Question

The real numbers ${x_1},{x_2},{x_3}$ satisfying the equation ${x^3} - {x^2} + \beta x + \gamma = 0$ are in AP. Find the interval in which $\beta {\text{ and }}\gamma $ lie?
A. $\beta \in \left( { - \infty ,\dfrac{1}{3}} \right]{\text{ and }}\gamma \in \left[ { - \dfrac{1}{{27}},\infty } \right)$
B. $\beta \in \left( { - \infty ,\dfrac{1}{3}} \right){\text{and }}\gamma \in \left[ { - \dfrac{1}{{27}},\infty } \right)$
C. $\beta \in \left( { - \infty , - \dfrac{1}{3}} \right]{\text{and }}\gamma \in \left( { - \dfrac{1}{{27}},\infty } \right)$
D. none of these.

Answer 1

Hint: Here we can proceed by letting the three roots so that they are in AP. Now we can easily form the equations from the given equations of the sum of the roots, sum of roots taken two at a time and also for the product of the roots. If we have the cubic equation $a{x^3} + b{x^2} + cx + d = 0$
Sum of roots$ = - \dfrac{b}{a}$
Product of roots$ = - \dfrac{d}{a}$
Sum when taken two at a time$ = \dfrac{c}{a}$

Complete step-by-step answer:
Here we are given that ${x_1},{x_2},{x_3}$ satisfying the equation ${x^3} - {x^2} + \beta x + \gamma = 0$ are its roots and are in AP. So we know that the progression in which there is a common difference in each consecutive term is known as AP. So here if these three are in AP, we can let:
${x_1} = a - d,{x_2} = a,{x_3} = a + d$
If we have the cubic equation of the form $a{x^3} + b{x^2} + cx + d = 0$ then we can say:
Sum of roots$ = - \dfrac{b}{a}$
So we can write it as:
$
\Rightarrow a - d + a + a + d = \dfrac{{ - ( - 1)}}{1} = 1 \\
\Rightarrow 3a = 1 \\
\Rightarrow a = \dfrac{1}{3} \\
 $
So we get three roots as $\dfrac{1}{3} - d,\dfrac{1}{3},\dfrac{1}{3} + d$
We also know that:
Sum of roots when taken two at a time$ = \dfrac{c}{a}$
\[{x_1}{x_2} + {x_2}{x_3} + {x_3}{x_1} = \dfrac{c}{a}\]
$
\Rightarrow \dfrac{1}{3}\left( {\dfrac{1}{3} - d} \right) + \dfrac{1}{3}\left( {\dfrac{1}{3} + d} \right) + \left( {\dfrac{1}{3} - d} \right)\left( {\dfrac{1}{3} + d} \right) = \beta \\
\Rightarrow \dfrac{1}{9} - \dfrac{d}{3} + \dfrac{1}{9} + \dfrac{d}{3} + \dfrac{1}{9} - {d^2} = \beta \\
\Rightarrow \dfrac{3}{9} - {d^2} = \beta \\
\Rightarrow \dfrac{1}{3} - {d^2} = \beta \\
 $
We also know that:
The value of ${d^2} \geqslant 0$ as a square of any number is always non-negative.
So we can say that:
$\Rightarrow$ $\dfrac{1}{3} - {d^2} = \beta $
As ${d^2} \geqslant 0$
$\Rightarrow$ $\dfrac{1}{3} - {d^2} \leqslant \dfrac{1}{3}$
So $\beta \leqslant \dfrac{1}{3}$$ - - - - - (1)$
Now we can take the product of the roots:
$\left( {\dfrac{1}{3} - d} \right)\dfrac{1}{3}\left( {\dfrac{1}{3} + d} \right) = \left( {\dfrac{1}{9} - \dfrac{d}{3}} \right)\left( {\dfrac{1}{3} + d} \right)$$ = \dfrac{{ - \gamma }}{1}$
Now simplifying it we will get:
$\dfrac{1}{{27}} + \dfrac{d}{9} - \dfrac{d}{9} - \dfrac{{{d^2}}}{3} = - \gamma $
$
\Rightarrow \dfrac{1}{{27}} - \dfrac{{{d^2}}}{3} = - \gamma \\
\Rightarrow \gamma = \dfrac{1}{3}\left( {{d^2} - \dfrac{1}{9}} \right) \\
 $
Also we know that ${d^2} \geqslant 0$
So we can say that:
$
\Rightarrow \gamma = \dfrac{1}{3}\left( {{d^2} - \dfrac{1}{9}} \right) \geqslant \Rightarrow\dfrac{1}{3}\left( { - \dfrac{1}{9}} \right) \\
\Rightarrow \gamma \geqslant - \dfrac{1}{{27}} - - - - - - (2) \\
 $
From equation (1) and (2) we get that:
$\beta \in \left( { - \infty ,\dfrac{1}{3}} \right]{\text{ and }}\gamma \in \left[ { - \dfrac{1}{{27}},\infty } \right)$
Hence A is the correct option.

Note: Here the student must keep in mind all the formulas that are used in relation to the cubic equation as given above because without them, the problem cannot be solved easily. Hence one must remember that when we are given the cubic equation of the form $a{x^3} + b{x^2} + cx + d = 0$
Sum of roots$ = - \dfrac{b}{a}$
Product of roots$ = - \dfrac{d}{a}$
Sum when taken two at a time$ = \dfrac{c}{a}$