First slide

Theory of expressions

Question

If $a_{0}, a_{1}, a_{2}, a_{3}$ are all positive, then $4 a_{0} x^{3} + 3 a_{1} x^{2} + 2 a_{2} x + a_{3} = 0$ has at least one root in (-1, 0) if

Difficult

A

$a_{0} + a_{2} = a_{1} + a_{3} and 4 a_{0} + 2 a_{2} > 3 a_{1} + a_{3}$
B

$4 a_{0} + 2 a_{2} < 3 a_{1} + a_{3}$
C

$4 a_{0} + 2 a_{2} = 3 a_{1} + a_{3} and a_{0} + a_{2} < a_{1} + a_{3}$
D

None of these

Solution

$P (x) = 4 a_{0} x^{3} + 3 a_{1} x^{2} + 2 a_{2} x + a_{3}$ is a polynomial, so it is continuous for all x
$\begin{array}{l} P (x) = 0 has a root in (- 1, 0) \\ P (- 1) P (0) < 0 \\ now, P (0) = a_{3} > 0 \\ \Rightarrow P (- 1) = - 4 a_{0} + 3 a_{1} - 2 a_{2} + a_{3} < 0 \\ \Rightarrow 4 a_{0} + 2 a_{2} > 3 a_{1} + a_{3} \end{array}$
Also using Rolle's theorem,
$Q (x) = \int p (x) dx = a_{0} x^{4} + a_{1} x^{3} + a_{2} x^{2} + a_{3} x + a$
For $Q^{'} (x) = p (x) = 0$
for at least one root in
$\begin{array}{ll} Q (- 1) = Q (0) \\ \Rightarrow a_{0} - a_{1} + a_{2} - a_{3} + d = d \\ or a_{0} + a_{2} = a_{1} + a_{3} \end{array}$

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