Question

if x is equal to root 7 plus root 3 , show that x - 4/x = 2 root 3​

Answer 1

Answer:

answer in attachment dear

Answer 2

Aim:-

We know that, $x=\sqrt{7}+\sqrt{3}$. With the given information, we need to prove that $x-\frac{4}{x}=2\sqrt{3}$.

Procedure of proving:-

• We know that, any fraction with an irrational number in the denominator needs to be rationalised. So, $\frac{4}{x}$ needs to be rationalised as $x$ is an irrational number.
• After rationalising, the value of  $x-\frac{4}{x}$ is found and is proved to be equal to $2\sqrt{3}$.

Proof:-

Given that, $x=\sqrt{7}+\sqrt{3}$

So, $\frac{4}{x}=\frac{4}{\sqrt{7} +\sqrt{3} }$

On rationalising $\frac{4}{\sqrt{7} +\sqrt{3} }$,

⇒ $\frac{4}{\sqrt{7} +\sqrt{3} }\times\frac{\sqrt{7} -\sqrt{3} }{\sqrt{7} -\sqrt{3} }$

⇒ $\frac{4(\sqrt{7} -\sqrt{3} )}{(\sqrt{7}+\sqrt{3})(\sqrt{7}-\sqrt{3}) }$

⇒ $\frac{4(\sqrt{7} -\sqrt{3} )}{(\sqrt{7})^2-(\sqrt{3})^2 }$ $($∵ $(a+b)(a-b)=(a)^2-(b)^2)$

⇒  $\frac{4(\sqrt{7} -\sqrt{3} )}{7-3 }$

⇒ $\frac{4(\sqrt{7} -\sqrt{3} )}{4 }$

Therefore,  $\boxed{\frac{4}{x} =\sqrt{7}-\sqrt{3}}$

$x-\frac{4}{x}=2\sqrt{3}$

$(\sqrt{7}+\sqrt{3})-(\sqrt{7}-\sqrt{3})=2\sqrt{3}$

$\sqrt{7}+\sqrt{3}-\sqrt{7}+\sqrt{3}=2\sqrt{3}$

So, $2\sqrt{3} =2\sqrt{3}$

Therefore, $LHS=RHS$

Conclusion:-

Therefore, $x-\frac{4}{x} =2\sqrt{3}$ is proved.