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655-705 (Hard)|   Algebra|      
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Bunuel
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If a = 1/(2^(1/2) - 1) - 2, what is a^2 + 2a + 4? 29 Mar 2023, 07:45
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Difficulty:

65% (hard)

Question Stats:

60% (02:13) correct 40% (02:44) wrong based on 241 sessions

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If \(a = \frac{1}{\sqrt{2}-1}-2,\) what is \(a^2 + 2a + 4\)?

A. \(2\)

B. \(3\)

C. \(4\)

D. \(5\)

E. \(6\)
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Official Answer
D
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gmatophobia
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If a = 1/(2^(1/2) - 1) - 2, what is a^2 + 2a + 4? 29 Mar 2023, 11:18
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Bunuel
If \(a = \frac{1}{\sqrt{2}-1}-2,\) what is \(a^2 + 2a + 4\)?

A. \(2\)

B. \(3\)

C. \(4\)

D. \(5\)

E. \(6\)
Show more

Not sure what am I doing wrong here -

\(a = \frac{1}{\sqrt{2}-1}-2\)

\(a = \frac{1* (\sqrt{2}+1)}{(\sqrt{2}-1) (\sqrt{2}+1)}-2\)

\(a = \frac{1* (\sqrt{2}+1)}{2-1}-2\)

\(a = (\sqrt{2}+1) -2\)

\(a + 2= (\sqrt{2}+1) -2 + 2\)

\((a + 2)^2= (\sqrt{2}+1)^2\)

\(a^2 + 2a + 4= 3 + 2\sqrt{2}\)

:dontknow:
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Nabneet
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If a = 1/(2^(1/2) - 1) - 2, what is a^2 + 2a + 4? 30 Mar 2023, 08:40
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gmatophobia
Bunuel
If \(a = \frac{1}{\sqrt{2}-1}-2,\) what is \(a^2 + 2a + 4\)?

A. \(2\)

B. \(3\)

C. \(4\)

D. \(5\)

E. \(6\)

Not sure what am I doing wrong here -

\(a = \frac{1}{\sqrt{2}-1}-2\)

\(a = \frac{1* (\sqrt{2}+1)}{(\sqrt{2}-1) (\sqrt{2}+1)}-2\)

\(a = \frac{1* (\sqrt{2}+1)}{2-1}-2\)

\(a = (\sqrt{2}+1) -2\)

\(a + 2= (\sqrt{2}+1) -2 + 2\)

\((a + 2)^2= (\sqrt{2}+1)^2\)

\(a^2 + 2a + 4= 3 + 2\sqrt{2}\)

:dontknow:
Show more

a has two parts , u just cannot simplify one part of it. Its wrong
If u want to simplify then make it a single value and then u can multiply in num. & deno.

take example,
a= (1/root 2) - 2

is it same as the value , (root2/2) -2 ?



Solution:

a^2 + 2a + 4 = (a+2)^2 -2a

put the value of a in the eqn and solve

Answer we will get is 5.
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Bunuel
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If a = 1/(2^(1/2) - 1) - 2, what is a^2 + 2a + 4? 30 Mar 2023, 08:48
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gmatophobia
Bunuel
If \(a = \frac{1}{\sqrt{2}-1}-2,\) what is \(a^2 + 2a + 4\)?

A. \(2\)

B. \(3\)

C. \(4\)

D. \(5\)

E. \(6\)

Not sure what am I doing wrong here -

\(a = \frac{1}{\sqrt{2}-1}-2\)

\(a = \frac{1* (\sqrt{2}+1)}{(\sqrt{2}-1) (\sqrt{2}+1)}-2\)

\(a = \frac{1* (\sqrt{2}+1)}{2-1}-2\)

\(a = (\sqrt{2}+1) -2\)

\(a + 2= (\sqrt{2}+1) -2 + 2\)

\((a + 2)^2= (\sqrt{2}+1)^2\)

\(a^2 + 2a + 4= 3 + 2\sqrt{2}\)

:dontknow:
Show more

\(a = \frac{1}{\sqrt{2}-1}-2= \frac{\sqrt{2}+1}{(\sqrt{2}-1)(\sqrt{2}+1)}-2=\frac{\sqrt{2}+1}{2-1}-2=(\sqrt{2}+1)-2=\sqrt{2}-1\).

\(a^2 + 2a + 4=(a+1)^2 + 3= (\sqrt{2}-1 + 1)^2+3=2+3=5\).

Answer: D.
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If a = 1/(2^(1/2) - 1) - 2, what is a^2 + 2a + 4? 30 Mar 2023, 08:54
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Nabneet


a has two parts , u just cannot simplify one part of it. Its wrong
If u want to simplify then make it a single value and then u can multiply in num. & deno.

take example,
a= (1/root 2) - 2

is it same as the value , (root2/2) -2 ?



Solution:

a^2 + 2a + 4 = (a+2)^2 -2a

put the value of a in the eqn and solve

Answer we will get is 5.
Show more

Thanks for your reply Nabneet.

While I tried to grasp the part in which you mentioned that the rationalization part was incorrect I couldn't follow it completely. My apologies for the same.

