Question

3)
If the sum of the roots of the quadratic equation (1/x+p)+ (1/x+q) is zero.Show that the product of the roots is p^2+q^2/2​

Answer 1

$\bf\large\underline\pink{Solution:-}$

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 :

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,

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Answer 2

$\huge\star\mathfrak\green{{Answer:-}}$

My calculations say that the product of roots = -(p^2+q^2)/2. Please comment if you find a mistake.

1/(x+p) + 1/(x+q) = 1/r

Taking LCM of denominators and simplifying:

(x+q+x+p)/((x+p)(x+q)) = 1/r

2xr + pr + qr = x^2 + (p+q)x + pq

x^2 + (p+q-2r)x + pq-r(p+q) = 0

This is a quadratic equation in x

ax^2 + bx + c = 0 is a general form of quadratic equation where

Sum of roots = -b/a

Product of roots = c/a

Now let us assume that the roots of our equation are x1 & -x1.

Now sum of roots = 0

=> p+q-2r = 0 => p+q = 2r

Product of roots = c/a

=> pq-r(p+q)

=> pq - ((p+q)(p+q))/2 {using above result}

=> (2pq - (p^2+q^2+2pq))/2

=> - (p^2+q^2)/2