Question

If p, q, r be in H.P. and p & r be different having same sign, then the root of equation px² + 2qx + r = 0 will be-

• A. Real
• B. Equal
• C. Imaginary
• D. None

Answer 1

Answer: (C)

Imaginary

Answer 2

Here $q = \frac{2pr}{p + r}$

Now, D = 4q² – 4pr

= – 4 [pr – q²]

= $- 4\left\lbrack pr - \frac{4p^{2}r^{2}}{(p + r)^{2}} \right\rbrack$

= $- 4pr\left\lbrack \frac{(p - r)^{2}}{(p + r)^{2}} \right\rbrack$

since pr > 0, p ¹ r

So, D < 0