Question

if 4x =√5+2 then find the value of 4x-1/16x is​

Answer 1

The answer is $\frac{3\sqrt{5} + 10}{4}$

Step-by-step explanation:

Given that $4x = \sqrt{5} + 2$, then we have to calculate the value of $4x - \frac{1}{16x}$.

Now, $4x - \frac{1}{16x}$

= $(\sqrt{5} + 2 ) - \frac{1}{4(\sqrt{5} + 2 )}$

= $(\sqrt{5} + 2 ) - \frac{\sqrt{5} - 2}{4}$ { Rationalizing the denominator }

= $\frac{4\sqrt{5} + 8 - \sqrt{5} + 2}{4}$

= $\frac{3\sqrt{5} + 10}{4}$ ( Answer )

Answer 2

Given: If $4x=\sqrt{5}+2$

To Find: What is the value of $4x-\frac{1}{16x}$.

Step-by-step explanation:

Here,

        $4x=\sqrt{5}+2$

∴ $4x-\frac{1}{16x}$

$=4x-\frac{1}{4\times 4x}$

Plug the value of $4x$ in the above equation,

$=\sqrt{5}+2-\frac{1}{4\times (\sqrt{5}+2)}$

$=\sqrt{5}+2-\frac{(\sqrt{5}-2)}{4\times (\sqrt{5}+2)(\sqrt{5}-2)}$ (Multiply Numerator and Denominator with $(\sqrt{5}-2)$)      

$=\sqrt{5}+2-\frac{(\sqrt{5}-2)}{4\times ((\sqrt{5})^{2} -2^{2} )}$

$=\sqrt{5}+2-\frac{(\sqrt{5}-2)}{4\times (5-4)}$

$=\sqrt{5}+2-\frac{(\sqrt{5}-2)}{4}$

$=\frac{4\sqrt{5}+8- (\sqrt{5}-2)}{4}$

$=\frac{4\sqrt{5}+8- \sqrt{5}+2}{4}$

$=\frac{3\sqrt{5}+10}{4}$

Therefore, The value of $4x-\frac{1}{16x}$ is $\frac{3\sqrt{5}+10}{4}$.