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Let there be a cubic equation $f(x)=ax³+bx²+cx+d=0$, and the coefficients of the equation be related by $-a+b-c+d=3$ and $8a+4b+2c+d=6$. How can I show that the quadratic equation $3ax²+2bx+c-1$ has a root in $(-1,2)$?

I believe that if we apply Rolle's Theorem for the cubic and we show that $f(-1)=f(2)$ (already the cubic is continuous in $[-1,2]$ and differentiable in $(-1,2)$), we can prove that $f'(x)$ has a root in (-1,2). But the given quadratic isn't exactly equal to f'(x).

Can anybody help?

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Take $g(x)=f(x)-x$ and use Rolle.

Because $g(-1)=g(2)=4.$

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  • $\begingroup$ Thanks a lot! It was really silly of me to have not considered/spotted this. $\endgroup$ – Arka Seth Aug 1 '18 at 14:05
  • $\begingroup$ You are welcome! $\endgroup$ – Michael Rozenberg Aug 1 '18 at 14:05
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By Lagrange's theorem, for some $x\in(-1, 2)$: $$f(2)-f(-1)=(2-(-1))f'(x) \implies f'(x)=1$$

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    $\begingroup$ Thanks for the help! $\endgroup$ – Arka Seth Aug 1 '18 at 14:06

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