Multiple choice

If $\displaystyle x^{2}-2r\times p_{r}x+r=0;:r=1,2,3$ are the quadratc equations of which each pair has exaclty one root common then the number of solutions of the triplet $(p_{1},p_{2},p_{3})$ is

  1. $2$
  2. $1$
  3. $9$
  4. $27$
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A Correct answer
Explanation

For two quadratics to have exactly one common root, the condition is (c1a2 - c2a1)^2 = (a1b2 - a2b1)(b1c2 - b2c1). Applying this to the pairs of equations provided leads to a system that restricts the possible values of p1, p2, p3.

AI explanation

The given equations are x^2 - 2p_1x + 1 = 0, x^2 - 4p_2x + 2 = 0, and x^2 - 6p_3x + 3 = 0. For each pair of equations to have exactly one root in common, we can use the condition for a common root between x^2 - ax + r = 0 and x^2 - bx + s = 0, which leads to x = (s - r)/(a - b). For the first two equations, the common root is (2 - 1)/(-2p_1 - (-4p_2)) = 1/(4p_2 - 2p_1). For the second and third equations, the common root is (3 - 2)/(-4p_2 - (-6p_3)) = 1/(6p_3 - 4p_2). Assuming the common root is the same across the system, setting these expressions and their counterparts for the first and third equations leads to two distinct symmetric sets of roots. This provides exactly 2 unique solutions for the triplet.

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