Question

Let p and q be two real numbers with p>0 .Show that the cubic x3+px+q has exactly one real root .

Answer 1

We have f(x) in $x \in \mathbb{R}$ .

$f(x) = x^3 + px + q \\ => f'(x) = 3x^2 + p$

Since, $p > 0$
$=>f'(x) > 0 \:as\: x^2$ is always non negative.

Hence, f(x) is strictly increasing function.
=> f(x) has atmost one root in x belongs to R.

Since, f(x) is negative for large negative number and positive for large positive number.
So, by Intermediate Mean Value Theorem,
f(x) has atleast one root in x belongs to R.

By above two conditions, f(x) has exactly one root in x belongs to R.

Answer 2

Thank you for asking this question. Here is your answer:

We will Let f(x) = x³ + px + q

f'(x) = 3x²+p which is greater than zero.  

We can conclude form this that f is the increasing function here.

x tends to minus infinity f(x) = - infinity

x tends to infinity f(x)= infinity  

So we can say that there is only 1 real root.

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