Question

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

Answer 1

Answe1

Secondary School Math 5 points

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

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Answers

jainparul1973

jainparul1973 Ambitious

First find the discriminant of the given quadratic equation by formula (b^2 -4ac)

also d=0

so the equation will come.

then solve the equation...

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Comments Report

mysticd

mysticd Genius

Answer:

\green {\frac{1}{p}+\frac{1}{r}=\frac{2}{q}}

Step-by-step explanation:

Given:

The roots of the quadratic equation :

p(q-r)x²+q(r-p)x+r(p-q)=0 are equal.

To prove :

\frac{1}{p}+\frac{1}{r}=\frac{2}{q}

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