In the equation $x^2 + kx + 1 = 0$ , $x$ is a variable and $k$ is a constant. If the quadratic equation has two distinct real roots, which of the following could be true?
I. $k = 0$
II. $k = -1$
III. $k = -3$

Answer Choices

Correct answer marked below

I only

III only

Correct

I and II only

II and III only

I and III only

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Comments

HarshalaAug 4

Discriminant has to be positive, so k^2 - 4 > 0. Only one of the three options clears that bar. Took me a second to remember the a=1, c=1 setup.

ColinAug 13

The c=1 is what saves time here.

AmelSep 19

This felt more like a 650 to me, pretty quick once you set up k^2 - 4 > 0.

TriJun 9

Quick question — for k = -1, the discriminant comes out to -3, so that's no real roots at all, right? Just want to make sure I'm not missing something.

GaganJun 12

Right, and k=0 gives -4, also out.

RaghavSep 28

Nice little inequality trap. II looks tempting because it's a clean integer, but it fails. Only III works.

What this question tests

What it tests

Your fluency with the order of operations, fractions, decimals, and basic number sense — the foundation every quant question leans on.

Common trap

Applying the order of operations out of sequence or rounding intermediate values before the final step.

In the equation x2+kx+1=0x^2 + kx + 1 = 0x2+kx+1=0, xxx is a variable and kkk is a constant. If the quadratic equation has two distinct real roots, which of the following could be true?I. k=0k = 0k=0II. k=−1k = -1k=−1III. k=−3k = -3k=−3