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

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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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Answer 2

p and q be two real numbers with p > 0
Let f(x) = x³ + px + q
differentiate f(x) , with respect to x,
f'(x) = 3x² + p , here p >0
so, f'(x) > 0 for all real value of x.
means, f(x) is strictly increasing function.

now we also see,
if $x\to\infty$ , f(∞) = ∞³ + p∞ + q = ∞ > 0
and if $x\to-\infty$, f(-∞) = -∞³ -p∞ + q = -∞ < 0
here, we get , f(∞) > 0 and f(-∞) < 0 and also f(x) is strictly increasing function.
so, we conclude that f(x) has exactly one real number where f(x) = 0,

hence, x³ + px + 1 has exactly one real root .