First slide

Theory of expressions

Question

Let $a > 0, b > 0$ and $c > 0$ . Then both the roots of the equation
$2 a x^{2} + 3 b x + 5 c = 0$ (1)

Moderate

A

are real and negative
B

have negative real parts
C

have positive real parts
D

none of these

Solution

The discriminant $D$ of the equation (1) is given by
$D = (3 b)^{2} - 4 (2 a) (5 c) = 9 b^{2} - 40 a c$
If $D \geq 0$ , then both the roots of (1) are real.
Also, since $a c > 0, D < 9 b^{2}$
$\Rightarrow \sqrt{D} < 3 b [∵ b > 0]$
Therefore, in this case both the roots $\frac{- 3 b - \sqrt{D}}{4 a}$ and
$\frac{- 3 b + \sqrt{D}}{4 a}$ are negative.
If $D < 0$ , both the roots of (1) are imaginary and are given by
$x = \frac{- 3 b \pm i \sqrt{40 a c - 9 b^{2}}}{2}$
Both these roots have negative real parts.

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