2020 AMC 10 A

Complete problem set with solutions and individual problem pages

Problem 14 Medium

Real numbers x and y satisfy x+y=4 and x \cdot y=-2. What is the value of x+\frac{x^{3}}{y^{2}}+\frac{y^{3}}{x^{2}}+y?

  • A.

    360

  • B.

    400

  • C.

    420

  • D.

    440

  • E.

    480

Answer:D

Solution 1: x+\frac{x^{3}}{y^{2}}+\frac{y^{3}}{x^{2}}+y=x+\frac{x^{3}}{y^{2}}+y+\frac{y^{3}}{x^{2}}=\frac{x^{3}}{x^{2}}+\frac{y^{3}}{x^{2}}+\frac{y^{3}}{y^{2}}+\frac{x^{3}}{y^{2}} Continuing to combine \frac{x^{3}+y^{3}}{x^{2}}+\frac{x^{3}+y^{3}}{y^{2}}=\frac{\left(x^{2}+y^{2}\right)\left(x^{3}+y^{3}\right)}{x^{2} y^{2}}=\frac{\left(x^{2}+y^{2}\right)(x+y)\left(x^{2}-x y+y^{2}\right)}{x^{2} y^{2}} From the givens, it can be concluded that x^{2} y^{2}=4. Also, (x+y)^{2}=x^{2}+2 x y+y^{2}=16_{\text {This means }} that x^{2}+y^{2}=20. Substituting this information into \frac{\left(x^{2}+y^{2}\right)(x+y)\left(x^{2}-x y+y^{2}\right)}{x^{2} y^{2}}, we have \frac{(20)(4)(22)}{4}=20 \cdot 22=(\text{D}) 440.

Solution 2: As above, we need to calculate \frac{\left(x^{2}+y^{2}\right)\left(x^{3}+y^{3}\right)}{x^{2} y^{2}}. Note that x, y, are the roots of x^{2}-4 x-2 and so x^{3}=4 x^{2}+2 x and y^{3}=4 y^{2}+2 y. Thus x^{3}+y^{3}=4\left(x^{2}+y^{2}\right)+2(x+y)=4(20)+2(4)=88 where x^{2}+y^{2}=20 and x^{2} y^{2}=4 as in the previous solution. Thus the \frac{(20)(88)}{4}=(\mathbf{D}) 440