Quadratic 4

Algebra Level 4

Let a , b a,b and c c be real numbers with a ≠ 0 a\ne 0 . If α \alpha is a root of the equation a 2 x 2 + b x + c = 0 a^2 x^2 + bx + c = 0 , and β \beta is a root of a 2 x 2 − b x − c = 0 a^2x^2-bx-c= 0 and 0 < α < β 0<\alpha <\beta , then the equation a 2 x 2 + 2 b x + 2 c = 0 a^2 x^2 + 2bx + 2c = 0 has a root γ \gamma that satisfies:

γ = α γ = α + β α < γ < β γ =2 α + β

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Radhesh Sarma by Radhesh Sarma , 22, India

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1 solution

James Wilson
Jan 12, 2021

Solving the quadratic equation that α \alpha is a root of, namely, a 2 x 2 + b x + c = 0 a^2x^2+bx+c=0 , I get:

x = − b ± b 2 − 4 a 2 c 2 a 2 x = \frac{-b\pm\sqrt{b^2-4a^2c}}{2a^2}

First note: b 2 − 4 a 2 c ≥ 0. b^2-4a^2c\geq 0. (Note that the other discriminants should be nonnegative as well. I'm just not considering them at the moment.)

Assume b = 0. b=0.

⇒ − 4 a 2 c ≥ 0 ⇒ c ≤ 0. \Rightarrow -4a^2c\geq 0\Rightarrow c\leq 0.

This implies α = − 4 a 2 c 2 a 2 \alpha=\frac{\sqrt{-4a^2c}}{2a^2} because α > 0 \alpha>0 (i.e., we cannot have α = − − 4 a 2 c 2 a 2 \alpha=-\frac{\sqrt{-4a^2c}}{2a^2} ),

and β = 4 a 2 c 2 a 2 \beta=\frac{\sqrt{4a^2c}}{2a^2} (because β > 0 \beta>0 as well).

If c = 0 , c=0, then α = β = 0 , \alpha=\beta=0, which contradicts 0 < α < β . 0<\alpha<\beta.

If c < 0 , c<0, then β \beta is imaginary, which also contradicts 0 < α < β . 0<\alpha<\beta.

Therefore, b ≠ 0. b\neq 0.

Assume b > 0. b>0.

Case 1: α = − b − b 2 − 4 a 2 c 2 a 2 . \alpha=\frac{-b-\sqrt{b^2-4a^2c}}{2a^2}.

α > 0 \alpha >0 leads to:

− b − b 2 − 4 a 2 c > 0 -b-\sqrt{b^2-4a^2c}>0

⇔ − b > b 2 − 4 a 2 c , \Leftrightarrow -b>\sqrt{b^2-4a^2c},

which is impossible. So, α ≠ − b − b 2 − 4 a 2 c 2 a 2 . \alpha \neq \frac{-b-\sqrt{b^2-4a^2c}}{2a^2}.

Case 2: α = − b + b 2 − 4 a 2 c 2 a 2 . \alpha=\frac{-b+\sqrt{b^2-4a^2c}}{2a^2}.

α > 0 \alpha > 0

⇔ − b > − b 2 − 4 a 2 c \Leftrightarrow -b>-\sqrt{b^2-4a^2c}

⇔ b 2 < b 2 − 4 a 2 c \Leftrightarrow b^2 < b^2-4a^2c

⇔ c < 0. \Leftrightarrow c < 0.

I conclude that for b > 0 , b>0, we must have α = − b + b 2 − 4 a 2 c 2 a 2 \alpha =\frac{-b+\sqrt{b^2-4a^2c}}{2a^2} and c < 0 c<0 .

Still assuming b > 0 , b>0, examine the two roots of the equation that β \beta is a root of (namely, a 2 x 2 − b x − c = 0 a^2x^2-bx-c=0 ):

x = b ± b 2 + 4 a 2 c 2 a 2 . x = \frac{b\pm\sqrt{b^2+4a^2c}}{2a^2}.

Assume β = b + b 2 + 4 a 2 c 2 a 2 . \beta = \frac{b+\sqrt{b^2+4a^2c}}{2a^2}.

