Question

Find the roots if the following equation: 1\x+4 - 1\x-7=11\30, x is not equal to -4,7.

Answer 1

Answer:

1 Or 2 roots of given Quadratic equation.

Explanation:

Given

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

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

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

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

Do the cross multiplication, we get

=> 30 = 11(x²-3x-28)/(-11)

=> 30 = -(x²-3x-28)

=> x²-3x-28+30 =0

=> x²-3x+2=0

Splitting the middle term, we get

=> x²-1x-2x+2 =0

=> x(x-1)-2(x-1)=0

=> (x-1)(x-2)=0

=>x-1 = 0 or x-2 = 0

=> x = 1 or x = 2

Therefore,

Roots of the given quadratic equation are 1 , 2

•••

Answer 2

Answer:

The values of x are 1 and 2 hence it must not be equal to -4 and 7. Thus, the assumption is true.

Solution:

Given expression is,

$\begin{array} { c } { \frac { 1 } { x + 4 } - \frac { 1 } { x - 7 } = \frac { 11 } { 30 } } \\\\ { \frac { ( x - 7 ) - ( x + 4 ) } { ( x + 4 ) ( x - 7 ) } = \frac { 11 } { 30 } } \\\\ { \frac { x - 7 - x - 4 } { ( x + 4 ) ( x - 7 ) } = \frac { 11 } { 30 } } \end{array}$

$\begin{array} { c } { \frac { - 11 } { x ^ { 2 } + 4 x - 7 x - 28 } = \frac { 11 } { 30 } } \\\\ { - 30 = \left( x ^ { 2 } - 3 x - 28 \right) } \\\\ { x ^ { 2 } - 3 x - 28 + 30 = 0 } \\\\ { x ^ { 2 } - 3 x + 2 = 0 } \end{array}$

$\begin{array} { c } { x ^ { 2 } - 2 x - x + 2 = 0 } \\\\ { x ( x - 2 ) - 1 ( x - 2 ) = 0 } \\\\ { ( x - 1 ) ( x - 2 ) = 0 } \end{array}$

Either  

$(x-1)=0\\\\x=1$

Or,

$(x-2)=0\\\\x=2$

Therefore, the values of x are 1 and 2 respectively which must not be equal to -4 and 7. Thus, the given assumption is true.