PS arithmetic

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PS arithmetic

by kyuhunl » Tue Jan 29, 2019 11:14 pm

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$$\left(1+\sqrt{3}\right)\sqrt{2+\sqrt{3}}=?$$

A. $$\sqrt{2}\left(2-\sqrt{3}\right)$$
B. $$\sqrt{2}\left(2+\sqrt{2}\right)$$
C. $$\sqrt{2}\left(2+\sqrt{3}\right)$$
D. $$\sqrt{2}\left(3+\sqrt{3}\right)$$
E. $$\sqrt{3}\left(2+\sqrt{3}\right)$$

OA: C

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kyuhunl wrote:$$\left(1+\sqrt{3}\right)\sqrt{2+\sqrt{3}}=?$$

$$A. \sqrt{2}\left(2-\sqrt{3}\right)\,\,\,\,\,\,\,B. \sqrt{2}\left(2+\sqrt{2}\right)\,\,\,\,\,\,\,C. \sqrt{2}\left(2+\sqrt{3}\right)\,\,\,\,\,\,\,D. \sqrt{2}\left(3+\sqrt{3}\right)\,\,\,\,\,\,\,E. \sqrt{3}\left(2+\sqrt{3}\right)$$
Beautiful problem, kyuhunl. Congrats!

$$? = \left( {1 + \sqrt 3 } \right)\sqrt {2 + \sqrt 3 } = x\,\,\,\,\,\left( {x > 0} \right)$$
$${x^2} = {\left( {1 + \sqrt 3 } \right)^2}\left( {2 + \sqrt 3 } \right) = \left( {1 + 2\sqrt 3 + 3} \right)\left( {2 + \sqrt 3 } \right)$$
$${x^2} = 2\left( {2 + \sqrt 3 } \right)\left( {2 + \sqrt 3 } \right)\,\,\,\,\,\mathop \Rightarrow \limits^{x\,\, > \,\,0} \,\,\,\,\,? = \sqrt 2 \left( {2 + \sqrt 3 } \right)\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left( {\rm{C}} \right)$$


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by GMATGuruNY » Wed Jan 30, 2019 8:10 am
kyuhunl wrote:$$\left(1+\sqrt{3}\right)\sqrt{2+\sqrt{3}}=?$$

A. $$\sqrt{2}\left(2-\sqrt{3}\right)$$
B. $$\sqrt{2}\left(2+\sqrt{2}\right)$$
C. $$\sqrt{2}\left(2+\sqrt{3}\right)$$
D. $$\sqrt{2}\left(3+\sqrt{3}\right)$$
E. $$\sqrt{3}\left(2+\sqrt{3}\right)$$
An alternate approach is to BALLPARK.
√2 ≈ 1.4
√3 ≈ 1.7

(1+√3) * √(2 + √3) ≈ (2.7)(√3.7) = (2.7)(a bit less than 2) = a bit less than 5.4.

When we evaluate the answer choices, only C is a bit less than 5.4:
√2(2 + √3) ≈ (1.4)(3.7) = 5.18.

The correct answer is C.
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by Scott@TargetTestPrep » Fri Feb 01, 2019 5:21 pm
kyuhunl wrote:$$\left(1+\sqrt{3}\right)\sqrt{2+\sqrt{3}}=?$$

A. $$\sqrt{2}\left(2-\sqrt{3}\right)$$
B. $$\sqrt{2}\left(2+\sqrt{2}\right)$$
C. $$\sqrt{2}\left(2+\sqrt{3}\right)$$
D. $$\sqrt{2}\left(3+\sqrt{3}\right)$$
E. $$\sqrt{3}\left(2+\sqrt{3}\right)$$

OA: C
Recall that if x > 0, then x = √(x^2).

Since 1 + √3 > 0, 1 + √3 = √[( 1+ √3)^2] = √[1 + 2√3 + 3] = √(4 + 2√3).

Therefore,

(1 + √3)√(2 + √3)

√(4 + 2√3)√(2 + √3)

√[(4 + 2√3)(2 + √3)]

√[2(2 + √3)(2 + √3)]

√[2(2 + √3)^2]

√2(2 + √3)

Alternate Solution:

Let's let (1 + √3)√[2 + √3] = x and square each side:

(1 + √3)√[2 + √3] = x

[(1 + √3)^2](2 + √3) = x^2

(1 + 2√3 + 3)(2 + √3) = x^2

(4 + 2√3)(2 + √3) = x^2

2(2 + √3)(2 + √3) = x^2

2(2 + √3)^2 = x^2

Taking the square root of each side, we conclude that x = ±√2(2 + √3). Going back to the original expression that was given to us, we observe that both factors of (1 + √3)√[2 + √3] are positive; therefore, we can eliminate the negative root and conclude that x = √2(2 + √3).

Answer: C

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