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

Question

Solve for x.
$\frac{16}{\text{x}}-1=\frac{15}{\text{x}+1},$ $\text{x}\neq0,-1$

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Answer

We have been given,
$\frac{16}{x}-1=\frac{15}{x+1}, x \neq 0,-1$
Now we solve the above equation as follows,
$\Rightarrow(16-x)(x+1)=15 x$
$\Rightarrow 16 x+16-x^2-x=15 x$
$\Rightarrow 15 x+16-x^2-15 x=0$
$\Rightarrow 16-x^2=0$
$\Rightarrow x^2-16=0$
Now we also know that for an equation $a x^2+b x+c=0$, the discriminant is given by the following equation,
$D=b^2-4 a c$
Now, according to the equation given to us we have, $a=1, b=0$ and $c=-16$
Therefore, the discriminant is given as,
$D=(0)^2-4(1)(-16)$
$D=64$
Now, the roots of an equation is given by the following equation,
$x=\frac{-b \pm \sqrt{D}}{2 a}$
Therefore, the roots of the equation are given as follows,
$x=\frac{-0 \pm \sqrt{64}}{2(1)}$
$x=\frac{ \pm 8}{2}$
$x= \pm 4$
Therefore, the value of $x= \pm 4$

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