α < β \alpha < \beta

⇔ − b + b 2 − 4 a 2 c < b + b 2 + 4 a 2 c \Leftrightarrow -b +\sqrt{b^2-4a^2c} < b+\sqrt{b^2+4a^2c}

⇔ b 2 − 4 a 2 c < 2 b + b 2 + 4 a 2 c \Leftrightarrow \sqrt{b^2-4a^2c} < 2b + \sqrt{b^2+4a^2c}

⇔ b 2 − 4 a 2 c < 4 b 2 + b 2 + 4 a 2 c + 4 b b 2 + 4 a 2 c \Leftrightarrow b^2-4a^2c < 4b^2+b^2+4a^2c+4b\sqrt{b^2+4a^2c}

⇔ − 8 a 2 c − 4 b 2 < 4 b b 2 + 4 a 2 c \Leftrightarrow -8a^2c -4b^2 < 4b\sqrt{b^2+4a^2c}

⇔ − b 2 − 2 a 2 c < b b 2 + 4 a 2 c \Leftrightarrow -b^2-2a^2c < b\sqrt{b^2+4a^2c}

Invoking b 2 + 4 a 2 c ≥ 0 , b^2+4a^2c\geq 0, we find − b 2 − 2 a 2 c ≤ 2 a 2 c < 0. -b^2-2a^2c\leq 2a^2c < 0.

That means the inequality α < β \alpha < \beta is true.

Now, assume β = b − b 2 + 4 a 2 c 2 a 2 \beta=\frac{b-\sqrt{b^2+4a^2c}}{2a^2}

α < β \alpha < \beta

⇔ − b + b 2 − 4 a 2 c < b − b 2 + 4 a 2 c \Leftrightarrow -b+\sqrt{b^2-4a^2c}<b-\sqrt{b^2+4a^2c}

⇔ b 2 − 4 a 2 c < 2 b − b 2 + 4 a 2 c \Leftrightarrow \sqrt{b^2-4a^2c}<2b-\sqrt{b^2+4a^2c}

⇔ b 2 − 4 a 2 c < 4 b 2 + b 2 + 4 a 2 c − 4 b b 2 + 4 a 2 c \Leftrightarrow b^2-4a^2c < 4b^2+b^2+4a^2c-4b\sqrt{b^2+4a^2c}

⇔ − b 2 − 2 a 2 c < − b b 2 + 4 a 2 c \Leftrightarrow -b^2-2a^2c < -b\sqrt{b^2+4a^2c}

⇔ b 2 + 2 a 2 c > b b 2 + 4 a 2 c \Leftrightarrow b^2+2a^2c > b\sqrt{b^2+4a^2c}

⇔ b 2 + 4 a 2 b 2 c + 4 a 4 c 2 > b 4 + 4 a 2 b 2 c \Leftrightarrow b^2+4a^2b^2c+4a^4c^2 > b^4 + 4a^2b^2c

⇔ a 4 c 2 > 0 , \Leftrightarrow a^4c^2>0,

which is always true. Therefore, this combination of α \alpha and β \beta also works in the case b > 0. b>0.

It is time to compare γ \gamma to α \alpha and β \beta for the cases A) α = − b + b 2 − 4 a 2 c 2 a 2 \alpha=\frac{-b+\sqrt{b^2-4a^2c}}{2a^2} and β = b + b 2 + 4 a 2 c 2 a 2 \beta=\frac{b+\sqrt{b^2+4a^2c}}{2a^2} and B) α = − b + b 2 − 4 a 2 c 2 a 2 \alpha=\frac{-b+\sqrt{b^2-4a^2c}}{2a^2} and β = b − b 2 + 4 a 2 c 2 a 2 \beta=\frac{b-\sqrt{b^2+4a^2c}}{2a^2} .

Case A: α = − b + b 2 − 4 a 2 c 2 a 2 \alpha=\frac{-b+\sqrt{b^2-4a^2c}}{2a^2} and β = b + b 2 + 4 a 2 c 2 a 2 \beta=\frac{b+\sqrt{b^2+4a^2c}}{2a^2}

We must show that one of the roots of a 2 x 2 + 2 b x + 2 c = 0 a^2x^2+2bx+2c=0 is such that α < x = γ < β . \alpha<x=\gamma<\beta.

