How many real roots does x^4 - 2x^2 + 1 = 0 have?

Question

A. 0
B. 1
C. 2
D. 3
E. 4

chetan2u · Answer

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

Nipungupta9081 · Answer

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

ShankSouljaBoi · Answer

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

MathRevolution · Answer

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

Kinshook · Answer

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

13swiftness · Answer

i dont think you can take x^2 common as the third term doesnt have x^2

How many real roots does x^4 - 2x^2 + 1 = 0 have?

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How many real roots does x^4 - 2x^2 + 1 = 0 have?

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