Question

X = (√5+1)/(√5-1) Y = (√5-1)/(√5+1)then, What is the value of (x²+3xy+y²)/(x²-3xy+y²)​

Answer 1

The value of (x²+3xy+y²)/(x²-3xy+y²) is $(\sqrt{5} - 1)$ .

• We have X = (√5+1)/(√5-1) and Y = (√5-1)/(√5+1).
• We have to find the value of (x²+3xy+y²)/(x²-3xy+y²).

We simplify the expression first,

$\frac{x^{2} + 3xy + y^{2} }{x^{2} - 3xy + y^{2}}$

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

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

• We have to find the value of $\frac{{(x+y)}^2 + xy }{{(x-y)}^2 - xy}$

• Now we calculate the value of xy , ${(x+y)}^{2}$ and ${(x-y)}^{2}$.

xy = ($\frac{\sqrt{5}+1}{\sqrt{5}-1}$)($\frac{\sqrt{5}-1}{\sqrt{5}+1}$)

xy = 1

• Now, ${(x+y)}^{2}$ = ($\frac{\sqrt{5}+1}{\sqrt{5}-1}$) + ($\frac{\sqrt{5}-1}{\sqrt{5}+1}$)

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

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

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

${(x+y)}^{2}$ = $\frac{12}{4}$ = 3

${(x+y)}^{2}$ = 3

• Similarly, ${(x-y)}^{2}$ = ($\frac{\sqrt{5}+1}{\sqrt{5}-1}$) - ($\frac{\sqrt{5}-1}{\sqrt{5}+1}$)

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

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

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

${(x-y)}^{2}$ = $\frac{4\sqrt{5}}{4}$ = $\sqrt{5}$

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

• Now, $\frac{{(x+y)}^2 + xy }{{(x-y)}^2 - xy}$ = $\frac{3 + 1 }{\sqrt{5} - (1)}$

$\frac{{(x+y)}^2 + xy }{{(x-y)}^2 - xy}$ = $\frac{4}{\sqrt{5} - 1}$

Now, we normalize the denominator,

$\frac{{(x+y)}^2 + xy }{{(x-y)}^2 - xy}$ = $\frac{4}{\sqrt{5} - 1}$ × $\frac{\sqrt{5} + 1}{\sqrt{5} + 1}$

$\frac{{(x+y)}^2 + xy }{{(x-y)}^2 - xy}$ = $\frac{4 (\sqrt{5} - 1)}{(\sqrt{5} - 1)(\sqrt{5} + 1)}$

$\frac{{(x+y)}^2 + xy }{{(x-y)}^2 - xy}$ = $\frac{4 (\sqrt{5} - 1)}{{(\sqrt{5} )}^{2} - 1^{2} }$

$\frac{{(x+y)}^2 + xy }{{(x-y)}^2 - xy}$ = $\frac{4 (\sqrt{5} - 1)}{4}$

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