Question

Let $p,q \in \{ 1,2,3,4\} $. The number of equations of the form $p{x^2} + qx + 1 = 0$ having real roots is:
A. 15
B. 9
C. 7
D. 8

Answer 1

Hint: Here we need to find the total number of real roots for the given quadratic equation $p{x^2} + qx + 1 = 0$ by using the discriminant condition for real root i.e., $D \geqslant 0$.

Complete step-by-step answer:
The equation will have real roots, when it’s discriminant $D \geqslant 0$.
The given equation is $p{x^2} + qx + 1 = 0$.
Discriminant for given equation:
$
  D = {q^2} - 4p \geqslant 0 \\
   \Rightarrow {q^2} \geqslant 4p \\
$
Now put q=1 in the ${q^2} \geqslant 4p$
$1 \geqslant 4p \Rightarrow p \leqslant \dfrac{1}{4}$
Doesn’t have any pair of (p, q) which have real roots.

Now put q=2 in the ${q^2} \geqslant 4p$
$4 \geqslant 4p \Rightarrow p \leqslant 1$
Pair is (1, 2).

Now put q=3 in the ${q^2} \geqslant 4p$
$9 \geqslant 4p \Rightarrow p \leqslant \dfrac{9}{4}$
Pairs are (1,3), (2,3).

Now put q=4 in the ${q^2} \geqslant 4p$
$16 \geqslant 4p \Rightarrow p \leqslant 4$
Pairs are (1, 4), (2, 4), (3, 4), (4, 4).

Now count all the pairs which have the real roots of the given equation, there are 7 pairs of (p, q).
The correct option is C.

Note: By using discriminant rule we can easily find the pair of (x, y) which has real roots. The value of the discriminant decides the nature of the roots.