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18 Apr 2019, 05:51
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B
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Difficulty:
95%
(hard)
Question Stats:
38% (02:26) correct
62%
(02:27)
wrong
based on 103
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\(a\) and \(b\) are two numbers such that:
How many distinct values can \(a\) assume?
A. 1
B. 2
C. 3
D. 4
E. None
\(2a^4 -4ab + b^2 + 2 = 0\)
How many distinct values can \(a\) assume?
A. 1
B. 2
C. 3
D. 4
E. None
20 Apr 2019, 08:03
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lucajava
\(2a^4 -4ab + b^2 + 2 = 0\).....\(2a^4-4a^2+4a^2 -4ab + b^2 + 2 = 0.....2(a^4-2a^2+1)+(4a^2-4ab+b^2)=0....2(a^2-1)^2+(2a-b)^2=0\)...
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
21 Apr 2019, 13:18
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Another way is considering \(a\) a fixed parameter. Hence, you can consider:
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:
In conclusion, we derive \(a^2 = t = 1\). Hence, \(a = ± 1\). Two solutions, pick B.
\(b^2 -4ab +2a^4 + 2 = 0\)
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.
