An algebra problem by A Former Brilliant Member

Algebra Level 3

If $a$ , $b$ , and $c$ are real numbers such that $a+b+c=0$ , then the equation $3ax^2+2bx+c=0$ can not have any root in $[0,1]$ .

True or false?

False True

3 solutions

Aaghaz Mahajan
May 6, 2019

Let $f\left(x\right)=ax^3+bx^2+cx$

Observe that $f\left(x\right)$ has roots 0 and 1.

Now, simply apply Rolle's Theorem on the function to get the answer.

Nicely written, I didn't expect Rolle's to pop up here.

Pi Han Goh - 2 years, 1 month ago

Thanks Sir!!! And a bigger THANKS for helping me learn $\LaTeX$

Aaghaz Mahajan - 2 years, 1 month ago

My pleasure. I've been enjoying many of your recent solutions lately. They are quite enjoyable.

Pi Han Goh - 2 years, 1 month ago

Great solution, really elegant. Note that it implies that, in fact, $3ax^2+2bx+c=0$ always has a root in the interval $[0,1]$ when $a+b+c=0$ (which makes me wonder if that is what the question should have been).

Chris Lewis - 2 years, 1 month ago

Chew-Seong Cheong
May 7, 2019

Let $(a, b, c) = \left(0, \dfrac 12, - \frac 12\right)$ . Then $a+b+c = 0$ . And

$\begin{aligned} 3ax^2+2bx+ c & = 0 \\ x - \frac 12 & = 0 \\ \implies x & = \frac 12 \in [0,1] \end{aligned}$

False , the equation is a root in $[0, 1]$ .

Dan Czinege
May 6, 2019

If c = 0 and b=-a. Our equation will look like this: $3ax^2-2ax=0$ and this is equivalent with $x(3ax-2a)=0$ and this equation has got two roots: x=0 and $x=2/3$ . And these are from interval [0,1] so the statement is false.

An algebra problem by A Former Brilliant Member

3 solutions

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