Question

State true or false:

If $x=2-\sqrt{6}$, $y=2+\sqrt{6}$, then the value of $x^2+y^2+xy$ is $323$.

• A. True
• B. False

Answer 1

The correct option is B False

$x^2={(2-\sqrt{6})}^2=4+6-4\sqrt{6}=10-4\sqrt{6}{y}^2={(2+\sqrt{6})}^2=4+6+4\sqrt{6}=10+4\sqrt{6}xy=(2-\sqrt{6})(2+\sqrt{6})=4-6=-2$

Putting these values in $x^2+y^2+xy$ we get

$10-4\sqrt{6}+10+4\sqrt{6}-2=18$

So the given statement is false.