With respect to the example provided in the explanation, let me work that here with you -

Quote:

take example,
a= (1/root 2) - 2

is it same as the value , (root2/2) -2 ?
Show more

\(a = \frac{1}{\sqrt{2}} - 2 \)

By PEDMAS rule, we first solve \(\frac{1}{\sqrt{2}}\) = \(\frac{1}{1.41421356237}\)

\(a = 0.70710678118 - 2 \) = -1.29289321881

\(a = \frac{\sqrt{2}}{2} - 2 \)

By PEDMAS rule, we first solve \(\frac{\sqrt{2}}{2}\) = \(\frac{1.41421356237}{2}\)

\(a = 0.70710678118 - 2 \) = -1.29289321881

Aren't they the same :think:
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If a = 1/(2^(1/2) - 1) - 2, what is a^2 + 2a + 4? 30 Mar 2023, 09:00
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Bunuel
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Bunuel
If \(a = \frac{1}{\sqrt{2}-1}-2,\) what is \(a^2 + 2a + 4\)?

A. \(2\)

B. \(3\)

C. \(4\)

D. \(5\)

E. \(6\)

Not sure what am I doing wrong here -

\(a = \frac{1}{\sqrt{2}-1}-2\)

\(a = \frac{1* (\sqrt{2}+1)}{(\sqrt{2}-1) (\sqrt{2}+1)}-2\)

\(a = \frac{1* (\sqrt{2}+1)}{2-1}-2\)

\(a = (\sqrt{2}+1) -2\)

\(a + 2= (\sqrt{2}+1) -2 + 2\)

\((a + 2)^2= (\sqrt{2}+1)^2\)

\(a^2 + 2a + 4= 3 + 2\sqrt{2}\)

:dontknow:

\(a = \frac{1}{\sqrt{2}-1}-2= \frac{\sqrt{2}+1}{(\sqrt{2}-1)(\sqrt{2}+1)}-2=\frac{\sqrt{2}+1}{2-1}-2=(\sqrt{2}+1)-2=\sqrt{2}-1\).

\(a^2 + 2a + 4=(a+1)^2 + 3= (\sqrt{2}-1 + 1)^2+3=2+3=5\).

Answer: D.
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:facepalm_man:

I messed up \((a+2)^2\) - Where should I go and hide :cry:

Thanks Bunuel :)
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If a = 1/(2^(1/2) - 1) - 2, what is a^2 + 2a + 4? 30 Mar 2023, 09:07
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gmatophobia
:facepalm_man:

I messed up \((a+2)^2\) - Where should I go and hide :cry:

Thanks Bunuel :)
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Errare humanum est. It happens to the best of us.
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If a = 1/(2^(1/2) - 1) - 2, what is a^2 + 2a + 4? 28 Sep 2025, 18:07
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hey could you tell did you come with the (a+1)sqr +3


Bunuel


\(a = \frac{1}{\sqrt{2}-1}-2= \frac{\sqrt{2}+1}{(\sqrt{2}-1)(\sqrt{2}+1)}-2=\frac{\sqrt{2}+1}{2-1}-2=(\sqrt{2}+1)-2=\sqrt{2}-1\).

\(a^2 + 2a + 4=(a+1)^2 + 3= (\sqrt{2}-1 + 1)^2+3=2+3=5\).

Answer: D.
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If a = 1/(2^(1/2) - 1) - 2, what is a^2 + 2a + 4? 28 Sep 2025, 18:07
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hey could you tell did you come with the (a+1)sqr +3


Bunuel


\(a = \frac{1}{\sqrt{2}-1}-2= \frac{\sqrt{2}+1}{(\sqrt{2}-1)(\sqrt{2}+1)}-2=\frac{\sqrt{2}+1}{2-1}-2=(\sqrt{2}+1)-2=\sqrt{2}-1\).

\(a^2 + 2a + 4=(a+1)^2 + 3= (\sqrt{2}-1 + 1)^2+3=2+3=5\).

Answer: D.
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If a = 1/(2^(1/2) - 1) - 2, what is a^2 + 2a + 4? 28 Sep 2025, 22:50
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hey could you tell did you come with the (a+1)sqr +3



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Vaidehi06surya

When you expand \((a+1)^2\), you get:
\((a+1)^2 = a^2 + 2a + 1\)

But we need \(a^2 + 2a + 4\)

So: \(a^2 + 2a + 4 = (a^2 + 2a + 1) + 3 = (a+1)^2 + 3\)

Why This Transformation Helps:

Once we have \((a+1)^2 + 3\) and know that \(a = \sqrt{2} - 1\):
- \(a + 1 = \sqrt{2} - 1 + 1 = \sqrt{2}\) (much simpler!)
- \((a+1)^2 = (\sqrt{2})^2 = 2\)
- Final answer: \(2 + 3 = 5\)

Without this transformation, you'd have to calculate \((\sqrt{2}-1)^2 + 2(\sqrt{2}-1) + 4\), which involves expanding and simplifying multiple radical terms - much more tedious!

Strategic Pattern Recognition:

When you see expressions like \(a^2 + 2a + \text{constant}\), always check if you can complete the square:
- If the coefficient of \(a\) is \(2\), you're looking for \((a+1)^2 = a^2 + 2a + 1\)
- If the coefficient of \(a\) is \(4\), you're looking for \((a+2)^2 = a^2 + 4a + 4\)
- If the coefficient of \(a\) is \(6\), you're looking for \((a+3)^2 = a^2 + 6a + 9\)

This technique saves significant time on the GMAT, especially when dealing with irrational numbers like \(\sqrt{2}\).

You can practice similar algebraic manipulation questions here (you'll find a lot of OG questions) - select Algebra under Problem Solving and choose Medium level questions based on your current understanding.
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