  \Rightarrow x = 2 \text{   or   }  x = -1

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Question 3: Find the quadratic equation whose solution set is {-2, 3}.

Answer:

Since, solution set \displaystyle  = \{ -2, 3 \}

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

\displaystyle  \Rightarrow x+2=0, \text{   or   }  x- 3 = 0

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

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

\displaystyle  \Rightarrow x^2 - x - 6 = 0   which is the required quadratic equation.

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Question 4: Use the substitution \displaystyle  x = 3y + 1 to solve for y , if 5(3y+1)^2 + 6(3y+1) - 8 = 0

Answer:

\displaystyle  5(3y+1)^2 + 6(3y+1) - 8 = 0

\displaystyle  \text{Putting } x = 3y+1

\displaystyle  \Rightarrow 5x^2 + 6x - 9 = 0

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

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

\displaystyle  \Rightarrow \text{When } x = - 2, \text{ then } 3y+1 = - 2  \Rightarrow y = -1

\displaystyle  \Rightarrow \text{ and when } x = \frac{4}{5}, \text{ then } 3y+1 = \frac{4}{5} \Rightarrow y = \frac{-1}{15}

\displaystyle  \Rightarrow \text{ Therefore } y = -1 \text{   or   }  y =  \frac{-1}{15}

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Question 5: Without solving the quadratic equation 3x^2 - 2x- 1 = 0, find whether x = 1 is a solution (root) of this equation or not.

Answer:

Substituting x = 1 in the given equation 3x^2 - 2x - 1 = 0, we get :

3(1)^2 - 2(1) -1 = 0

\Rightarrow 3 -2 -1 = 0

Which is true.

Therefore x = 1 is a solution of the given equation 3x^2 - 2x- 1 = 0

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Question 6:  Without solving the quadratic equation x^2 - x + 1 = 0, find whether x = -1 is a solution (root) of this equation or not.

Answer:

Substituting x = -1 in the given equation x^2 - x + 1 = 0, we get :

(-1)^2 -(-1)+1 = 0 

\Rightarrow 1+1+1 = 0

Which is not true.

Therefore x =- 1 is a solution of the given equation x^2 - x + 1 = 0

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Question 7: Find the value of k for which x = 2 is a root (solution) of equation kx^2 + 2x - 3 = 0.

Answer:

Substituting x = 2 in the given equation kx^2 + 2x - 3 = 0, we get :

k(2)^2 + 2 \times 2 - 3 = 0

\Rightarrow 4k + 4 - 3 = 0 

\Rightarrow 4k = -1 

\displaystyle \Rightarrow k = \frac{-1}{4} 

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Question 8: If x = 2 and x = 3 are roots of the equation 3x^2 - 2mx + 2n = 0, find the values of m and n.

Answer:

x=2 is a root of the equation 3x^2 - 2mx + 2n = 0

\Rightarrow 3(2)^2 - 2m \times 2 + 2 n = 0

\Rightarrow 12 - 4m + 2n = 0

\Rightarrow - 4m + 2n =- 12

\Rightarrow 2m - n = 6  \text{   ... ... ... ... ... i)}

x=3 is a root of the equation 3x^2 - 2mx + 2n = 0

\Rightarrow 3(3)^2 - 2m \times 3 + 2 n = 0

\Rightarrow 27 - 6m + 2n = 0

\Rightarrow - 6m + 2n =- 27

\Rightarrow 6m - 2n = 27  \text{   ... ... ... ... ... ii)}

On solving equations I and II, we get m = 7.5 and n = 9

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Question 9: If one root of the quadratic equation 2x^2 + ax - 6 = 0   is 2 , find the value of a. Also, find the other root.

