Question

if the equation 2xsquare-6x+p=0 has real and different roots then the values of p are given by ​

Answer 1

Answer:

==> p < 36/8 or p < 4 or -∞ < p < 4

Explanation:

given 2x^2 - 6x + p = 0 also given it has real and different roots.

that mean D>0 i.e;

b^2 - 4ac > 0 comparing the coefficients with ax^2 + bx + c we get:

b = -6 ; a = 2; c = p

-> (-6)^2 - 4(2)(p) >0

-> 36-8p > 0

-> 8p < 36

==> p < 36/8 or p < 4 or -∞ < p < 4