Question

what is
the value of p for which the equation 9x2-px+4=0 has real roots​

Answer 1

Answer:

P = ±12

Step-by-step explanation:

For real roots , D = 0.

Let , a = 9 ,b = -p , c = 4

by using rule discriminant rule .

D = b ^ 2 - 4 ac

=> (-p) ^2 - 4.9.4 = 0

=> p^2 - 144 = 0

=> p^2 = 144

=> p = √144

=> p = $±12$

Case -2-

when D > 0

= b^2 - 4.a.c > 0

= (-p)^2 - 4.9.4 > 0

= (p)^2 - 144 > 0

= p^2 > 144

= P > √144

= p > ±12

p is greater than ±12.

For ,more see attachment.