Question

find value of p for which equation x2-px+p+3=0 have real and distinct root​

Answer 1

Answer:

Any p < -2 or p > 6.

Step-by-step explanation:

The equation has real and distinct roots when the discriminant is greater than zero.

The discriminant of ax²+bx+c is b²-4ac.

So the discriminant of x²-px+(p+3) is

 p² - 4 ( p + 3 ) = p² - 4p - 12 = ( p - 6 ) ( p + 2 ).

This is positive when p < -2 and when p > 6.

Hope this helps!