Question

if the roots of the equation p(q-r)x^+q(r-p)x+(p-q)=0 are equal ,show that 1/p+1/r=2/q​

Answer 1

Answer:

1/p+1/r=1/q

Step-by-step explanation:

The roots of the quadratic equation :

p(q-r)x²+q(r-p)x+r(p-q)=0 are equal.

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

qr/pqr+pr/pqr-2pr/pqr=0

1/p+1/r-2/q=0

=1/p+1/r=1/q