Question

simplify sqrt(4-sqrt(7))+sqrt(8+3sqrt(7))

Answer 1

\[\sqrt{4-\sqrt{7}}+\sqrt{8+3\sqrt{7}}\]

Answer 2

Is that the problem?

Answer 3

ya

Answer 4

its answer is given as sqrt(2)(sqrt(7)+1)
bt how its so???

Answer 5

I'm going to have to work on it a little bit.

Answer 6

k,thnks in advance

Answer 7

From Wolf:
\[\sqrt{4-\sqrt{7}}=\frac{\sqrt{7}-1}{\sqrt{2}}\]

Answer 8

And
\[\sqrt{8+3\sqrt{7}}=\frac{3+\sqrt{7}}{\sqrt{2}}\]

Answer 9

But I haven't been able to figure out how they get that. Sorry.

Answer 10

I hope you come back and check for a solution:
\[\sqrt{4-\sqrt{7}}=\sqrt{\frac{8-2\sqrt{7}}{2}}=\sqrt{\frac{7-2\sqrt{7}+1}{2}}=\]

Answer 11

\[\sqrt{\frac{(\sqrt{7}-1)^2}{2}}=\frac{\sqrt{7}-1}{\sqrt{2}}\]

Answer 12

Similarly\[\sqrt{8+3\sqrt{7}}=\sqrt{\frac{16+6\sqrt{7}}{2}}=\sqrt{\frac{9+6\sqrt{7}+7}{2}}=\]

Answer 13

\[\sqrt{\frac{(3+\sqrt{7})^2}{2}}=\frac{3+\sqrt{7}}{\sqrt{2}}\]

Answer 14

And so adding the two:
\[\frac{\sqrt{7}-1}{\sqrt{2}}+\frac{3+\sqrt{7}}{\sqrt{2}}=\frac{2\sqrt{7}+2}{\sqrt{2}}=\]

Answer 15

\[\frac{2^1(\sqrt{7}+1)}{2^{\frac{1}{2}}}=2^{\frac{1}{2}}(\sqrt{7}+1)=\sqrt{2}(\sqrt{7}+1)\]

Answer 16

Hope this helps @Arunp