Question

for cube polynomial t(x)=px3+px2+rx+s, which of the following is always true​

Answer 1

Answer:

Correct option is

A

If p

2

−2q<0, then the equation has one real and two imaginary root.

Let f(x)=x

3

+px

2

+qx+r

∴f

′

(x)=3x

2

+2px+q

Disc =4p

2

−12q=4(p

2

−3q)

=4(p

2

−2q−q)

∴ If p

2

<2q⇒p

2

<3q

So, the equation f(x)=0 has one real and two imaginary roots.

Step-by-step explanation: