Question
If $$p{x^2} + qx + r = 0$$ has no real roots and $$p, q, r$$ are real such that $$p + r > 0$$ then
A.
$$p - q + r < 0$$
B.
$$p - q + r >0$$
C.
$$p + r = q$$
D.
all of these
Answer :
$$p - q + r >0$$
Solution :
Let $$\alpha + i\beta ,\alpha - i\beta $$ be the roots. Then $${\alpha ^2} + {\beta ^2} = \frac{r}{p} > 0.$$ So, $$p, r$$ are of the same sign. Also $$p + r > 0.$$ So, $$p, r$$ are both positive.
If $$q < 0, p - q + r > 0.$$
If $$q > 0,{\left( {p + r} \right)^2} - {\left( {p - r} \right)^2} = 4pr \geqslant {q^2}\left( {\because \,\,{\text{roots are non - real}}} \right).$$
$$\eqalign{
& \therefore \,\,{\left( {p + r} \right)^2} \geqslant {q^2} + {\left( {p - r} \right)^2} \geqslant {q^2} \cr
& \therefore \,\,p + r > q. \cr} $$