Answer:

Since, x=2   is a root of the given equation  2x^2 + ax - 6 = 0

\Rightarrow 2(2)^2 + a \times 2 - 6 = 0

\Rightarrow 8 + 2a - 6 = 0

\Rightarrow a = - 1

Substituting a = -1, we get :

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

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

\Rightarrow 2x^2 - 4x + 3x - 6 = 0

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

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

\Rightarrow x-2 = 0 \text{   or   } 2x + 3 = 0

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

\displaystyle \Rightarrow \text{The other root is } \frac{-3}{2}

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Question 10:  Solve each of the following equations by using the formula:

(i) 5x^2 - 2x - 3 = 0      (ii) x^2 = 18x - 77     (iii) \sqrt{3} x^2 + 11 x + 6\sqrt{3} =0

Answer:

\displaystyle \text{(i) Comparing } 5x^2 - 2x - 3 = 0 \text{ with } ax^2 + bx + c = 0, \\ \\ \text{ we get: } a = 5, b = -2  \text{ and } c= -3

\displaystyle \text{We know } x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Substituting the values we get

\displaystyle x = \frac{2 \pm \sqrt{(-2)^2 - 4(5)(-3)}}{2(5)} = \frac{2 \pm \sqrt{64}}{10} = \frac{2\pm8}{10}

\displaystyle \Rightarrow x = 1 \text{ and } x = \frac{-3}{5}

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\displaystyle \text{(ii) Comparing } x^2 = 18x - 77 \text{ with } ax^2 + bx + c = 0, \\ \\ \text{ we get: } a = 1, b = -18  \text{ and } c= 77

\displaystyle \text{We know } x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Substituting the values we get

\displaystyle x = \frac{18 \pm \sqrt{(-18)^2 - 4(1)(77)}}{2(1)} = \frac{18 \pm \sqrt{16}}{2} = \frac{18\pm4}{2}

\displaystyle \Rightarrow x = 11 \text{ and } x = 7

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\displaystyle \text{(ii) Comparing } \sqrt{3} x^2 + 11 x + 6\sqrt{3} =0 \text{ with } ax^2 + bx + c = 0, \\ \\ \text{ we get: } a = \sqrt{3}, b = 11  \text{ and } c= 6\sqrt{3}

\displaystyle \text{We know } x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Substituting the values we get

\displaystyle x = \frac{-11 \pm \sqrt{(11)^2 - 4(\sqrt{3})(6\sqrt{3})}}{2(\sqrt{3})} = \frac{-11 \pm \sqrt{49}}{2\sqrt{3}} = \frac{-11\pm 7}{2\sqrt{3}}

\displaystyle \Rightarrow x = \frac{-2\sqrt{3}}{3} \text{ and } x = -3\sqrt{3}

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Question 11: Witt out solving, examine the nature of the roots of the equations :

(i) 5x^2 - 6x +7 = 0      (ii) x^2+6x+9=0      (iii) 2x^2+6x+3=0

Answer:

\displaystyle \text{(i) Comparing } 5x^2 - 6x +7 = 0 \text{ with } ax^2 + bx + c = 0, \\ \\ \text{ we get: } a = 5, b = -6  \text{ and } c= 7

Therefore Discriminant = b^2 - 4ac = (-6)^2 - 4 (5)(7) = 36 - 140 = - 104;  which is negative.

\displaystyle \text{Since } a, b,  \text{ and }  c \text{ are real numbers, }  a \neq 0 \text{ and } b^2 - 4ac < 0 

Therefore the roots are not real i.e the roots are imaginary.

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\displaystyle \text{(ii) Comparing } x^2+6x+9=0 \text{ with } ax^2 + bx + c = 0, \\ \\ \text{ we get: } a = 1, b = 6  \text{ and } c= 9

Therefore Discriminant = b^2 - 4ac = (6)^2 - 4 (1)(9) = 36-36 = 0; 

\displaystyle \text{Since } a, b,  \text{ and }  c \text{ are real numbers, }  a \neq 0 \text{ and } b^2 - 4ac = 0 

Therefore the roots are equal.

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\displaystyle \text{(iii) Comparing } 2x^2+6x+3=0 \text{ with } ax^2 + bx + c = 0, \\ \\ \text{ we get: } a = 2, b = 6  \text{ and } c= 3

Therefore Discriminant = b^2 - 4ac = (6)^2 - 4 (2)(3) = 36-24 = 12;  which is positive.

\displaystyle \text{Since } a, b,  \text{ and }  c \text{ are real numbers, }  a \neq 0 \text{ and } b^2 - 4ac > 0 

Therefore the roots are real and unequal.

