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Let $a,b,c \in \mathbb R$ such that no two of them are equal and satisfy $$\det\begin{bmatrix}2a&b&c\\b&c&2a\\c&2a&b\end{bmatrix} = 0 ,$$ then the equation $24ax^2 + 4bx +c=0$ has:

a) atleast one root in $[0,\frac 12]$

b) at least one root in $[-\frac 12, 0)$

c) at least one root in $[1,2]$

d) none of these

On solving the determinant equation I have obtained $2a+b+c =0$ but I am not sure how that is of any help. because I am uanble to transform the given equation in $x$ to $2a+b+c= 0$ for any $x$. How do I proceed?

Edit:

Using @AnotherJohnDoe's suggestion, if we let $c=0$, we get $b=-2a$ after solving the determinant equation.

On substituting these values into the quadratic equation, we get:

$-12bx^2 + 4bx = 0$

which has roots $\dfrac 13$ and $0$ so we get option A/D. But there doesn't seem to be a way to eliminate option D. So there must be another solution to this question?

Paul Frost
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Archer
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    $2a+b+c$ is the trace of that matrix, not its determinant, which is $-8 a^3 + 6 a b c - b^3 - c^3$. – lhf Sep 20 '18 at 12:17
  • @Ihf $-8a^3 -b^3 -c^3 = -6abc \implies 2a + b+ c =0$ – Archer Sep 20 '18 at 12:26
  • @Ihf I never said that $2a+b+c$ is the value of the determinant. – Archer Sep 20 '18 at 12:26
  • Could you explain how you got $2a+b+c=0$? WA tells me that $-8 a^3 + 6 a b c - b^3 - c^3=-(2 a + b + c) (4 a^2 - 2 a b - 2 a c + b^2 - b c + c^2)$ – lhf Sep 20 '18 at 12:41
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    @lhf This quadratic factor is irreducible over the reals... It only remains $2a+b+c =0$. – Alan Muniz Sep 20 '18 at 12:44
  • Let $A=2a$, then the determinant has the one factor $A+b+c$, the other being $A^2+b^2+c^2-Ab-bc-cA\ge 0$, equality when $A=b=c$, excluded... (There is also a $(-1)$ factor.) – dan_fulea Sep 20 '18 at 12:44
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    We have an explicit formula for the roots in terms of $a,b$, $c$ being given by $-b-2a$. Then we can further norm by homogenity $a=1$. It remains to examine the function $b\to (-b\pm\sqrt{b^2+6b+12})/12$. (Or something similar, it was a quick computation. Letting $b\to 0_+$ excludes the case (c). Letting $b\to\pm\infty$...) – dan_fulea Sep 20 '18 at 12:52
  • If this holds for arbitrary values of $a,b,c$, why not let $c=0?$ –  Sep 20 '18 at 13:09
  • @AnotherJohnDoe See my edit. – Archer Sep 20 '18 at 13:32
  • Isn't option $A$ then the answer? –  Sep 20 '18 at 13:37
  • @AnotherJohnDoe it does not exclude the possibility of option D. – Archer Sep 20 '18 at 13:50
  • Hmm, the question reads as - If $a,b,c$ satisfy $2a+b+c=0$, then it satisfies one of options $A$ through $D$. Doesn't it then follow that $A$ must be the solution? –  Sep 20 '18 at 13:56
  • The quadratic polynomial $f(x)= 24 a x^2+4 b x+c$ has both roots greater than $1/2$ if and only if $a f(1/2)>0 $ and the sum of the roots is greater than $1$. Similarly, it has two negative roots if and only if $a f(0)>0 $ and the sum of the roots is negative. Show neither is possible so $f$ must have a root between $0$ and $1/2$ – Lozenges Sep 20 '18 at 14:14
  • @AnotherJohnDoe We have just checked for c=0 . How can we claim Option A is correct for all possible combinations of a b and c? – Archer Sep 20 '18 at 14:15
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    I was saying that the phrasing of the question suggested one of the four options was true for all valid values of $a,b,c$. Since it was true for a particular case, it would them imply it being true for all valid values –  Sep 20 '18 at 15:14

1 Answers1

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Let $f(x)=24ax^2+4bx+c$

Suppose $f(0)\gt0$. Then it follows $c\gt0$. Now, $f(\frac 12)=6a+2b+c=2a-c$. If $f(\frac 12)\gt0,$ then $2a\gt c\gt 0.$

Then $f(\frac 14)=\frac{24}{16}a+b+c=-\frac 12a<0.\\$

Similarly, if $f(0)\lt 0$, then $c\lt 0$. If $f(\frac 12)=2a-c\lt0$, then $2a\lt c\lt 0\Rightarrow f(\frac 14)=-\frac 12\gt 0$.