Question

Under what condition will be root of equation px^+qx+r=0be real and
equal?

(a)q^-4pq=0 (b)p^-4qr=0 (c)q^-4qr=0 (d)q^-4pr=0​

Answer 1

Answer:

$\large\boxed{\sf{(d){q}^{2}-4pr=0}}$

Step-by-step explanation:

Given a quadratic equation:

$\sf{p {x}^{2} + qx + r = 0}$

• We know that , the descriminant must be equal to zero for the roots to be real and equal.

But, descriminant is given by formula,

$\sf{D = {b}^{2} - 4ac}$

But, here we have,

a = p

b = q

c = r

Therefore, we have,

$\sf{= > D =0 }\\ \\ \sf{= > {q}^{2} - 4pr = 0 }\\ \\ \sf{ = > {q}^{2} = 4pr}$

Hence, option (d) is Correct Answer.