How many real roots does X^4 - 2x^2 - 1 = 0 have? 0 1 2 3 4

Question

How many real roots does X^4 - 2x^2 - 1 = 0 have? 

0
1
2
3
4


please show work and reasoning! :panel

BumblebeeMan · Answer

Is it 0?

x^4 - 2x^2 - 1 = 0
x^2(x^2 - 2) = 1

I can't find any factors that make x^2(x^2 - 2) equal to one.

Am I missing something here?

laxieqv · Answer

Let Y be X^2 (  Y>=0 )

We have a quadratic equation now
Y^2-2Y-1=0
Delta= (-2)^2-4(-1)*1= 8
The two roots are Y1= (2+2sqrt2)/2 and Y2= (2-2sqrt2)/2
Y2< 0 ----> this root is not accepted
We have now : X^2= Y1 >0 , which has 2 roots.
Thus, 2.

duttsit · Answer

2 roots.

laxieqv mathod is very good.
as per fundamental theorem of algebra: polynomial with degree 4 will have 4 roots.

out of these, as laxieqv shown, 2 are real, 2 imaginary.

BG · Answer

as already done X^2=Y then Y^2-2Y-1=0 solving this gives Y1=1+sqrt2 and Y2=1-sqrt2 or (X^2)1=1+sqrt2 and (X^2)2=1-sqrt2 now second doesn't make sense cause 1-sqrt2 is <0 which is not true since X is squared . So the equation has only 2 real roots

How many real roots does X^4 - 2x^2 - 1 = 0 have? 0 1 2 3 4

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards