Using the properties of proportion solve for x given x^4+1 2 x^2 … - Vidyadip

Question

Using the properties of proportion solve for x given $\frac{x^4+1}{2 x^2}=\frac{17}{8}$

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Answer

$\frac{x^4+1}{2 x^2}=\frac{17}{8}$
Applying componendo and dividendo we get
$\frac{x^4+1+2 x^2}{x^4+1-2 x^2}=\frac{17+8}{17-8}$
$ \Rightarrow \frac{\left(x^2\right)^2+(1)^2+2 \times x^2+1}{\left(x^2\right)^2+(1)^2-2 \times x^2 \times 1}=\frac{25}{9}$
$\Rightarrow \frac{\left(x^2+1\right)^2}{\left(x^2-1\right)^2}=\frac{5^2}{3^2} $
$ \Rightarrow\left(\frac{x^2+1}{x^2-1}\right)^2=\left(\frac{5}{3}\right)^2$
$\Rightarrow \frac{x^2+1}{x^2-1}=\frac{5}{3}$
Applying componeddo and diividendo we get
$\frac{x^2+1+x^2-1}{x^2+1-x^2+1}=\frac{5+3}{5-3} $
$ \Rightarrow \frac{2 x^2}{2}=\frac{8}{2}$
$ \Rightarrow x^2=4 $
$ \Rightarrow x= \pm 2$

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