Take γ = − 2 b + 4 b 2 − 8 a 2 c 2 a 2 . \gamma = \frac{-2b+\sqrt{4b^2-8a^2c}}{2a^2}.

Otherwise, γ = − 2 b − 4 b 2 − 8 a 2 c 2 a 2 < 0 \gamma =\frac{-2b-\sqrt{4b^2-8a^2c}}{2a^2}<0 (which contradicts γ > 0 \gamma>0 ).

α < γ \alpha<\gamma

⇔ − b + b 2 − 4 a 2 c < − 2 b + 4 b 2 − 8 a 2 c \Leftrightarrow -b + \sqrt{b^2-4a^2c} < -2b + \sqrt{4b^2-8a^2c}

⇔ b + b 2 − 4 a 2 c < 4 b 2 − 8 a 2 c \Leftrightarrow b+\sqrt{b^2-4a^2c} < \sqrt{4b^2-8a^2c}

⇔ b 2 + b 2 − 4 a 2 c + 2 b b 2 − 4 a 2 c < 4 b 2 − 8 a 2 c \Leftrightarrow b^2+b^2-4a^2c+2b\sqrt{b^2-4a^2c}<4b^2-8a^2c

⇔ b b 2 − 4 a 2 c < b 2 − 2 a 2 c \Leftrightarrow b\sqrt{b^2-4a^2c} < b^2-2a^2c

(since c < 0 , c<0, I can square both sides)

⇔ b 4 − 4 a 2 b 2 c < b 4 − 4 a 2 b 2 c + 4 a 4 c 2 \Leftrightarrow b^4-4a^2b^2c < b^4 -4a^2b^2c+4a^4c^2

⇔ 0 < a 4 c 2 , \Leftrightarrow 0 < a^4c^2,

which is always true. Therefore, α < γ \alpha <\gamma when b > 0. b>0.

γ < β \gamma < \beta

⇔ − 2 b + 4 b 2 − 8 a 2 c < b + b 2 + 4 a 2 c \Leftrightarrow -2b+\sqrt{4b^2-8a^2c}<b+\sqrt{b^2+4a^2c}

⇔ 4 b 2 − 8 a 2 c < 3 b + b 2 + 4 a 2 c \Leftrightarrow \sqrt{4b^2-8a^2c}<3b+\sqrt{b^2+4a^2c}

⇔ 4 b 2 − 8 a 2 c < 9 b 2 + b 2 + 4 a 2 c + 6 b b 2 + 4 a 2 c \Leftrightarrow 4b^2-8a^2c<9b^2+b^2+4a^2c+6b\sqrt{b^2+4a^2c}

⇔ − b 2 − 2 a 2 c < b b 2 + 4 a 2 c \Leftrightarrow -b^2-2a^2c < b\sqrt{b^2+4a^2c}

Since − b 2 − 2 a 2 c ≤ 2 a 2 c < 0 -b^2-2a^2c\leq 2a^2c<0 (by b 2 + 4 a 2 c ≥ 0 b^2+4a^2c\geq 0 ), − b 2 − 2 a 2 c < b b 2 + 4 a 2 c -b^2-2a^2c < b\sqrt{b^2+4a^2c} is true.

Therefore, γ < β \gamma <\beta .

Case B: α = − b + b 2 − 4 a 2 c 2 a 2 \alpha=\frac{-b+\sqrt{b^2-4a^2c}}{2a^2} and β = b − b 2 + 4 a 2 c 2 a 2 \beta=\frac{b-\sqrt{b^2+4a^2c}}{2a^2}

α < γ \alpha<\gamma has already been done above.

