Which of the following is equal to $$^{\left(\sqrt{16+4\sqrt{15}}+\sqrt{16-4\sqrt{15}}\right)^2}$$

A. 10
B. 12
C. 24
D. 36
E. 40

Hi Max@Math Revolution,
Let's take a look at your question.

$$\left(\sqrt{16+4\sqrt{15}}+\sqrt{16-4\sqrt{15}}\right)^2$$
Evaluating the square,
$$=\left(\sqrt{16+4\sqrt{15}}\right)^2+2\left(\sqrt{16+4\sqrt{15}}\right)\left(\sqrt{16-4\sqrt{15}}\right)+\left(\sqrt{16-4\sqrt{15}}\right)^2$$ $$=16+4\sqrt{15}+2\left(\sqrt{\left(16\right)^2+\left(4\right)^2\left(15\right)}\right)+16-4\sqrt{15}$$
$$=16+2\left(\sqrt{256-\left(16\right)\left(15\right)}\right)+16$$
$$=32+2\left(\sqrt{16}\right)$$
$$=32+2\left(4\right)$$
$$=32+8=40$$
Therefore, Option E is correct.

Hope it helps.
I am available if you'd like any follow up.

$$\left(\sqrt{16+4\sqrt{15}}+\sqrt{16-4\sqrt{15}}\right)^2$$
= $$\left(\sqrt{16+2\sqrt{60}}+\sqrt{16-2\sqrt{60}}\right)^2$$
= $$\left(\sqrt{10}+\sqrt{6}+\sqrt{10}-\sqrt{6}\right)^2$$
= $$\left(2\sqrt{10}\right)^2$$
$$=40\cdot10$$
=40

Therefore, the answer is E.
Answer: E

$$^{(a+b)^2}$$ = $$^{a^2}$$ + $$^{b^2}$$ + 2ab
and $$^{a^2}$$ = 16 + 4 $$\sqrt{15}$$
and $$^{b^2}$$ = 16 - 4 $$\sqrt{15}$$
and 2ab = 2 x 4 = 8 (from the difference of 2 squares)
So 16 + 16 + 8 = 40
Answer = E

Max@Math Revolution wrote:[GMAT math practice question]

Which of the following is equal to $$^{\left(\sqrt{16+4\sqrt{15}}+\sqrt{16-4\sqrt{15}}\right)^2}$$

A. 10
B. 12
C. 24
D. 36
E. 40

We can BALLPARK.

16 + 4√15 = 16 + 4(a bit less than 4) ≈ 16 + (a bit less than 16) ≈ 32.
16 - 4√15 = 16 - 4(a bit less than 4) ≈ 16 - (a bit less than 16) ≈ 0.5.

The expression becomes:
(√32 + √0.5)².

Since (a + b)² = a² + b² + 2ab, we get:
(√32 + √0.5)²

= 32 + 0.5 + 2(√32)(√0.5)

≈ 32 + 0.5 + 2√16

= 32 + 0.5 + 8

= 40.5.

The correct answer is E.