+0

0
25
2

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

Jan 5, 2021

#1
+289
+2

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

$$(√2006+√2007)+(√2006-√2007)$$

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

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

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

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

8026

Jan 5, 2021
#2
+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