AMC10 2015 A

AMC10 2015 A · Q16

AMC10 2015 A · Q16. It mainly tests Systems of equations, Manipulating equations.

If $y+4=(x-2)^2$, $x+4=(y-2)^2$, and $x\ne y$, what is the value of $x^2+y^2$?

如果 $y+4=(x-2)^2$，$x+4=(y-2)^2$，且 $x\ne y$，那么 $x^2+y^2$ 的值是多少？

(A) 10 10

(B) 15 15

(D) 25 25

(E) 30 30

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): Expanding the binomials and subtracting the equations yields $x^2-y^2=3(x-y)$. Because $x-y\ne 0$, it follows that $x+y=3$. Adding the equations gives $x^2+y^2=5(x+y)=5\cdot 3=15$. Note: The two solutions are $(x,y)=\left(\frac{3}{2}+\frac{\sqrt{21}}{2},\frac{3}{2}-\frac{\sqrt{21}}{2}\right)$ and $\left(\frac{3}{2}-\frac{\sqrt{21}}{2},\frac{3}{2}+\frac{\sqrt{21}}{2}\right)$.

答案（B）：展开二项式并将两式相减得到 $x^2-y^2=3(x-y)$。因为 $x-y\ne 0$，所以 $x+y=3$。将两式相加得到 $x^2+y^2=5(x+y)=5\cdot 3=15$。注：两个解为 $(x,y)=\left(\frac{3}{2}+\frac{\sqrt{21}}{2},\frac{3}{2}-\frac{\sqrt{21}}{2}\right)$ 和 $\left(\frac{3}{2}-\frac{\sqrt{21}}{2},\frac{3}{2}+\frac{\sqrt{21}}{2}\right)$。

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