γ < β \gamma<\beta

⇔ − 2 b + 4 b 2 − 8 a 2 c < b − b 2 + 4 a 2 c \Leftrightarrow -2b+\sqrt{4b^2-8a^2c}<b-\sqrt{b^2+4a^2c}

⇔ 4 b 2 − 8 a 2 c < 3 b − b 2 + 4 a 2 c \Leftrightarrow \sqrt{4b^2-8a^2c}<3b-\sqrt{b^2+4a^2c}

(Note that we cannot have 3 b − b 2 + 4 a 2 c ≤ 0 3b - \sqrt{b^2+4a^2c} \leq 0 because c < 0 c<0 , so we can continue with equivalence)

⇔ 4 b 2 − 8 a 2 c < 9 b 2 + b 2 + 4 a 2 c − 6 b b 2 + 4 a 2 c \Leftrightarrow 4b^2-8a^2c < 9b^2+b^2+4a^2c-6b\sqrt{b^2+4a^2c}

⇔ b b 2 + 4 a 2 c < b 2 + 2 a 2 c \Leftrightarrow b\sqrt{b^2+4a^2c} <b^2+2a^2c

⇔ b 4 + 4 a 2 b 2 c < b 4 + 4 a 2 b 2 c + 4 a 4 c 2 \Leftrightarrow b^4+4a^2b^2c < b^4 + 4a^2b^2c+4a^4c^2

⇔ 0 < a 4 c 2 , \Leftrightarrow 0<a^4c^2,

which is a true statement. Therefore, γ < β . \gamma<\beta.

Now, consider b < 0. b<0.

Case 1: α = − b − b 2 − 4 a 2 c 2 a 2 . \alpha = \frac{-b-\sqrt{b^2-4a^2c}}{2a^2}.

α > 0 \alpha > 0

⇔ − b > b 2 − 4 a 2 c \Leftrightarrow -b > \sqrt{b^2-4a^2c}

⇔ b 2 > b 2 − 4 a 2 c \Leftrightarrow b^2 > b^2-4a^2c

⇔ c > 0. \Leftrightarrow c >0.

Subcase I: β = b − b 2 + 4 a c 2 a 2 . \beta = \frac{b-\sqrt{b^2+4a^c}}{2a^2}.

α < β \alpha < \beta

⇔ − b + b 2 − 4 a 2 c < b − b 2 + 4 a 2 c \Leftrightarrow -b+\sqrt{b^2-4a^2c}<b-\sqrt{b^2+4a^2c}

⇔ b 2 − 4 a 2 c < 2 b − b 2 + 4 a 2 c \Leftrightarrow \sqrt{b^2-4a^2c} < 2b-\sqrt{b^2+4a^2c}

b 2 − 4 a 2 c < 2 b − b 2 + 4 a 2 c \sqrt{b^2-4a^2c} < 2b-\sqrt{b^2+4a^2c} is impossible, so we cannot have this combination of α \alpha and β . \beta.

Subcase II: β = b + b 2 + 4 a 2 c 2 a 2 . \beta = \frac{b+\sqrt{b^2+4a^2c}}{2a^2}.

α < β \alpha < \beta

⇔ − b − b 2 − 4 a c < b + b 2 + 4 a 2 c \Leftrightarrow -b-\sqrt{b^2-4a^c} < b+\sqrt{b^2+4a^2c}

⇔ − b 2 − 4 a 2 c < 2 b + b 2 + 4 a 2 c \Leftrightarrow -\sqrt{b^2-4a^2c} < 2b+\sqrt{b^2+4a^2c}

Since − 3 b 2 < 4 a 2 c -3b^2<4a^2c is true (by noting c > 0 c>0 ) and is equivalent to 2 b + b 2 + 4 a 2 c > 0 , 2b+\sqrt{b^2+4a^2c}>0, we can conclude α < β \alpha<\beta .

Take γ = − b + 4 b 2 − 8 a 2 c 2 a 2 . \gamma = \frac{-b+\sqrt{4b^2-8a^2c}}{2a^2}.

α < γ \alpha<\gamma

⇔ − b − b 2 − 4 a 2 c < − 2 b + 4 b 2 − 8 a 2 c \Leftrightarrow -b-\sqrt{b^2-4a^2c}<-2b+\sqrt{4b^2-8a^2c}

⇔ − b 2 − 4 a 2 c < − b + 4 b 2 − 8 a 2 c , \Leftrightarrow -\sqrt{b^2-4a^2c}<-b+\sqrt{4b^2-8a^2c},

which is a true statement (since b < 0 b<0 ). That is, α < γ \alpha<\gamma has been shown.

