Number of roots (part - 3)

Algebra Level 4

If a a is a constant, then the equation x 2 + a ∣ x ∣ + 1 = 0 x^2 + a|x| + 1 = 0 has four distinct roots when __________ \text{\_\_\_\_\_\_\_\_\_\_} .

ϕ \phi a < − 2 a < -2 a = − 2 a = -2 None of these − 2 < a < 0 -2 < a < 0 a ≥ 0 a \geq 0

Akhil Bansal by Akhil Bansal , 22, India

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3 solutions

Yashas Ravi
Sep 2, 2020

The answer is a < − 2 a < -2 and not a > 2 a > 2 because ∣ x ∣ > 0 |x|>0 . a ∣ x ∣ = − 1 − x 2 a|x| = -1 - x^2 and we know that − 1 − x 2 > 0 -1 - x^2 > 0 for all real x x . This means that a a has to be negative, so a < − 2 a < -2 .

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We must have a 2 ≥ − 4 a^2\ge -4 for real roots. But to have distinct roots a 2 > 4. ⟹ a^2>4.\ \implies x< - 2 or x>2. But the answer option has given only one part. That being the best of option answer..

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Avi Agrawal
Feb 19, 2016

Its very simple..

For distinct real roots.. D > 0 a^2 - 4(1) > 0 |a| > 2 a > 2 and a < -2

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It's a bit confusing, because the right answer is just part of the correct solution.

Imi Mali - 5 years, 3 months ago

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The actual ans. is a < (-2) and a > 2... but as it is not given in the options.. we have to ans. it according to the options..

Avi Agrawal - 5 years, 3 months ago

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No , the answer is a < 2 a < 2 .

The option is perfectly right.!

Ankit Kumar Jain - 4 years, 2 months ago

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