Question

Find the root of the following equations :

$\frac{1}{x + 4} - \frac{1}{x - 7} = \frac{11}{30}$
x is not equal to -4 , 7

Find the root by using Quadratic formula

​

Answer 1

Please mark me brainiest

Answer 2

${\underline { \huge{ \bold \red{ Solution:-}}}}$

${ \bold{ \frac{1}{x + 4} - \frac{1}{x - 7} = \frac{11}{30}}}$

▪ taking LCM

${ \bold{ \implies{ \frac{(x - 7) - (x + 4)}{(x + 4)(x - 7)} = \frac{11}{30} }}}$

${ \bold{ \implies{ \frac{x - 7 - x - 4}{ {x}^{2} - 7x + 4x - 28} = \frac{11}{30} }}}$

${ \bold{ \implies{ \frac{ - 11}{ {x}^{2} - 3x - 28} = \frac{11}{30} }}}$

${ \bold{ \implies{ \frac{ - 1}{ {x}^{2} - 3x - 28} = \frac{1}{30} }}}$

▪ cross multiplying both the sides

${ \bold{ \implies{ {x}^{2} - 3x - 28 = - 30}}}$

${ \bold{ \implies{ {x}^{2} - 3x - 28 + 30 = 0}}}$

${ \bold{ \implies{ {x}^{2} - 3x + 2 = 0}}}$

${ \underline{ \bold{ \orange{quadratic \: formula}}}}$

${ \boxed{ \bold{ \red{x = \frac{ - b + - \sqrt{ {b}^{2} - 4ac} }{2a}}}}}$

in the equation--->

${ \bold{ {x}^{2} - 3x + 2 = 0}}$

${ \bold{ a = 1}} \\ { \bold{ b = - 3}} \\ { \bold{c = 2}}$

☆ putting the values of a, b and c in the quadratic formula......

${ \bold{x = \frac{ - ( - 3) + - \sqrt{ { - 3}^{2} - 4 \times 1 \times 2 } }{2 \times 1} }}$

${ \bold{ \implies{x = \frac{3 + - \sqrt{9 - 8} }{2} }}}$

${ \bold{ \implies{x = \frac{3 + - \sqrt{1} }{2} }}}$

● first root value of x

${ \bold{ \rightarrow{x = \frac{3 + \sqrt{1} }{2} }}}$

${ \bold{ \implies{x = \frac{3 + 1}{2} = \frac{4}{2} }}}$

${ \boxed{ \bold{ \red{ \: \: \: x = 2 \: \: \: }}}}$

● second root value of x

${ \bold{ \rightarrow{x = \frac{3 - \sqrt{1} }{2} }}}$

${ \bold{ \implies{x = \frac{3 - 1}{2} = \frac{2}{2} }}}$

${ \boxed{ \bold{ \red{ \: \: x = 1 \: \: }}}}$

therefore,

${ \bold{ \huge{ \orange{x = 1 \: and \: 2}}}}$