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

 Jan 5, 2021
 #1
avatar+289 
+2

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

 Jan 5, 2021
 #2
avatar+27 
0

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

 Jan 5, 2021

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