Question

If ${x^2} - 2r*{p_r}x + r = 0;r = 1,2,3$ are the quadratic equations 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: We are given a equation ${x^2} - 2r*{p_r}x + r = 0;r = 1,2,3$ and it is said that each pair has one common root and let's assume the roots for the set of equations. By using the concept of relationship between the sum and product of the roots and the coefficients we can find the value of the roots . And with the values of the roots we can find the number of solutions.

Complete step-by-step answer:
We are given the equation is ${x^2} - 2r{p_r}x + r = 0;r = 1,2,3$
And we are also given that each pair has one common root and let the common roots be $\alpha ,\beta ,\gamma $
So now ,
$\alpha ,\beta $ be the roots of the equation ${x^2} - 2(1){p_{(1)}}x + 1 = {x^2} - 2{p_1}x + 1 = 0$………..(1)
$\beta ,\gamma $ be the roots of the equation ${x^2} - 2(2){p_{(2)}}x + 2 = {x^2} - 4{p_2}x + 2 = 0$………(2)
$\alpha ,\gamma $ be the roots of the equation ${x^2} - 2(3){p_{(3)}}x + 3 = {x^2} - 6{p_3}x + 3 = 0$………(3)
We know that the relationship of the coefficients of the equation and the roots of a quadratic equation $a{x^2} + bx + c = 0$ where $\alpha ,\beta $ are its roots, then
Sum of the roots = $\alpha + \beta = \frac{{ - b}}{a}$
Product of the roots =$\alpha \beta = \frac{c}{a}$
Using the above concept,
From the equation (1)
Sum of the roots = $\alpha + \beta = \frac{{2{p_1}}}{1} = 2{p_1}$………….(4)
Product of the roots =$\alpha \beta = \frac{1}{1} = 1$……………(5)
From the equation (2)
Sum of the roots = $\beta + \gamma = \frac{{4{p_2}}}{1} = 4{p_2}$………….(6)
Product of the roots =$\beta \gamma = \frac{2}{1} = 2$……………(7)
From the equation (3)
Sum of the roots = $\alpha + \gamma = \frac{{6{p_3}}}{1} = 6{p_3}$………….(8)
Product of the roots =$\alpha \gamma = \frac{3}{1} = 3$……………(9)
Now we need the solve the above equations to find the values of $\alpha ,\beta ,\gamma $
Now let's multiply (7) and (9) and divide by equation (5)
$
   \Rightarrow \frac{{\left( {\beta \gamma } \right)\left( {\gamma \alpha } \right)}}{{\left( {\alpha \beta } \right)}} = \frac{{2*3}}{1} \\
   \Rightarrow {\gamma ^2} = 6 \\
   \Rightarrow \gamma = \pm \sqrt 6 \\
$
From lets substitute the value of $\gamma $ in equation (7)
...$
   \Rightarrow \beta \gamma = 2 \\
   \Rightarrow \beta ( \pm \sqrt 6 ) = 2 \\
   \Rightarrow \beta = \pm \frac{2}{{\sqrt 6 }} \\
$...
From lets substitute the value of $\gamma $ in equation (9)
$
   \Rightarrow \alpha \gamma = 3 \\
   \Rightarrow \alpha ( \pm \sqrt 6 ) = 3 \\
   \Rightarrow \alpha = \pm \frac{3}{{\sqrt 6 }} \\
$
Now we see that there are two values for $\alpha ,\beta ,\gamma $
Therefore we will have two sets of solution for the triplet$\left( {{p_1},{p_2},{p_3}} \right)$
The correct option is A.

Note: Additional information:
If in case a cubic equation is given then
A cubic equation is an equation which can be represented in the form $a{x^3} + b{x^2} + cx + d = 0$, where a,b,c,d are complex numbers and a is non-zero. The fundamental theorem of the algebra cubic equation always has 3 roots, some of which might be equal.
Relation between coefficients and roots:
For a cubic equation $a{x^3} + b{x^2} + cx + d = 0$, let p,q and r be its roots, then the following holds:

Root expression	Equals to
$p + q + r$	$\frac{{ - b}}{a}$​
$pq + qr + rp$	.$\frac{c}{a}$.
$pqr$	$\frac{{ - d}}{a}$