Question

Hey guys...Solve this please...Best Answer will be marked brainliest. Detailed answer required....☺️

The answer is given as (A) and (D). Anyone please explain how?
Kindly don't spam.❌​

Answer 1

Here is ur answer,

Let the two roots of the quadratic equation x^2+px+q=0 be a and b and p^2>4q.....(i)

And let the roots of the equation x^2-rs+x be a^4 And b^4

And, a+b=-p, ab=q, and a^4+b^4= (a^2+b^2)-2a^2b^2

                                                     => [(a+b)^2-2ab)]-2(ab)^2

r= (p^2-2q)^2-2q^2= p^4-4p^2q+4q^2-2q^2= p^4-4p^2q+2q^2

=> 2q^2-r= p^2(4q-p^2)<0, since p^2>0 and 4q-p^2<0[from (i)]

Therefore, x^2-4px+2q^2-r=0

For the above eqn to have real and distinct roots, => D>0

D= 16p^2-4(2q^2-r)=  16p^2-4(negative value)

Therefore, D>0..........