Question
Solve the following quadratic equations by factorization:
$\frac{16}{\text{x}}-1=\frac{15}{\text{x}+1},$ $\text{x}\neq0,-1$
$\frac{16}{\text{x}}-1=\frac{15}{\text{x}+1},$ $\text{x}\neq0,-1$
✓
Answer
$\frac{16}{\text{x}}-1=\frac{15}{\text{x}+1}$
$\Rightarrow\frac{16-\text{x}}{\text{x}}-\frac{15}{\text{x}+1}$
$ \Rightarrow 16 x+16-x^2-x=15 x $
$ \Rightarrow-x^2+16+15 x=15 x $
$ \Rightarrow-x^2+16=0 $
$ \Rightarrow x^2-16=0 $
$ \Rightarrow(x-4)(x+4)=0 $
$ \Rightarrow x-4=0 \text { or } x+4=0 $
$ \Rightarrow x=4 \text { or } x=-4$
Hence, the factors are 4 and -4
$\Rightarrow\frac{16-\text{x}}{\text{x}}-\frac{15}{\text{x}+1}$
$ \Rightarrow 16 x+16-x^2-x=15 x $
$ \Rightarrow-x^2+16+15 x=15 x $
$ \Rightarrow-x^2+16=0 $
$ \Rightarrow x^2-16=0 $
$ \Rightarrow(x-4)(x+4)=0 $
$ \Rightarrow x-4=0 \text { or } x+4=0 $
$ \Rightarrow x=4 \text { or } x=-4$
Hence, the factors are 4 and -4
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