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Question 12: Find the value of 'p' is the roots of the following quadratic equation are equal: (p-3)x^2 + 6x + 9 = 0

Answer:

\displaystyle \text{Comparing } (p-3)x^2 + 6x + 9 = 0 \text{ with } ax^2 + bx + c = 0, \\ \\ \text{ we get: } a = (p-3), b = 6  \text{ and } c= 9

Since the roots are equal, the discriminant is 0

\therefore b^2 - 4ac = 0

\Rightarrow (6)^2 - 4 ( p-3) (9) = 0

\Rightarrow 36 - 36 p + 108 = 0

\Rightarrow p = 4

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Question 13: Find the value of 'm' is the roots of the following quadratic equation are equal: (4+m)x^2 + ( m+1)x + 1 = 0

Answer:

\displaystyle \text{Comparing } (4+m)x^2 + ( m+1)x + 1 = 0 \text{ with } ax^2 + bx + c = 0, \\ \\ \text{ we get: } a = (4+m), b = (m+1)  \text{ and } c= 1

Since the roots are equal, the discriminant is 0

\therefore b^2 - 4ac = 0

\Rightarrow (m+1)^2 - 4 ( 4+m) (1) = 0

\Rightarrow m^2 + 2m + 1 - 16 - 4m = 0

\Rightarrow m^2 - 2m - 15 = 0

\Rightarrow (m-5)(m+3) = 0

Therefore m = 5     or    m = - 3

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Question 14: Solve each of the following equations for x and give, in each case, your answer correct to 2 decimal places.

(i) x^2 - 10x + 6= 0      (ii) 3x^2+5x-9=0 

Answer:

\displaystyle \text{(i) Comparing } x^2 - 10x + 6= 0 \text{ with } ax^2 + bx + c = 0, \\ \\ \text{ we get: } a = 1, b = -10  \text{ and } c= 6

\therefore b^2 - 4ac = (-10)^2 - 4 (1)(6) = 100 - 24 = 76

\Rightarrow \sqrt{b^2-4ac} = \sqrt{76} = 8.718

\displaystyle \therefore x = \frac{10 \pm 8.718}{2(1)}

\displaystyle \Rightarrow x = 9.359  \text{ and } x = 0.641

or correct to 2 decimal places:  \displaystyle x = 9.36    \text{ and } x = 0.64

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\displaystyle \text{(i) Comparing } 3x^2+5x-9=0 \text{ with } ax^2 + bx + c = 0, \\ \\ \text{ we get: } a = 3, b = 5  \text{ and } c= -9

\therefore b^2 - 4ac = (5)^2 - 4 (5)(-9) = 25+108 = 133

\Rightarrow \sqrt{b^2-4ac} = \sqrt{133} = 11.533

\displaystyle \therefore x = \frac{-5 \pm 11.533}{2(3)}

\displaystyle \Rightarrow x = 1.089  \text{ and } x =-2.756

or correct to 2 decimal places:  \displaystyle x = 1.09    \text{ and } x = -2.76

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\displaystyle \text{Question 15: Solve the following equation: } x - \frac{18}{x} = 6.  

Give your answer correct to two significant figures. [ICSE Board 2011]

Answer:

\displaystyle x - \frac{18}{x} = 6 \hspace{1.0cm} \Rightarrow x^2 - 18 = 6x \hspace{1.0cm} \Rightarrow x^2 - 6x - 18 = 0

\displaystyle \text{Comparing } x^2 - 6x - 18 = 0 \text{ with } ax^2 + bx + c = 0, \\ \\ \text{ we get: } a = 1, b = -6  \text{ and } c= -18

\displaystyle \text{We know } x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Substituting the values we get

\displaystyle x = \frac{6 \pm \sqrt{(6)^2 - 4(1)(-18)}}{2(1)} = \frac{6 \pm 10.392}{2} 

\displaystyle \Rightarrow x = 8.196 \text{ and } x = -2.196

or correct to 2 decimal places:  \displaystyle x = 8.2    \text{ and } x = -2.2

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Question 16: Solve: 

