Question

If ${x^2} - 2r \times {p_r}x + r = 0;r = 1,2,3$ are the quadratic equation of which each pair has exactly one root common then the number of solutions of the triplet $\left( {{p_1},{p_2},{p_3}} \right)$ is
A) 2
B) 1
C) 9
D) 27

Answer 1

Hint:
It is given in the question that if ${x^2} - 2r \times {p_r}x + r = 0;r = 1,2,3$ are the quadratic equation of which each pair has exactly one root common then the number of solutions of the triplet $\left( {{p_1},{p_2},{p_3}} \right)$ is
We are given the equation is ${x^2} - 2r \times {p_r}x + r = 0;r = 1,2,3$ and it is said that each pair has one common root.
First, we assume the roots for the set of the equation. Then after using the concept of relationship between the sum and product of the roots and the coefficient we can find the value of roots.
Finally, with the values of the roots we can find the number of solutions.

Complete step by step solution:
It is given in the question that if ${x^2} - 2r \times {p_r}x + r = 0;r = 1,2,3$ are the quadratic equation of which each pair has exactly one root common then the number of solutions of the triplet $\left( {{p_1},{p_2},{p_3}} \right)$ is
Since, we are given the equation is ${x^2} - 2r \times {p_r}x + r = 0;r = 1,2,3$
And we have also given that each pair has one common root.
So, let the common roots be $\alpha ,\beta ,\gamma $
Now,
 $\alpha ,\beta $ be the roots of the equation,
 ${x^2} - 2\left( 1 \right) \times {p_1}x + 1 = {x^2} - 2{p_1}x + 1 = 0$ (I)
 $\beta ,\gamma $ be the roots of the equation,
${x^2} - 2\left( 2 \right) \times {p_2}x + 2 = {x^2} - 4{p_2}x + 2 = 0$ (II)
$\alpha ,\gamma $be the roots of the equation,
 ${x^2} - 2\left( 3 \right) \times {p_3}x + 3 = {x^2} - 6{p_3}x + 3 = 0$ (III)
Since, we know the relation of the coefficient of the equation and the roots of the quadratic equation $a{x^2} + bx + c = 0$ where $\alpha ,\beta $ are its roots, then
Sum of the roots, $\alpha + \beta = \dfrac{{ - b}}{a}$
Product of the roots, $\alpha \beta = \dfrac{c}{a}$
Now, using the above concept,
From the equation (I), we get,
Sum of the roots, $\alpha + \beta = \dfrac{{2{p_1}}}{1} = 2{p_1}$ (IV)
   Product of the roots, $\alpha \beta = \dfrac{1}{1} = 1$ (V)
Now, from equation (II), we get,
Sum of the roots, $\beta + \gamma = \dfrac{{4{p_2}}}{1} = 4{p_2}$ (VI)
    Product of the roots, $\beta \gamma = \dfrac{2}{1} = 2$ (VII)
Now, from equation (III), we get,
Sum of the roots, $\alpha + \gamma = \dfrac{{6{p_3}}}{1} = 6{p_3}$ (VIII)
    Product of the roots, $\alpha \gamma = \dfrac{3}{1} = 3$ (IX)
Now, we have to solve the above equations to get the value of $\alpha ,\beta ,\gamma $
Let us multiply equation (VII) and (IX) and divide it by equation (V), we get,
 $ \Rightarrow \dfrac{{\left( {\beta \gamma } \right)\left( {\gamma \alpha } \right)}}{{\left( {\alpha \beta } \right)}} = \dfrac{{2 \times 3}}{1}$
 $
   \Rightarrow {\gamma ^2} = 6 \\
   \Rightarrow \gamma = \pm 6 \\
 $
Now, substitute the value of $\gamma $ in equation (VII), we get,
 $
   \Rightarrow \beta \gamma = 2 \\
   \Rightarrow \beta \left( { \pm 6} \right) = 2 \\
   \Rightarrow \beta = \pm \dfrac{2}{{\sqrt 6 }} \\
 $
Now, substitute the value of $\gamma $ in equation (IX), we get,
 $
   \Rightarrow \alpha \gamma = 3 \\
   \Rightarrow \alpha \left( { \pm \sqrt 6 } \right) = 3 \\
   \Rightarrow \alpha = \pm \dfrac{3}{{\sqrt 6 }} \\
 $
Here, we see that there are two values of $\alpha ,\beta ,\gamma $

$\therefore $we will have two sets of solution for the triplet.

Note:
Sum of the roots: The sum of the roots of the quadratic equation is equal to the negation of the coefficient of the second term, divided by the leading coefficient.
Sum of the roots, $\alpha + \beta = \dfrac{{ - b}}{a}$

Product of the roots: The product of the roots of a quadratic equation is equal to the constant term
(the third term), divided by the leading coefficient.
Product of the roots, $\alpha \beta = \dfrac{c}{a}$