Solve the following quadratic equation by factorization:
$\frac{16}{x}\ –\ 1\ =\ \frac{15}{(x\ +\ 1)},\ x\ ≠\ 0,\ -1$

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 by factorization. 

Solution:

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

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

$(x+1)(16-x)=15(x)$

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

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

$x^2-16=0$

$(x+4)(x-4)=0$

$x+4=0$ or $x-4=0$

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

The roots of the given quadratic equation are $-4$ and $4$. 

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Updated on: 10-Oct-2022

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