Question

If roots of the quadratic equation px2-2√ 5 px+15 = 0 are real, then find the value of p.

Answer 1

Answer:

For real roots, discriminant must be greater than 0 or equal to 0.

Discriminant, for ax^2 + bx + c = 0, is given by b^2 - 4ac.

Here, b = -2

$\sqrt5 p$

c = 15

a = p

= > discrimiant > 0

= > (-2$\sqrt5$)^2 - 4(15)p > 0

= > 20 - 60p > 0

= > 1/3 > p

Solution set for the values of p is { ...0,1/9...1/4} or accurately, (-infinity,1/3 ).