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Difficulty:
45%
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Question Stats:
60% (01:10) correct
40%
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wrong
based on 217
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How many real roots does x^4 - 2x^2 + 1 = 0 have?
A. 0
B. 1
C. 2
D. 3
E. 4
_________________
A. 0
B. 1
C. 2
D. 3
E. 4
_________________
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Bunuel
\(x^4-2x^2+1=0.......(x^2)^2-2*1*x^2+1^2=0......(x^2-1)^2=0\)
So \(x^2=1\), meaning x can be 1 or -1.
Thus, two values
C
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Nipungupta9081
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Taking X^2 common we get = X^2 ( x^2 - 2 + 1 ) = 0
= X^2 ( X^2 - 1 ) = 0
= X^2 = 0 or X^2 = 1
X = +1 or X = -1.
Answer C
= X^2 ( X^2 - 1 ) = 0
= X^2 = 0 or X^2 = 1
X = +1 or X = -1.
Answer C
ShankSouljaBoi
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Lets x^2 be y then y^2 -2y +1 =0 So , y =1 or x^2 = 1---- x = +1 or -1 ... So only two real roots to the equation. C
GMAT 1: 760 Q51 V42
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Descartes’s rule of signs can give you the numbers of possible real roots, both positive and negative:
Positive real roots: For the number of positive real roots, look at the polynomial, written in descending order, and count how many times the sign changes from term to term
=> x^4 - x^2 + 1 = 0 - Sign change 2 times.
Negative real roots: For the number of negative real roots, find f(–x) and count again
=> (-x)^4 - (-x)^2 + 1 = 0
=> x^4 - x^2 + 1 = 0 - Sign change 2 times.
Hence, there are two real roots.
Answer C
Positive real roots: For the number of positive real roots, look at the polynomial, written in descending order, and count how many times the sign changes from term to term
=> x^4 - x^2 + 1 = 0 - Sign change 2 times.
Negative real roots: For the number of negative real roots, find f(–x) and count again
=> (-x)^4 - (-x)^2 + 1 = 0
=> x^4 - x^2 + 1 = 0 - Sign change 2 times.
Hence, there are two real roots.
Answer C
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Asked: How many real roots does x^4 - 2x^2 + 1 = 0 have?
Let t = xˆ2
tˆ2 - 2t + 1 = 0
(t-1)ˆ2 = 0
t= 1
xˆ2 = 1
x = {-1,1}: 2 real roots
IMO C
Let t = xˆ2
tˆ2 - 2t + 1 = 0
(t-1)ˆ2 = 0
t= 1
xˆ2 = 1
x = {-1,1}: 2 real roots
IMO C
Posts: 3
i dont think you can take x^2 common as the third term doesnt have x^2
Nipungupta9081
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