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Let $a,b,c,d$ are non-zero real numbers such that $6a+4b+3c+3d=0$. Then the equation $ax^3+bx^2+cx+d=0$ has:

(A) At least one root in $[-2,0]$
(B) At least one root in $[0,2]$
(C) At least two roots in $[-2,2]$
(D) No root in $[-2,2]$

Let $f(x)=ax^3+bx^2+cx+d$

$f(x)$ has at least one root in [-2,0] if $f(-2)f(0)<0$: $$(-8a+4b-2c+d)d<0$$

$f(x)$ has at least one root in [0,2] if $f(2)f(0)<0$: $$(8a+4b+2c+d)d<0$$

$f(x)$ has at least two roots in [-2,2] if $f(2)f(0)>0$: $$(-8a+4b-2c+d)(8a+4b+2c+d)>0$$

Am I right uptil here? I am stuck from hereon.

Community
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user1557
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    "At least one root in an interval [a, b]" is not equivalent to f (a)f (b)<0 (think to the case of 2 roots) – Jean Marie Apr 07 '16 at 09:02
  • Are you simply asked to explore whether each of (A), (B), (C), (D) are always true, always false or sometimes true and sometimes false? Or is this a multiple choice question where exactly one of (A), (B), (C), (D) is supposed to be always true? – almagest Apr 07 '16 at 10:05

2 Answers2

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We show that (B) must be true. We are given $d\ne0$, so switching the signs of all of $a,b,c,d$ if necessary (which does not affect the existence of roots or the relation $6a+4b+3c+3d=0$) we can assume $d>0$ and hence $f(0)>0$. We show that $f(1)>0$ and $f(2)>0$ leads to a contradiction. We have:

$d>0$ (1); $a+b+c+d>0$ (2); $8a+4b+2c+d>0$ (3); and $6a+4b+3c+3d=0$ (4)

(4)-3(2): $3a+b<0$

(1)+(2)+(3)-(4): $3a+b>0$

Contradiction. So we must have either $f(1)<0$ or $f(2)<0$ and either is sufficient to give (B) true.

It is not hard to construct examples where (A), (C), (D) are false.

almagest
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This question can be solved by using Rolle's theorem.

Let's consider a function

$$f(x)=\frac{ax^4}{4}+\frac{bx^3}{3}+\frac{cx^2}{2}+dx=0$$

[Considered so that on differentiating it, we would get our orignal equation ]

$$f(0)=0$$

$$f(2)=4a+\frac{8b}{3}+2c+2c$$

$$ =\frac{2}{3}(6a+4b+3c+3d)$$

$$=0$$

Thus according to Rolle' theorem ,

$$f'(x)=0$$for atleast one value of x between (0,2)

$$f'(x)=ax^3+bx^2+cx+d=0$$will have atleast one real solution between (0,2)

Option (b)

Spandan Kukade
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