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

Given:

Given quadratic equation is $\frac{16}{x}-1=\frac{15}{x+1}, x≠0, -1$.

To do:

We have to solve the given quadratic equation.

Solution:

$\frac{16}{x}-1=\frac{15}{x+1}, x≠0, -1$

$\frac{16}{x}-1=\frac{15}{x+1}$

$\frac{16-x}{x}=\frac{15}{x+1}$

$(16-x)(x+1)=15(x)$   (On cross multiplication)

$16x+16-x^2-x=15x$

$x^2+(15-15)x-16=0$

$x^2-16=0$

Comparing the given quadratic equation with the standard form of the quadratic equation $ax^2+bx+c=0$, we get,

$a=1, b=0$ and $c=-16$.

Therefore, the roots of the given equation are

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-(0)\pm \sqrt{(0)^2-4(1)(-16)}}{2(1)}$ 

$x=\frac{0\pm \sqrt{0+64}}{2}$ 

$x=\frac{\pm \sqrt{64}}{2}$ 

$x=\frac{\pm 8)}{2}$ 

$x=\frac{8}{2}$ or $x=\frac{-8}{2}$

$x=4$ or $x=-4$

The values of $x$ are $-4$ and $4$.

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Updated on: 10-Oct-2022

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