If roots are equal of quadratic …
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Gaurav Seth 4 years, 9 months ago
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 get
a = 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
=> (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 = 0
Divide each term by pqr , we get
Therefore,
2Thank You