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If $x = \sqrt{2006}$ and $y = \sqrt{2007}$, what is the simplified value of $(x + y)^2 + (x - y)^2$?

Guest Jan 5, 2021

hihihi · Answer 1 · 2021-01-05T05:27:03

Answer:

$(x+y)^2+(x-y)^2$

$(√2006+√2007)+(√2006-√2007)$

$\left(3\sqrt{223}+\sqrt{2006}\right)^2+\left(\sqrt{2006}-\sqrt{2007}\right)^2$

$\left(3\sqrt{223}+\sqrt{2006}\right)^2+\left(\sqrt{2006}-3\sqrt{223}\right)^2$

$4013+6\sqrt{447338}+\left(\sqrt{2006}-3\sqrt{223}\right)^2$

$4013+6\sqrt{447338}+4013-6\sqrt{447338}$

8026

GR0001 · Answer 2 · 2021-01-05T05:35:03

hihihi's solution is good, but this one may help you solve the problem quicker and easier

(x+y)² = x²+2xy+y²

(x-y)²=x²-2xy+y²

As such, when you add them together, you get 2x²+2y²

When you plug in x and y, this equals to 2(2006)+2(2007)=4012+4014=8026