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Question: Let $p$, $q$ and $r$ be real numbers ($p \neq q, r \neq 0$), such that the roots of the equation $\f...

Let pp, qq and rr be real numbers (p≠q,r≠0p \neq q, r \neq 0), such that the roots of the equation 1x+p+1x+q=1r\frac{1}{x+p} + \frac{1}{x+q} = \frac{1}{r} are equal in magnitude but opposite in sign, then the sum of squares of these roots is equal to:

A

p2+q22\frac{p^2+q^2}{2}

B

p2+q2p^2 + q^2

C

2(p2+q2)2(p^2 + q^2)

D

p2+q2+r2p^2 + q^2 + r^2

Answer

p2+q2p^2 + q^2

Explanation

Solution

To find the sum of squares of the roots, we first need to convert the given equation into a standard quadratic form Ax2+Bx+C=0Ax^2 + Bx + C = 0.

The given equation is: 1x+p+1x+q=1r\frac{1}{x+p} + \frac{1}{x+q} = \frac{1}{r}

Combine the terms on the left side: (x+q)+(x+p)(x+p)(x+q)=1r\frac{(x+q) + (x+p)}{(x+p)(x+q)} = \frac{1}{r} 2x+p+qx2+px+qx+pq=1r\frac{2x + p + q}{x^2 + px + qx + pq} = \frac{1}{r} 2x+p+qx2+(p+q)x+pq=1r\frac{2x + p + q}{x^2 + (p+q)x + pq} = \frac{1}{r}

Cross-multiply: r(2x+p+q)=1(x2+(p+q)x+pq)r(2x + p + q) = 1(x^2 + (p+q)x + pq) 2rx+r(p+q)=x2+(p+q)x+pq2rx + r(p+q) = x^2 + (p+q)x + pq

Rearrange the terms to form a standard quadratic equation: x2+(p+q)x−2rx+pq−r(p+q)=0x^2 + (p+q)x - 2rx + pq - r(p+q) = 0 x2+(p+q−2r)x+(pq−rp−rq)=0x^2 + (p+q-2r)x + (pq - rp - rq) = 0

Let the roots of this quadratic equation be α\alpha and β\beta. According to the problem statement, the roots are equal in magnitude but opposite in sign. This means α=−β\alpha = -\beta.

From Vieta's formulas for a quadratic equation Ax2+Bx+C=0Ax^2 + Bx + C = 0:

  1. Sum of roots: α+β=−BA\alpha + \beta = -\frac{B}{A}
  2. Product of roots: αβ=CA\alpha \beta = \frac{C}{A}

In our equation, A=1A=1, B=(p+q−2r)B=(p+q-2r), and C=(pq−rp−rq)C=(pq - rp - rq).

Since the roots are equal in magnitude but opposite in sign (α=−β\alpha = -\beta), their sum must be zero: α+β=0\alpha + \beta = 0 So, −(p+q−2r)=0-(p+q-2r) = 0 p+q−2r=0p+q-2r = 0 p+q=2rp+q = 2r (This is a crucial relation derived from the given condition)

Now, let's find the sum of squares of these roots, which is α2+β2\alpha^2 + \beta^2. Since β=−α\beta = -\alpha, we have β2=(−α)2=α2\beta^2 = (-\alpha)^2 = \alpha^2. Therefore, the sum of squares is α2+α2=2α2\alpha^2 + \alpha^2 = 2\alpha^2.

From the product of roots: αβ=pq−rp−rq\alpha \beta = pq - rp - rq Substitute β=−α\beta = -\alpha: α(−α)=pq−rp−rq\alpha(-\alpha) = pq - rp - rq −α2=pq−rp−rq-\alpha^2 = pq - rp - rq α2=−(pq−rp−rq)\alpha^2 = -(pq - rp - rq) α2=rp+rq−pq\alpha^2 = rp + rq - pq

Now, substitute the relation r=p+q2r = \frac{p+q}{2} (derived from p+q=2rp+q=2r) into the expression for α2\alpha^2: α2=p(p+q2)+q(p+q2)−pq\alpha^2 = p\left(\frac{p+q}{2}\right) + q\left(\frac{p+q}{2}\right) - pq α2=p2+pq2+pq+q22−pq\alpha^2 = \frac{p^2+pq}{2} + \frac{pq+q^2}{2} - pq α2=p2+pq+pq+q22−pq\alpha^2 = \frac{p^2+pq+pq+q^2}{2} - pq α2=p2+2pq+q22−pq\alpha^2 = \frac{p^2+2pq+q^2}{2} - pq α2=(p+q)22−pq\alpha^2 = \frac{(p+q)^2}{2} - pq α2=p2+2pq+q2−2pq2\alpha^2 = \frac{p^2+2pq+q^2 - 2pq}{2} α2=p2+q22\alpha^2 = \frac{p^2+q^2}{2}

Finally, the sum of squares of the roots is 2α22\alpha^2: Sum of squares =2×(p2+q22)= 2 \times \left(\frac{p^2+q^2}{2}\right) Sum of squares =p2+q2= p^2+q^2