Question

It is given that the polynomial P(x) = x + ax +bx+c and P(x) = 0 has three distinct positive integral roots and P(14) = 231. Let Q(x) = x2 - 2x + 232. It is also given that the polynomial equation P(Q(x))= 0 has no real roots. Then | a | =​

Answer 1

Answer:

Explanation:

If a

2

−2b<0, then the equation has one real and two imaginary roots

Let y=f(x)= x

3

+ax

2

+bx+c=0

dy/dx =f

′

(x)= 3x

2

+2ax+b

Discriminant of f'(x) = 4a

2

−12b=4(a

2

−3b)

If a

2

−3b<0, Then f'(x) has unreal roots and there are no points of local maxima or minima.

Therefore, the graph of f(x) is always increasing and you have only one real root. By Rolle's Theorem.

a

2

−2b<0=>a

2

<2b is a subset of a

2

−3b<0.

Therefore, for a

2

−2b<0, the equation has one real and two imaginary roots.