Solve quadratic equation ^@\dfrac {x} { x + 1 } + \dfrac { x + 1 | Math Question and Answer | Edugain Brazil

Login/Register Edugain for School

Brazil

Solve quadratic equation $\dfrac {x} { x + 1 } + \dfrac { x + 1 } { x } = \dfrac { 25 } { 12 }$ ( $x \neq 0$ and $x \neq -1$ ).

Answer:

$x = 3$ , or $x = -4$

Step by Step Explanation:

On adding two fractions on LHS,
$\begin{align} &\dfrac { x^2 + (x + 1)^2 } { x(x + 1) } = \dfrac { 25 } { 12 } \\ \implies &\dfrac { x^2 + (x^2 + 2x + 1) } { x(x + 1) } = \dfrac { 25 } { 12 } \end{align}$
$\begin{align} & 12 (2x^2 + 2x + 1) = 25 (x^2 + x) \\ \implies & 24x^2 + 24x + 12 = 25x^2 + 25x \\ \implies & 25x^2 + 25x - 24x^2 - 24x - 12 = 0 \\ \implies & x^2 + x - 12 = 0 \\ \implies & x^2 + 4 x - 3 x - 12 = 0 \\ \implies & x(x + 4) - 3(x + 4) = 0 \\ \implies & (x - 3) (x + 4) \\ \implies & x = 3 \space or \space -4 \end{align}$
Hence, $x = 3$ , or $x = -4$ .

You can reuse this answer
Creative Commons License