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

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...
MARK MY ANSWER BRAINIEST TOO

Answer 2

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

Divide each term by pqr , we get

$\implies\frac{qr}{pqr}+\frac{pq}{pqr}-\frac{2pr}{pqr}=0$

$\implies \frac{1}{p}+\frac{1}{r}-\frac{2}{q}=0$

Therefore,

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

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