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

Question

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

✓

Answer

We have been given,
$\frac{16}{\text{x}}-1=\frac{15}{\text{x}+1},$ $\text{x}\neq0,-1$
Now we solve the above equation as follows,
$\Rightarrow (16 - x)(x + 1) = 15x$
$\Rightarrow 16x + 16 - x^2 - x = 15x$
$\Rightarrow 15x + 16 - x^2 - 15x = 0$
$\Rightarrow 16 - x^2 = 0$
$\Rightarrow x^2 - 16 = 0$
Now we also know that for an equation $ax^2 + bx + c = 0$, the discriminant is given by the following equation,
$D = b^2 - 4ac$
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,
$\text{x} = {-\text{b} \pm \sqrt{\text{D}} \over 2\text{a}}$
Therefore, the roots of the equation are given as follows,
$\text{x}=\frac{-0\pm\sqrt{64}}{2(1)}$
$\text{x}=\frac{\pm8}{2}$
$\text{x}=\pm4$
Therefore, the value of $\text{x}=\pm4$

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