Question

find the roots of x-1/x+2 +x-3/x-2= 11/8


by using quadratic formula​

Answer 1

Find the roots of the given expression -

$\sf{ \dfrac{ x - 1 }{ x + 2 } + \dfrac{ x - 3 }{ x - 2 } = \dfrac{ 11 }{ 8 } }$

Solution -

$\sf{ \dfrac{ x - 1 }{ x + 2 } + \dfrac{ x - 3 }{ x - 2 } = \dfrac{ 11 }{ 8 } } \\ \\ \sf{ \implies { \dfrac{ (x - 1 )( x - 2 ) + ( x + 2 )( x - 3 ) }{ ( x + 2 )( x - 2 ) } = \dfrac{ 11 }{ 8 } }} \\ \\ \sf{ \leadsto { \dfrac{ x^2 - 3x + 2 + x^2 - x - 6 }{ x^2 - 4 } = \dfrac{ 11 }{ 8 } }} \\ \\ \sf{ \mapsto { \dfrac{ 2x^2 - 4x - 4 }{ x ^ 2 - 4 } = \dfrac{ 11 }{ 8 } }} \\ \\ \sf{ \implies { \dfrac { x^2 - 2x - 2 }{ x^2 - 4 } = \dfrac{ 11 }{ 16 } }} \\ \\ \sf{ \leadsto { 11 ( x^2 - 4 ) = 16 ( x^2 - 2x - 2 ) }} \\ \\ \sf{ \mapsto { 11x^2 - 44 = 16x^2 - 32x - 32 }} \\ \\ \sf{ \implies { 5x^2 - 32x + 12 = 0 }} \\ \\ \sf{ \bold { Quadratic \: Formula \: - }} \\ \\ \sf{ \implies { x = \dfrac{ -b \pm \sqrt { b^2 - 4ac } }{ 2a } }} \\ \\ \sf {\bold { Substituting \: the \: given \: values \: - }} \\ \\ \sf{ \implies { x = \dfrac{ 32 \pm \sqrt{ 784 } }{ 10 } }} \\ \\ \sf{ \implies { x = \dfrac{ 32 \pm 28 }{ 10 } }} \\ \\ \sf{ \mapsto { x = 6 \: or \: 0.4 }} \\ \\ \sf{ \bold { This \: is \: the \: required \: answer . }}$

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Answer 2

Answer:

$x = 6 \: or \: x = \frac{2}{5}$

Given:

$\frac{x - 1}{x + 2} + \frac{x - 3}{x - 2} = \frac{11}{8}$

Explanation:

Taking the LCM of LHS, we get:

$\frac{(x - 1) \times (x - 2) + (x - 3) \times (x + 2)}{(x - 2) \times (x + 2)} = \frac{11}{8}$

$We \: know \: that \: (a + b) \times (a - b) \: gives \: {a}^{2} - {b}^{2}$

After multiplying everything, we get:

$\frac{ {x}^{2} - 2x- x+ 2 + {x}^{2} - 3x + 2x - 6 }{ {x}^{2} - 4 } = \frac{11}{8}$

Cancelling 2x and adding every like term, we get:

$\frac{2 {x}^{2} - 4x - 4 }{ {x}^{2} - 4 } = \frac{11}{8}$

Cross multiplying LHS and RHS, we get:

$16 {x}^{2} - 32x - 32 = 11 {x}^{2} - 44$

Bringing RHS to LHS, we get:

$5 {x}^{2} - 32x + 12 = 0$

Using Quadratic Formula or Sridharacharya Method:

$x = \frac{ - b \: \pm \: \sqrt{ {b}^{2} - 4ac} }{2a}$

$where \: a \: , \: b \: and \: c \: are \: three \: constants \: in \: the \: equation \:$

$a {x}^{2} + bx + c = 0$

$Here, \:$

$a=5 \: \: \: \: b=-32 \: \: \: \: </p><p>c=12 \: </p><p>$

Substituting the value of a, b and c in formula, we get:

$x = \frac{ - ( - 32) \: \pm \: \sqrt{( - 32 )^{2} - 4 \times 5 \times 12} }{2 \times 5}$

$x = \frac{32 \: \pm \: \sqrt{1024 - 240} }{10}$

$x = \frac{32 \: \pm \: \sqrt{784} }{10}$

$x = \frac{32 \: \pm \: 28}{10}$

$x = \frac{32 + 28}{10} \: or \: x = \frac{32 - 28}{10}$

$x = \frac{60}{10} \: or \: x = \frac{4}{10}$

$x = 6 \: or \: x = \frac{2}{5}$

$x = 6 \: or \: x = 0.4$

Additional Information:

In some questions, the square root is not perfect. So we will solve it by division method.

In the quadratic formula, we place the value of a, b and c as they are given in the quadratic equation.

If in the question, the value of x cannot be negative, so we neglect the negative value of x.

We can also use other methods to solve the equation.