If p,q and r are real numbers, then roots of the equation (x-p)(x-q) + (x-q)(x-r) + (x-p)(x-r) = 0 are equal if

Dear Student,(x-p)(x-q) + (x-q)(x-r) + (x-p)(x-r)=0x2-xq-xp+pq+x2-xr-xq+rq+x2-xr-xp+pr=03x2-x(q+p+r+q+r+p)+pq+pr+qr=03x2-2x(q+p+r)+pq+pr+qr=0Now, we know that the roots of the equation are equal if the discriminant is zero.i.e.4(p+q+r)2-12pq+pr+qr=04(p2+q2+r2+2pq+2qr+2pr)-12pq+pr+qr=04p2+4q2+4r2+8pq+8qr+8pr-12pq-12pr-12qr=04p2+4q2+4r2-4pq-4qr-4pr=02p2+q2+r2-pq-qr-pr=0p2+q2+r2-pq-qr-pr=0Regards

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