Question

Solve the following Quadratic Equation [Using Factorisation method]
$\\ \\ \: \: \: \: \: \: \boxed{\orange{ \bold {\tt \dfrac{1}{2a + b + 2x} = \dfrac{1}{2a} + \dfrac{1}{b} + \dfrac{1}{2x} }}}$
​

Answer 1

Required Answer:-

Given Equation:

$\dag\:\:\boxed{\tt \dfrac{1}{2a + b + 2x} = \dfrac{1}{2a} + \dfrac{1}{b} + \dfrac{1}{2x} }$

To Find:

• The values of x.

Solution:

Given,

$\tt \implies\dfrac{1}{2a + b + 2x} = \dfrac{1}{2a} + \dfrac{1}{b} + \dfrac{1}{2x}$

Moving 1/2x to left side, we get,

$\tt \implies\dfrac{1}{2a + b + 2x} - \dfrac{1}{2x} = \dfrac{1}{2a} + \dfrac{1}{b}$

$\tt \implies\dfrac{2x - (2a + b + 2x)}{2x(2a + b + 2x)} = \dfrac{1}{2a} + \dfrac{1}{b}$

$\tt \implies\dfrac{2x - 2a - b - 2x}{2x(2a + b + 2x)} = \dfrac{1}{2a} + \dfrac{1}{b}$

$\tt \implies\dfrac{2x - 2a - b - 2x}{2x(2a + b + 2x)} = \dfrac{1}{2a} + \dfrac{1}{b}$

$\tt \implies\dfrac{ - 2a - b }{2x(2a + b + 2x)} = \dfrac{2a + b}{2ab}$

$\tt \implies\dfrac{ -1(2a + b) }{2x(2a + b + 2x)} = \dfrac{2a + b}{2ab}$

$\tt \implies\dfrac{ -1}{2x(2a + b + 2x)} = \dfrac{1}{2ab}$

On transposing, we get,

$\tt \implies 2x(2a + b + 2x) = - 2ab$

$\tt \implies x(2a + b + 2x) = - ab$

$\tt \implies 2 {x}^{2} + (2a + b) x + ab = 0$

$\tt \implies 2 {x}^{2} + 2ax + bx + ab = 0$

$\tt \implies 2x(x +a)+ b(x + a) = 0$

$\tt \implies (2x + b)(x + a) = 0$

$\tt \implies Either \: (2x + b) = 0 \: or \: (x + a) = 0$

So,

$\tt \implies x = \dfrac{ - b}{2} , - a$

★ Hence, the values of x are - (-b/2) and (-a).

Answer:

• x = -b/2, -a.

•••♪

Answer 2

Question:-

Solve the following Quadratic Equation [Using Factorisation method]

$\: \\ \\{ \bold {\sf \dfrac{1}{2a + b + 2x} = \dfrac{1}{2a} + \dfrac{1}{b} + \dfrac{1}{2x} }}$

Required Answer:-

Given Equation:-

$\: \\ \\ {\sf \dfrac{1}{2a + b + 2x} = \dfrac{1}{2a} + \dfrac{1}{b} + \dfrac{1}{2x} }$

To Find:-

$\\ \sf{ \rightarrow{values \: of \: \bold{x}}}$

♡Solution:-

$\: \\ \\{\sf{ \bold{ \dfrac{1}{2a + b + 2x} = \dfrac{1}{2a} + \dfrac{1}{b} + \dfrac{1}{2x} }}}$

$\\ \sf{ \implies{ \frac{1}{2a + b + 2x} - \frac{1}{2x} = \frac{1}{2a} + \frac{1}{b} }}$

$\\ \sf{ \implies{ \frac{2x - (2a + b + 2x)}{2x(2a + b + 2x)} = \frac{1}{2a} }} + \frac{1}{b}$

$\\ \sf{ \implies{ \frac{2x - 2a - b - 2x}{2x(2a + 2b + 2x)} = \frac{b + 2a}{2ab} }}$

$\\ \sf{ \implies{ \frac{ - 2a - b}{2x(2a + b + 2x)} = \frac{2a + b}{2ab} }}$

$\\ \sf{ \implies{ \frac{ -( 2a + b)}{2x(2a + b + 2x)} = \frac{2a + b}{2ab} }}$

$\\ \sf{ \implies{ \frac{ - 1}{2x(2a + b + 2x)} = \frac{1}{2ab} }} \: \: \: \: \: \: \: \: \boxed{ \rm{dividing \: by \: (2a + b)}}$

$\\ \sf{ \implies{2x(2a + b + 2x) = - 2ab}}$

$\\\sf{ \implies{x(2a + b + 2x) = - ab}}$

$\\ \sf{ \implies{ 2ax + bx + 2 {x}^{2} = - ab}}$

$\\ \sf{ \implies{2 {x}^{2} + 2ax + bx + ab = 0}}$

$\\ \sf{ \implies{2x(x + a) + b(x + a) = 0}}$

$\\ \sf{ \implies{ \bold{(x + a)(2x + b) = 0}}}$

Now, Either,

$\\ \bold{x + a = 0}$

$\\ \sf{ \rightarrow{ \red{x = - a}}}$

Or,

$\\ \bold{2x + b = 0}$

$\\ \sf{ \rightarrow{2x = - b}}$

$\\ \sf{ \rightarrow{ \red{x = \frac{ - b}{x} }}}$

$\\ \\ \underline{ \boxed{ \rm{ \green{Hence, \: x = \bold{ - a, \: \frac{ - b}{2} }}}}}$