(i) 2x^4- 5x^2 + 3 = 0      (ii) (x^2+3x)^2 - ( x^2 + 3x) - 6 = 0 , x \in R    

Answer:

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

Let \displaystyle x^2 = y. Substituting:

\displaystyle 2y^2 - 5y + 3 = 0

\displaystyle \Rightarrow (y-1)(2y-3) = 0

\displaystyle \Rightarrow y = 1 \text{ and } y = \frac{3}{2}

\displaystyle \text{When } y = 1 \text{ then } x^2 = 1 \hspace{1.0cm}\Rightarrow x = \pm 1

\displaystyle \text{When } y = \frac{3}{2} \text{ then } x^2 = \frac{3}{2} \hspace{1.0cm}\Rightarrow x = \pm \sqrt{ \frac{3}{2} }

\displaystyle \text{Therefore the required solution is } 1, -1, \sqrt{ \frac{3}{2} }, - \sqrt{ \frac{3}{2} }

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(ii) (x^2+3x)^2 - ( x^2 + 3x) - 6 = 0

Let \displaystyle x^2+3x = y. Substituting:

\displaystyle y^2 - y -6 = 0

\displaystyle \Rightarrow (y-3)(2y+2) = 0

\displaystyle \Rightarrow y = 3 \text{ and } y = -2

\displaystyle \text{When } y = 3 \text{ then } x^2+3x = 3 \hspace{1.0cm}\Rightarrow x^2 + 3x - 3 =0

\displaystyle x = \frac{-3 \pm \sqrt{(3)^2 - 4(1)(-3)}}{2(1)} = \frac{-3 \pm \sqrt{21}}{2}

\displaystyle \text{When } y = -2 \text{ then } x^2+3x = -2 \hspace{1.0cm}\Rightarrow x^2 + 3x +2 =0

\displaystyle x = \frac{-3 \pm \sqrt{(3)^2 - 4(1)(2)}}{2(1)} = \frac{-3 \pm 1}{2} =-1  \text{  or  }  -2

\displaystyle \text{Therefore the required solution is } : \frac{-3 + \sqrt{21}}{2}, \frac{-3 - \sqrt{21}}{2} , -1 , -2

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\displaystyle \text{Question 17: Solve} : \sqrt{\frac{x}{1-x}} + \sqrt{\frac{1-x}{x}} = 2 \frac{1}{6}, x \neq 0 \text{ and } x \neq 1

Answer:

\displaystyle \text{Let } \sqrt{\frac{x}{1-x}} = y

\displaystyle \Rightarrow \sqrt{\frac{1-x}{x}} = \frac{1}{y}

Therefore Given equation reduces to:

\displaystyle y + \frac{1}{y} = \frac{13}{6}

\displaystyle \Rightarrow 6y^2 + 6 = 13y

\displaystyle \Rightarrow 6y^2 - 13y + 6 = 0

\displaystyle \Rightarrow ( 2y-3)(3y-2) = 0

\displaystyle \Rightarrow y = \frac{3}{2}  , \text{  or  } y = \frac{2}{3}

\displaystyle \text{When } y = \frac{3}{2} \hspace{1.0cm}\Rightarrow \sqrt{\frac{x}{1-x}} = \frac{3}{2} \hspace{1.0cm}\Rightarrow \frac{x}{1-x} = \frac{9}{4}

\displaystyle\Rightarrow 4x = 9 - 9x \hspace{1.0cm}\Rightarrow x = \frac{9}{13}

\displaystyle \text{When } y = \frac{2}{3} \hspace{1.0cm}\Rightarrow \sqrt{\frac{x}{1-x}} = \frac{2}{3} \hspace{1.0cm}\Rightarrow \frac{x}{1-x} = \frac{4}{9}

\displaystyle \Rightarrow 9x = 4 - 4x \hspace{1.0cm}\Rightarrow x = \frac{4}{13}

\displaystyle \text{Therefore the required solution is } : \frac{9}{13}, \frac{4}{13}

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Question 18: Find the solution set of the equation 3x^2 - 8x - 3 = 0; when 

(i) x \in Z (integers)     (ii) x \in Q ( Rational Numbers)

