Question

If x=√5-√2/ √5+√2 y=√5+√2 / √5-√2 then find the value x²+xy+y²

Answer 1

(x+y)²=x²+xy+y²
x²+xy+y²={(√5-√2÷√5+√2)+(√5+√2÷√5-√2)}²
x²+xy+y²={(√5-√2)(√5-√2)+(√5+√2)(√5+√2)÷(√5+√2)(√5-√2)}²
as we know that,
(a+b)²=(a+b)(a+b)
(a-b)²=(a-b)(a-b)
(a+b)(a-b)=a²-b²
so,x²+xy+y²={(√5-√2)²+(√5+√2)²÷(√5)²-(√2)²}²
x²+xy+y²=[{(√5)²-(√2)²}²÷5-3]²
x²+xy+y²={(5-2)²÷3}²
x²+xy+y²={(3)²÷3}²
x²+xy+y²={9÷3}²
x²+xy+y²=(3)²
x²+xy+y²=9

Answer 2

Answer:

$x^2+xy+y^2=\frac{14}{9}$

Step-by-step explanation:

$x=\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}$

$x=\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}} \times \frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}}$

$x=\frac{(\sqrt{5}-\sqrt{2})^2}{5-2}$

$x=\frac{(\sqrt{5}-\sqrt{2})^2}{3}$

$(x+y)^2=x^2+xy+y^2$

$x^2+xy+y^2=(\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}+\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}})^2$

$x^2+xy+y^2=(\frac{(\sqrt{5}-\sqrt{2})(\sqrt{5}-\sqrt{2})+(\sqrt{5}+\sqrt{2})(\sqrt{5}+\sqrt{2})}{(\sqrt{5}+\sqrt{2})(\sqrt{5}-\sqrt{2})})^2$

$x^2+xy+y^2=\frac{(\sqrt{5}-\sqrt{2})^2+(\sqrt{5}+\sqrt{2})^2}{(5-2)^2}$

$x^2+xy+y^2=\frac{(5+2-2\sqrt{5}\sqrt{2}+5+2+2\sqrt{5}\sqrt{2}}{(5-2)^2}$

$x^2+xy+y^2=\frac{14}{9}$

Hence $x^2+xy+y^2=\frac{14}{9}$