Question

Can anyone plz solve this quadratic equation.​

Answer 1

We have,

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

Standardising the equation,

$\implies \: \sf{ \frac{2x + 2}{x {}^{2} + 2x} = \frac{3}{4} } \\ \\ \implies \: \sf{8x + 8 = 3x {}^{2} + 6x } \\ \\ \implies \: \huge{\boxed{\sf{3x {}^{2} - 2x - 8 = 0}}}$

We obtain an equation,

3x² - 2x -8=0

By splitting the middle term,

→3x² -6x +4x -8=0

→3x(x-2)+4(x-2)=0

→(3x+4)(x-2)=0

→x= -4/3 or x=2

The roots of the equation are 2 and -4/3

Answer 2

$\huge\bf{Answer:-}$

$\begin{lgathered}\bf{ \frac{1}{x + 2} + \frac{1}{x} = \frac{3}{4}}\end{lgathered}$

$\begin{lgathered}\bf{ \frac{2x + 2}{x {}^{2} + 2x} = \frac{3}{4} }$

$=> \: {8x + 8 = 3x {}^{2} + 6x }$ $=>\bf{\frac{3x {}^{2} - 2x - 8 = 0}}}\end{lgathered}$

=> 3x² - 2x -8=0

Middle term

=3x² -6x +4x -8=0

=3x(x-2)+4(x-2)=0

=(3x+4)(x-2)=0

=x= -4/3 or x=2

2 and -4/3