Question

If x = root 3 + root2 upon root 3 -root 2 then find the value of x square

Answer 1

Answer:

x=root5 I think.Hope it helps you

Step-by-step explanation:

$x = \sqrt{3} + \sqrt{2} \div \sqrt{3} - \sqrt{2}$

Answer 2

The value of x² is $49+20\sqrt{6}$.

Step-by-step explanation:

It is given that

$x=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$

We need to find the value of x².

$x^2=x\cdot x=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times \dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$

$x^2=\dfrac{\left(\sqrt{3}+\sqrt{2}\right)^2}{\left(\sqrt{3}-\sqrt{2}\right)^2}$

$x^2=\dfrac{\left(\sqrt{3}\right)^2+2\sqrt{3}\sqrt{2}+\left(\sqrt{2}\right)^2}{\left(\sqrt{3}\right)^2-2\sqrt{3}\sqrt{2}+\left(\sqrt{2}\right)^2}$

$x^2=\dfrac{3+2\sqrt{6}+2}{3-2\sqrt{6}+2}$

$x^2=\dfrac{5+2\sqrt{6}}{5-2\sqrt{6}}$

After rationalizing the denominator we get

$x^2=\dfrac{5+2\sqrt{6}}{5-2\sqrt{6}}\times \dfrac{5+2\sqrt{6}}{5+2\sqrt{6}}$

$x^2=\dfrac{5^2+2\cdot \:5\cdot \:2\sqrt{6}+\left(2\sqrt{6}\right)^2}{(5)^2-(2\sqrt{6})^2}$

$x^2=\dfrac{49+20\sqrt{6}}{1}$

$x^2=49+20\sqrt{6}$

Therefore, the value of x² is $49+20\sqrt{6}$.

#Learn more

If x = root 3 + root2/root 3 - root 2 and y = root 3 - root 2/ root 3 + root 2 then find the value of x^2 + y^2​.

https://brainly.in/question/10268143