Let's see what you make of the Discriminant

Algebra Level 4

Let α \alpha and β \beta be the roots (of x x ) of the polynomial with real coefficients a x 2 + b x + c = 0 ax^2+bx+c=0 .

Given that α + β , α 2 + β 2 \alpha+\beta, \alpha^2+\beta^2 and α 3 + β 3 \alpha^3+\beta^3 are in a geometric progression (in that order), then what is the strictest conclusion that we can draw about the quadratic discriminant , D = b 2 − 4 a c D = b^2-4ac ?

D = 0 D=0 D > 0 D>0 D < 0 D<0 D ≥ 0 D\geq0 D ≤ 0 D\leq 0 D can be anything No conclusion can be drawn

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Skanda Prasad by Skanda Prasad , 20, India

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1 solution

Andy Hayes
Jan 3, 2017

It's given that α + β , \alpha+\beta, α 2 + β 2 , \alpha^2+\beta^2, and α 3 + β 3 \alpha^3+\beta^3 are in a geometric progression. Then,

α 2 + β 2 α + β = α 3 + β 3 α 2 + β 2 . \frac{\alpha^2+\beta^2}{\alpha+\beta}=\frac{\alpha^3+\beta^3}{\alpha^2+\beta^2}.

These ratios must be defined. Thus, α \alpha and β \beta cannot both be 0 , 0, α ≠ − β , \alpha \ne -\beta, and α ≠ ± i β . \alpha \ne \pm i \beta.

Without loss of generality, let β ≠ 0 \beta \ne 0 and let q = α β . q=\frac{\alpha}{\beta}. Then,

q 2 + 1 q + 1 = q 3 + 1 q 2 + 1 ( q 2 + 1 ) 2 = ( q 3 + 1 ) ( q + 1 ) q 4 + 2 q 2 + 1 = q 4 + q 3 + q + 1 q 3 − 2 q 2 − q = 0 q ( q − 1 ) 2 = 0 \begin{aligned} \frac{q^2+1}{q+1} &= \frac{q^3+1}{q^2+1} \\ \\ (q^2+1)^2 &=(q^3+1)(q+1) \\ \\ q^4+2q^2+1 &= q^4+q^3+q+1 \\ \\ q^3-2q^2-q &= 0 \\ \\ q(q-1)^2 &= 0 \end{aligned}

Then, q = 0 q=0 or q = 1. q=1.

Case 1 : q = 0. q=0. Then α = 0 , \alpha=0, and β \beta can be any number. The original polynomial equation can be expressed as a x 2 + b x = 0. ax^2+bx=0. Given that b b is a real number, the discriminant is greater than 0.

Case 2 : q = 1. q=1. Then α = β . \alpha=\beta. The discriminant is equal to 0.

Thus, D ≥ 0. D\ge 0.

3 Helpful 0 Interesting 0 Brilliant 0 Confused

(We have a GP if both terms are 0, which gives D=0. For all other cases, without loss of generality, β ≠ 0 \beta \neq 0 .) We could simplify the algebra by setting r = α β r = \frac{ \alpha } { \beta } . Then we are given that ( 1 + r 2 ) 2 = ( 1 + r ) ( 1 + r 3 ) ( 1 + r^2 )^2 = ( 1 + r) ( 1 + r^3) , or that 2 r 2 = r + r 3 2r^2 = r + r^3 . Thus r = 0 , 1 r = 0, 1 are solutions.

(Checking that we didn't introduce extraneous roots in the problem,) Notice that " α = 0 \alpha = 0 and β \beta being anything" and " α = β \alpha = \beta " does indeed satisfy the original conditions. Thus, we have D ≥ 0 D \geq 0 .

The error that you made in the solution was in dividing out by α β \alpha \beta , without checking that this term is non-zero.

Also, note that "discriminant implication on roots" only applies if the coefficients of the polynomial are real, which isn't stated in the question. Under this interpretation, we could have "All are possible", though I think the best way to fix it is to add the assumption in.

Calvin Lin Staff - 4 years, 5 months ago

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I've edited the problem for clarity. Can you make the corresponding edits to the problem?

Calvin Lin Staff - 4 years, 5 months ago

Ah, I had the thought of α + β \alpha+\beta being 0, but for some reason I didn't think of α β \alpha\beta being 0. I'll edit my solution so that it is correct.

Andy Hayes - 4 years, 5 months ago

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