Question

if x=5+2√6 ,y=1/x then find the value of x square and y square

Answer 1

GIVEN :

If $x=5+2\sqrt{6}$, $y=\frac{1}{x}$ then find the value of $x^2+y^2$

TO FIND :

The value of $x^2+y^2$

SOLUTION :

Given that the values are $x=5+2\sqrt{6}$, $y=\frac{1}{x}$

$x=5+2\sqrt{6}$

Squaring on both sides,

$x^2=(5+2\sqrt{6})^2$

By using the algebraic identity

$(a+b)^2=a^2+2ab+b^2$

$x^2=5^2+2(5)(2\sqrt{6})+(2\sqrt{6})^2$

$=25+20\sqrt{6}+2^2(\sqrt{6})^2$

By using the exponent property :

$(ab)^m=a^mb^m$

$=25+20\sqrt{6}+4(6)$

$=25+20\sqrt{6}+24$

∴ $x^2=49+20\sqrt{6}$

Now $y=\frac{1}{x}$

$y=\frac{1}{5+2\sqrt{6}}$

Squaring on both sides

$y^2=\frac{1}{x^2}=\frac{1}{(5+2\sqrt{6})^2}$

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

Now rationalise the denominator we get

$=\frac{1}{49+20\sqrt{6}}\times \frac{49-20\sqrt{6}}{49-20\sqrt{6}}$

$=\frac{49-20\sqrt{6}}{49^2-(20\sqrt{6})^2}$

$=\frac{49-20\sqrt{6}}{49^2-20^2(\sqrt{6})^2}$

$=\frac{49-20\sqrt{6}}{2401-400(6)}$

$=\frac{49-20\sqrt{6}}{2401-2400}$

$=\frac{49-20\sqrt{6}}{1}$

∴ $y^2=\frac{1}{x^2}=49-20\sqrt{6}$

Now we have to find the value of $x^2+y^2$

By substituting the values,

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

$=98$

∴ $x^2+y^2=98$

∴  the value of $x^2+y^2$ is 98

Answer 2

here is the ans

hope it helps you