Answer:

\displaystyle 3x^2 - 8x - 3 = 0

\displaystyle \Rightarrow 3x^2 - 9x + x - 3 = 0

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

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

\displaystyle \Rightarrow x = 3  \text{  or  } x = \frac{-1}{3}

\displaystyle \text{(i) when } x \in Z, \text{ the solution set } = \{ 3 \}

\displaystyle \text{(ii) when } x \in Q, \text{ the solution set } = \{ 3, -\frac{1}{3} \}

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Question 19: Solve: ( 2x-3)^2 = 25

Answer:

( 2x-3)^2 = 25

\Rightarrow 4x^2 - 12 x + 9 - 25 = 0

\Rightarrow 4x^2 - 12 x - 16 = 0

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

\Rightarrow (x-4)(x+1) = 0

\Rightarrow x = 4 \text{  or  } x = -1

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\displaystyle \text{Question 20: Solve for x } : 4 \Big( x - \frac{1}{x} \Big)^2 + 8 \Big(x + \frac{1}{x} \Big) = 29 , x \neq 0

Answer:

\displaystyle \text{Let } x + \frac{1}{x} = y

\displaystyle \Big( x - \frac{1}{x} \Big)^2 = \Big( x + \frac{1}{x} \Big)^2 - 4 = y^2 - 4

\displaystyle 4 \Big( x - \frac{1}{x} \Big)^2 + 8 \Big(x + \frac{1}{x} \Big) = 29

\displaystyle \Rightarrow 4(y^2 -4) + 8y = 29

\displaystyle \Rightarrow  4y^2 - 16 + 8y = 29

\displaystyle \Rightarrow  4y^2 + 8y - 45 = 0

\displaystyle \Rightarrow  ( 2y+9)(2y-5) = 0

\displaystyle \Rightarrow  y = \frac{-9}{2} \text{  or  } y = \frac{5}{2}

\displaystyle \text{When } y = \frac{-9}{2} \hspace{1.0cm} \Rightarrow x + \frac{1}{x} = \frac{-9}{2} \Rightarrow2x^2 + 9x + 2 = 0

\displaystyle x = \frac{-9 \pm \sqrt{(9)^2 - 4(2)(2)}}{2(2)} = \frac{-9 \pm \sqrt{65}}{4}

\displaystyle \text{When } y = \frac{5}{2} \hspace{0.5cm} \Rightarrow x + \frac{1}{x} = \frac{5}{2} \hspace{0.5cm} \Rightarrow2x^2 - 5x + 2 = 0  \hspace{0.5cm} \Rightarrow (x-2)(2x-1) = 0 \\ \\ \Rightarrow x = 2 \text{  or  } x = \frac{1}{2}

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\displaystyle \text{Question 21: Solve}: \frac{a}{ax-1} + \frac{b}{bx-1} = a + b , \text{ where } a + b \neq 0, ab \neq 0

Answer:

\displaystyle \frac{a}{ax-1} + \frac{b}{bx-1} = a + b

\displaystyle \Rightarrow \frac{a}{ax-1} - b + \frac{b}{bx-1} - a = 0

\displaystyle \Rightarrow \frac{a - abx + b}{ax-1} + \frac{b - abx + a}{bx-1} = 0

\displaystyle \Rightarrow ( a + b - abx) \Big( \frac{1}{ax-1} + \frac{1}{bx-1} \Big) = 0

\displaystyle \Rightarrow a+b- abx = 0 \ \ \ \ \ \text{  or  } \ \ \ \ \ \frac{1}{ax-1} + \frac{1}{bx-1} = 0

\displaystyle \Rightarrow -abx = - a- b \ \ \ \ \ \text{  or  } \ \ \ \ \ \frac{1}{ax-1} = -\frac{1}{bx-1}

\displaystyle \Rightarrow abx = a+b \ \ \ \ \ \text{  or  } \ \ \ \ \ bx-1 = -ax + 1

\displaystyle \Rightarrow x = \frac{a+b}{ab} \ \ \ \ \ \text{  or  } \ \ \ \ \ x = \frac{2}{a+b}