If the roots α, β of the equation px 2+ qx + r =0 are real and of opposite sign (where p , q , r are real coefficient), then the roots of the equation α(x-β)2+β(x-α)2=0 are

Question

If the roots α, β of the equation px 2+ qx + r =0 are real and of opposite sign (where p , q , r are real coefficient), then the roots of the equation α(x-β)2+β(x-α)2=0 are (A) positive (B) negative (C) real and of opposite sign (D) imaginary

Tardigrade Admin · Accepted Answer

The equation α( x -β)2+β( x -α)2=0 has the product of its roots given by (α β(α+β)/(α+β))=α β<0 text (Given) [(α+β) x2-2 x(α β+α β)+α β(α+β)=0] text i.e. (α+β) x2-4 α β x+α β(α+β)=0 Also the discriminant 16 α2 β2-4 α β(α+β)2=-4 α β[(α+β)2-4 α β]=-4 α β(α-β)2>0 So, the roots are of opposite sign and are real.

Q. If the roots α,β of the equation px2+qx+r=0 are real and of opposite sign (where p,q,r are real coefficient), then the roots of the equation α(x−β)2+β(x−α)2=0 are

positive

negative

real and of opposite sign

imaginary

Q. If the roots $α, β$ of the equation $p x^{2} + q x + r = 0$ are real and of opposite sign (where $p, q, r$ are real coefficient), then the roots of the equation $α (x - β)^{2} + β (x - α)^{2} = 0$ are