Question

x^2+y^2 =1 and x^4 + y^4 = 17/18 then xy=?​

Answer 1

$\textrm{The question gave us that -}$

• $x^{2}+y^{2}=1$
• $x^{4}+y^{4}=\dfrac{17}{18}$

$\textbf{Polynomial identity}$

$\boxed{(a+b)^{2}=a^{2}+2ab+b^{2}}$

$\textrm{Now, by applying this, -}$

$(x^{2}+y^{2})^{2}=1^{2}$

$x^{4}+2x^{2}y^{2}+y^{4}=1$

$2x^{2}y^{2}+\dfrac{17}{18}=1$

$2x^{2}y^{2}=\dfrac{1}{18}$

$x^{2}y^{2}=\dfrac{1}{36}$

$\therefore xy=\pm\dfrac{1}{6}$

So, the value of $xy$ is $\dfrac{1}{6}$ or $-\dfrac{1}{6}$.