Question

if X equal to root 2 + 1 divided by root 2 minus 1 and Y equal to root 2 minus 1 divided by root 2 + 1 show that X square + xy + Y square equal to 35

Answer 1

this may help u check this

Answer 2

Answer:

We have:

$x=\dfrac{\sqrt{2}+1}{\sqrt{2}-1}$

on rationalizing the denominator we have:

$x=\dfrac{\sqrt{2}+1}{\sqrt{2}-1}\times \dfrac{\sqrt{2}+1}{\sqrt{2}+1}\\\\\\x=\dfrac{(\sqrt{2}+1)^2}{(\sqrt{2})^2-1^2}\\\\x=\dfrac{2+1+2\sqrt{2}}{2-1}\\\\x=\dfrac{3+2\sqrt{2}}{1}\\\\x=3+2\sqrt{2}$

Similarly we are given:

$y=\dfrac{\sqrt{2}-1}{\sqrt{2}+1}\\\\y=\dfrac{\sqrt{2}-1}{\sqrt{2}+1}\times \dfrac{\sqrt{2}-1}{\sqrt{2}-1}\\\\y=\dfrac{(\sqrt{2}-1)^2}{2-1}\\\\y=\dfrac{2+1-2\sqrt{2}}{1}\\\\y=3-2\sqrt{2}$

Now,

we are asked to find the value of the expression:

$x^2+xy+y^2$

$=(3+2\sqrt{2})^2+(3+2\sqrt{2})(3-2\sqrt{2})+(3-2\sqrt{2})^2\\\\=9+8+12\sqrt{2}+9-8+9+8-12\sqrt{2}$

(

since,

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

)

so,

$x^2+y^2+xy=9+8+1+9+8=35$

since, the same term $12\sqrt{2}$ but with opposite signs are cancelled out.

Hence, the value of the expression:

$x^2+xy+y^2=35$