First slide

Theory of equations

Question

The equation $x^{3} - 6 x^{2} + 15 x + 3 = 0$ has

Moderate

A

only one positioe root
B

two positive and one negative roots
C

no positive root
D

none of these

Solution

Let $f (x) = x^{3} - 6 x^{2} + 15 x + 3 . T$ Then,
$\begin{aligned} f^{'} (x) = 3 x^{2} - 12 x + 15 = 3 (x^{2} - 4 x + 5) \\ \Rightarrow f^{'} (x) = 3 \{(x - 2)^{2} + 1\} > 0 for all x \in R \end{aligned}$
$\Rightarrow f (x)$ is strictly increasing on R.
Also $f (0) = 3 > 0$
Thus $f (x) > 0$ for all $x > 0$
Hence, f (x) has no positive root.

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