If roots are equal of quadratic equation p(q-r)x^2+q(r-p)x+r(p-q)=0 Prove that: 1/p+1/r= 2/q

If roots are equal of quadratic equation p(q-r)x^2+q(r-p)x+r(p-q)=0 Pr

1.	If roots are equal of quadratic equation p(q-r)x^2+q(r-p)x+r(p-q)=0 Prove that: 1/p+1/r= 2/q
Answer» Given:The roots of the quadratic equation :p(q-r)x²+q(r-p)x+r(p-q)=0 are equal.To prove :Solution:Compare given Quadratic equation with ax²+bx+c=0, we geta = p(q-r), b = q(r-p), c = r(p-q)Discreminant (D) = 0/* roots are equal given /=> b²-4ac=0=>[q(r-p)]²-4×p(q-r)×r(p-q)=0=>(qr-pq)²-4pr(q-r)(p-q)=0\xa0=> (qr)²+(pq)²-4(qr)(pq)-4pr(pq-q²-pr+qr)=0=> (qr)²+(pq)²-4pq²r-4p²qr+4prq²+4p²r²-4pqr²=0=> (qr)²+(pq)²+(-2pr)²+2pq²r-4p²qr-4pqr²=0=> (qr)²+(pq)²+(-2pr)²+2(qr)(pq)+2(pq)(-2pr)+2(-2pr)(qr)=0/ we know the algebraic identity//a²+b²+c²+2ab+2bc+2ca=(a+b+c)² */=> (qr+pq-2pr)² = 0=> qr+pq-2pr = 0Divide each term by pqr , we getTherefore,