Question

Let f (x) = px 2 + qx - 4 p, where p 0. Find the number of roots for the equation f (x) = 0 . Justify your answer.

Answer 1

f(x) = px² + qx - 4p , where p ≠ 0

we have to find number of roots for equation f(x) = 0

given, expression f(x) = px² + qx - 4p is a two degree polynomial function. so, maximum number of roots can be possible = 2.

here, you didn't mention about what is p and q , if we assume p and q are real numbers then, two real roots are possible.

i.e., px² + qx - 4p = 0

from formula,

x = $\frac{-q\pm\sqrt{q^2+16p^2}}{2p}$

so, there are two possible roots and these are $\frac{-q\pm\sqrt{q^2+16p^2}}{2p}$