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Algebra Difficulty 2.9 Junior Find the answer

If xx and yy are positive real numbers with 1x+y=1x1y\frac{1}{x+y}=\frac{1}{x}-\frac{1}{y}, what is the value of (xy+yx)2\left(\frac{x}{y}+\frac{y}{x}\right)^{2}?

A number or a short expression. Spacing and $ signs are ignored.

Solution

Starting with the given relationship between xx and yy and manipulating algebraically, we obtain successively 1x+y=1x1y\frac{1}{x+y}=\frac{1}{x}-\frac{1}{y} xy=(x+y)y(x+y)xxy=(x+y)y-(x+y)x xy=xy+y2x2xyxy=xy+y^{2}-x^{2}-xy x2+xyy2=0x^{2}+xy-y^{2}=0 x2y2+xy1=0\frac{x^{2}}{y^{2}}+\frac{x}{y}-1=0 where t=xyt=\frac{x}{y}. Since x>0x>0 and y>0y>0, then t>0t>0. Using the quadratic formula t=1±124(1)(1)2=1±52t=\frac{-1 \pm \sqrt{1^{2}-4(1)(-1)}}{2}=\frac{-1 \pm \sqrt{5}}{2}. Since t>0t>0, then xy=t=512\frac{x}{y}=t=\frac{\sqrt{5}-1}{2}. Therefore, $\left(\frac{x}{y}+\frac{y}{x}\right)^{2}=\left(\frac{\sqrt{5}-1}{2}+\frac{2}{\sqrt{5}-1}\right)^{2}=\left(\frac{\sqrt{5}-1}{2}+\frac{2(\sqrt{5}+1)}{(\sqrt{5}-1)(\sqrt{5}+1)}\right)^{2}=\left(\frac{\sqrt{5}-1}{2}+\frac{\sqrt{5}+1}{2}\right)^{2}=(\sqrt{5})^{2}=5

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