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Manager  G
Joined: 21 Feb 2019
Posts: 125
Location: Italy
a and b are two numbers such that 2a^4 -4ab + b^2 + 2 = 0...  [#permalink]

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3 00:00

Difficulty:   85% (hard)

Question Stats: 39% (02:25) correct 61% (02:08) wrong based on 28 sessions

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$$a$$ and $$b$$ are two numbers such that:

$$2a^4 -4ab + b^2 + 2 = 0$$

How many distinct values can $$a$$ assume?

A. 1
B. 2
C. 3
D. 4
E. None

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MEMENTO AUDERE SEMPER
Math Expert V
Joined: 02 Aug 2009
Posts: 7754
Re: a and b are two numbers such that 2a^4 -4ab + b^2 + 2 = 0...  [#permalink]

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1
lucajava wrote:
$$a$$ and $$b$$ are two numbers such that:

$$2a^4 -4ab + b^2 + 2 = 0$$

How many distinct values can $$a$$ assume?

A. 1
B. 2
C. 3
D. 4
E. None

$$2a^4 -4ab + b^2 + 2 = 0$$.....$$2a^4-4a^2+4a^2 -4ab + b^2 + 2 = 0.....2(a^4-2a^2+1)+(4a^2-4ab+b^2)=0....2(a^2-1)^2+(2a-b)^2=0$$...
Since sum of two squares will be 0 when each term is 0..
2(a^2-1)^2=0, that is a=-1 or a=1.
(2a-b)^2=0, that is a=b/2..
When a=-1.. $$2a^4 -4ab + b^2 + 2 = 0.....2*(-1)^4-4*(-1)b+b^2+2=0...2+4b+b^2+2=0....(b+2)^2=0$$..(a,b)=(-1,-2)..
When a=1.. $$2a^4 -4ab + b^2 + 2 = 0.....2*(1)^4-4*(1)b+b^2+2=0...2-4b+b^2+2=0....(b-2)^2=0$$..(a,b)=(1,2)..

So a has two distinct values -1 and 1..

B
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Manager  G
Joined: 21 Feb 2019
Posts: 125
Location: Italy
Re: a and b are two numbers such that 2a^4 -4ab + b^2 + 2 = 0...  [#permalink]

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Another way is considering $$a$$ a fixed parameter. Hence, you can consider:

$$b^2 -4ab +2a^4 + 2 = 0$$

a second-degree equation. Let's compute $$b_{12} = 2a ± \sqrt{4a^2 -2a^4 -2}$$.

Since GMAT math works on real numbers only, $$4a^2 -2a^4 - 2 ≥ 0$$. Hence, developing this, we get:

$$-a^4 +2a^2 -1 ≥ 0$$

$$a^4 -2a^2 +1 ≤ 0$$

Let be $$a^2 = t$$

$$t^2 -2t + 1 ≤ 0$$

$$t_{12} = 1 ± \sqrt{1 -1} = 1$$

In conclusion, we derive $$a^2 = t = 1$$. Hence, $$a = ± 1$$. Two solutions, pick B.
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MEMENTO AUDERE SEMPER Re: a and b are two numbers such that 2a^4 -4ab + b^2 + 2 = 0...   [#permalink] 21 Apr 2019, 13:18
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# a and b are two numbers such that 2a^4 -4ab + b^2 + 2 = 0...

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