Question

If equation PX =ax2 + bx + c an QX =-ax2 + bx + c then show that PX.QX= must have 2 real roots.

Answer 1

Hii brother!!!!!

I am here to help u

  Roots of equation P(x).Q(x) =0  will be the roots of equation i
e

     P(x)=0 and Q(x)=0
Let D1, be the discriminant of P(x)
let D2, be the discriminant of Q(x)

now. ..D1 +D2 = (b2 -4ac) + (b2 + 4ac) = 2b2 >=0

So at least one of D1 and D2 must be greater than or equal to zero.

So..... at least one of the equations P(x)=0, and Q(x)=0 has real roots.

......Hence the equation P(x).Q(x)=0 has at least two real roots.

Got it brother

Answer 2

heyyyy frnd

see here
if u take discriminent of both then add those two.
it will come 2b^2 >=0
hence the given equation will have to real roots