If \(x-y=\sqrt{5}\) and \(10+xy=\displaystyle\frac{13}{xy}\), then find the value of \(x^4+y^4+10\).
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 61
Write \(x^4+y^4\) through \(x^2+y^2\) and \(xy\): \(x^4+y^4=\left(x^2+y^2\right)^2-2(xy)^2\) and \(x^2+y^2=(x-y)^2+2xy\). The second condition gives \((xy)^2\) in terms of \(xy\), so \(xy\) cancels out.
Multiplying \(10+xy=\displaystyle\frac{13}{xy}\) by \(xy\): \((xy)^2=13-10xy\)
\(x^2+y^2=(x-y)^2+2xy=5+2xy\)
\(\left(x^2+y^2\right)^2=25+20xy+4(xy)^2=25+20xy+52-40xy=77-20xy\)
\(2(xy)^2=26-20xy\)
\(x^4+y^4=(77-20xy)-(26-20xy)=51\)
\(x^4+y^4+10=51+10=61\)