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Assume that $a,b,c,d$ are real numbers such that $12a+6b+4c+3d=0$.

Prove that $a+bx+cx^2+dx^3=0$ has a real solution in $(0,1)$.

Note : I have no idea ! How is the assumption even related to the statement that the question wants us to prove? I don't understand.

Second Note ( Edited ) : I know that if $x$ is too large, the equation goes to $+\infty$ and when $x$ is so much below $0$, the equation goes to $-\infty$. Then we can apply mean value theorem and say there exists some point such that on that point, the equation becomes zero. It's ok. But how to show that the root is in $(0,1)$ and what is the use of knowing $12a+6b+4c+3d=0$?

Thanks in advance.

xpaul
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Tina Ashtiani
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3 Answers3

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It turns out that $12a+6b+4c+3d=0$ is the condition that makes Rolle's Theorem applicable. Let $$ f(x)=\int_0^x(a+bt+ct^2+dt^3)dt=ax+\frac b2x^2+\frac b3x^3+\frac d4x^4.$$ Then clearly $f(x)$ is continuous in $[0,1]$, differentiable in $(0,1)$, $f(0)=0$, and $$f(1)=a+\frac b2+\frac b3+\frac d4=\frac{1}{12}(12a+6b+4c+3d)=0.$$ By Rolle's Theorem, there is $\xi\in(0,1)$ such that $f'(\xi)=0$, namely $a+bx+cx^2+dx^3=0$ has a root in $(0,1)$.

Andy Smith
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xpaul
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We know Every cubic equation $dx^3+cx^2+bx+a=0$ with real coefficients and $d\not= 0$ has three solutions (some of which may equal each other if they are real, and two of which may be complex non-real numbers) and at least one real solution. So now we put $d=0$ and we get : $$cx^2+bx+a=0$$ and $$12a + 6b+ 4c =0 \to 6a+3b+2c = 0 \to b = \frac{6a+2c}{-3}$$ Now we want to $\Delta_1 \ge0$ ( to has real root).

$\Delta_1 = b^2-4ac = \frac{36a^2+4c^2+24ac}{9} - 4ac = \frac{36a^2+4c^2-12ac}{9}\ge0\iff 9a^2-3ac+c^2\ge0$

Which it is obvious because $\Delta_2 = -27a^2c^2\le0$

Note : When $\Delta \lt 0 $ quadratic polynomial has same sign as coefficient of $x^2$.(In this case 9) and if $\Delta = 0$ value of quadratic polynomial will be zero at one point and in other points has same sign as coefficient of $x^2$.

S.H.W
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Here are a few things that might help guide you in the right direction:

Let $f(x)$ = $a+bx+cx^2+dx^3$

If $d\neq 0$ :

1) Is $f$ continuous?

2) What is $\lim_{x\to \infty}$$f(x)$? What is $\lim_{x\to -\infty}$$f(x)$?

3) What do we know about the signs of the above two limits, and what can this tell us about the existence of a zero/root?

If $d=0$ , $f$ has degree $\leq2$

How can we show a quadratic (or linear) function has a zero/root?

Andy Smith
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  • I understand what you're saying ... but doesn't it seem so simple ? what's the use of $12a+6b+4c+3d=0$ ? – Tina Ashtiani Dec 17 '16 at 23:21
  • Read my comment. – S.H.W Dec 17 '16 at 23:25
  • If you are asking whether the consequence of the Intermediate Value Theorem seems so simple/obvious, then you are not alone in thinking that. It does seem intuitive that if you have a continuous function that is at some point negative and at another point positive, that there must be a point in between the two such that f is 0 at that point. – Andy Smith Dec 17 '16 at 23:27
  • The additional constraint essentially says that perhaps d=0 (and maybe the other coefficients are zero as well). In the case that d is 0 but c is not 0, we cannot guarantee the existence of a zero using the argument we did for the cubic. This is because the end behavior of a quadratic has the same sign for both ends. – Andy Smith Dec 17 '16 at 23:32
  • If $d=0$, then $12f(\pm\frac1{\sqrt 3})=12a\pm 4\sqrt 3 b+4c=(\pm 4\sqrt 3-6)b$, one is $>0$, one is $<0$ (or both are $=0$) – Hagen von Eitzen Dec 17 '16 at 23:35