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# (8 sqrt2-4)^2

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(8 sqrt2-4)^2 [#permalink]  12 Feb 2020, 10:26
Expert's post
00:00

Question Stats:

81% (00:58) correct 18% (01:26) wrong based on 27 sessions
$$\left( \begin{array}{cc} \frac{8 \sqrt{2} - 4}{4} \end{array} \right) ^2 =$$

A. $$9-4 \sqrt{2}$$

B. $$36-16 \sqrt{2}$$

C. $$8$$

D. $$9$$

E. $$32 \sqrt{2}$$

Kudos for the right answer and explanation
[Reveal] Spoiler: OA

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GRE Instructor
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Kudos [?]: 4196 [2] , given: 67

Re: (8 sqrt2-4)^2 [#permalink]  12 Feb 2020, 10:37
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Expert's post
Carcass wrote:
$$\left( \begin{array}{cc} \frac{8 \sqrt{2} - 4}{4} \end{array} \right) ^2 =$$

A. $$9-4 \sqrt{2}$$

B. $$36-16 \sqrt{2}$$

C. $$8$$

D. $$9$$

E. $$32 \sqrt{2}$$

Kudos for the right answer and explanation

One approach to quickly arrive at the answer is to estimate the value of the given expression.

By test day, all students should have the following approximations memorized:
√2 ≈ 1.4
√3 ≈ 1.7
√5 ≈ 2.2

So, our approximation will go as follows: $$(\frac{8 \sqrt{2} - 4}{4})^2 ≈ (\frac{8(1.4)- 4}{4}) ^2$$

$$≈ (\frac{11- 4}{4}) ^2$$

$$≈ (\frac{7}{4}) ^2$$

$$≈ (\frac{49}{16})$$

$$≈ 3$$

Now let's estimate each answer choice....

A. $$9-4 \sqrt{2}≈9-4(1.4)≈9-6≈3$$ Nice!! KEEP

B. $$36-16 \sqrt{2}≈36-16(1.4)≈36-23≈13$$ Not close. ELIMINATE

C. $$8$$ Not close. ELIMINATE

D. $$9$$ Not close. ELIMINATE

E. $$32 \sqrt{2}≈32(1.4)$$ Not close. ELIMINATE
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Founder
Joined: 18 Apr 2015
Posts: 12529
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Kudos [?]: 3156 [2] , given: 11597

Re: (8 sqrt2-4)^2 [#permalink]  12 Feb 2020, 10:58
2
KUDOS
Expert's post
I think here the best approach is algebraical, if you do know what to do, you can solve it in 20 seconds

Inside the parentheses will become

$$\frac{8 \sqrt{2}}{4} -\frac{4}{4}$$ raised to the power of two

Simplify

$$(2 \sqrt{2} - 1 )^2$$

$$4 (\sqrt{2})^2+1-4 \sqrt{2}$$

$$8+1 -4 \sqrt{2}$$

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Manager
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Re: (8 sqrt2-4)^2 [#permalink]  13 Feb 2020, 23:03
1
KUDOS
Carcass wrote:
$$\left( \begin{array}{cc} \frac{8 \sqrt{2} - 4}{4} \end{array} \right) ^2 =$$

A. $$9-4 \sqrt{2}$$

B. $$36-16 \sqrt{2}$$

C. $$8$$

D. $$9$$

E. $$32 \sqrt{2}$$

Kudos for the right answer and explanation

I would go with simplifying by rewriting every number as 2 to some power:

$$((2 ^3 * 2 ^\frac{1}{2} - 2 ^2 )*2 ^(-2) ) ^2$$

That simplifies to:

$$(2 ^\frac{3}{2} - 1 ) ^2$$

That simplifies to:

$$2 ^3 - 2*2 ^\frac{3}{2} + 1$$ = $$9-4 \sqrt{2}$$

Manager
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Re: (8 sqrt2-4)^2 [#permalink]  27 Apr 2020, 03:33
1
KUDOS
My approach:

((8√2-4)*(8√2-4))/16

(128-64√2+16)/16

(16*(8-4√2+1))/16

8-4√2+1

9-4√1
Re: (8 sqrt2-4)^2   [#permalink] 27 Apr 2020, 03:33
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