γ < β \gamma<\beta

⇔ − 2 b + 4 b 2 − 8 a 2 c < b + b 2 + 4 a 2 c \Leftrightarrow -2b+\sqrt{4b^2-8a^2c}<b+\sqrt{b^2+4a^2c}

⇔ 4 b 2 − 8 a 2 c < 3 b + b 2 + 4 a 2 c \Leftrightarrow \sqrt{4b^2-8a^2c} < 3b+\sqrt{b^2+4a^2c}

⇔ 4 b 2 − 8 a 2 c < 9 b 2 + b 2 + 4 a 2 c + 6 b b 2 + 4 a 2 c \Leftrightarrow 4b^2-8a^2c<9b^2+b^2+4a^2c+6b\sqrt{b^2+4a^2c}

⇔ − 6 b 2 − 12 a 2 c < 6 b b 2 + 4 a 2 c \Leftrightarrow -6b^2-12a^2c<6b\sqrt{b^2+4a^2c}

⇔ b 2 + 2 a 2 c > − b b 2 + 4 a 2 c \Leftrightarrow b^2+2a^2c>-b\sqrt{b^2+4a^2c}

⇔ b 4 + 4 a 2 b 2 c + 4 a 4 c 2 > b 4 + 4 a 2 b 2 c \Leftrightarrow b^4+4a^2b^2c+4a^4c^2>b^4+4a^2b^2c

⇔ a 4 c 2 > 0 , \Leftrightarrow a^4c^2>0,

which is a true statement. Therefore, γ < β \gamma<\beta .

Case 2: α = − b + b 2 − 4 a 2 c 2 a 2 . \alpha = \frac{-b+\sqrt{b^2-4a^2c}}{2a^2}.

α > 0 \alpha>0

⇔ − b > − b 2 − 4 a 2 c , \Leftrightarrow -b > - \sqrt{b^2-4a^2c},

which is true, since b < 0. b<0.

Subcase 1: β = b − b 2 + 4 a c 2 a 2 . \beta = \frac{b-\sqrt{b^2+4a^c}}{2a^2}.

α < β \alpha < \beta

⇔ − b + b 2 − 4 a c < b − b 2 + 4 a 2 c \Leftrightarrow -b+\sqrt{b^2-4a^c}<b-\sqrt{b^2+4a^2c}

⇔ b 2 − 4 a 2 c < 2 b − b 2 + 4 a 2 c , \Leftrightarrow \sqrt{b^2-4a^2c}<2b-\sqrt{b^2+4a^2c},

which is impossible. Therefore, this pair of α \alpha and β \beta do no satisfy α < β \alpha < \beta (for b < 0 b<0 ).

Subcase 2: β = b + b 2 + 4 a c 2 a 2 . \beta=\frac{b+\sqrt{b^2+4a^c}}{2a^2}.

α < β \alpha<\beta

⇔ − b + b 2 − 4 a 2 c < b + b 2 + 4 a 2 c \Leftrightarrow -b+\sqrt{b^2-4a^2c}<b+\sqrt{b^2+4a^2c}

⇔ − 2 b + b 2 − 4 a 2 c < b 2 + 4 a 2 c \Leftrightarrow -2b+\sqrt{b^2-4a^2c}<\sqrt{b^2+4a^2c}

⇔ 4 b 2 + b 2 − 4 a 2 c − 4 b b 2 − 4 a c 2 < b 2 + 4 a 2 c \Leftrightarrow 4b^2+b^2-4a^2c-4b\sqrt{b^2-4ac^2}<b^2+4a^2c

⇔ − 4 b b 2 − 4 a 2 c < − 4 b 2 + 8 a 2 c \Leftrightarrow -4b\sqrt{b^2-4a^2c}<-4b^2+8a^2c

The inequality, − 4 b b 2 − 4 a 2 c < − 4 b 2 + 8 a 2 c , -4b\sqrt{b^2-4a^2c}<-4b^2+8a^2c, is impossible because the discriminant of a 2 x 2 + 2 b x + 2 c a^2x^2+2bx+2c must be nonnegative.

Therefore, I have shown that in every case, where the given conditions are satisfied, α < γ < β . \alpha<\gamma<\beta.

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