Solve for x :16 x -1=15 x+1 ; x 0,-1 - Vidyadip

Question

Solve for $x :\frac{16}{x}-1=\frac{15}{x+1} ; x \neq 0,-1$

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Answer

The given quadratic equation is
$\sqrt{3} x^2-2 \sqrt{2} x-2 \sqrt{3}=0$
So,
$a=\sqrt{3}, b=-2 \sqrt{2}, c=-2 \sqrt{3}$
The quadratic formula to find the root is
$x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$
$x=\frac{-(-2 \sqrt{2}) \pm \sqrt{(-2 \sqrt{2})^2-4 \times \sqrt{3} \times(-2 \sqrt{3})}}{2 \times \sqrt{3}}$
$x=\frac{2 \sqrt{2} \pm \sqrt{8+24}}{2 \sqrt{3}}$
$x=\frac{2 \sqrt{2} \pm \sqrt{32}}{2 \sqrt{3}}$
$x=\frac{2 \sqrt{2} \pm 4 \sqrt{2}}{2 \sqrt{3}}$
$x=\frac{2 \sqrt{2}+4 \sqrt{2}}{2 \sqrt{3}}, \frac{2 \sqrt{2}-4 \sqrt{2}}{2 \sqrt{3}}$
$x=\frac{6 \sqrt{2}}{2 \sqrt{3}}, \frac{-2 \sqrt{2}}{2 \sqrt{3}}$
$x=\frac{3 \sqrt{2}}{\sqrt{3}}, \frac{-\sqrt{2}}{\sqrt{3}}$

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