Question

If the equation whose roots are the squares of the roots of the cubic $x^3-ax^2+bx-1=0$ is identical with the given cubic equation, then

• A. a, b are roots of $x^2+x+2=0$
• B. a = b = 0
• C. a = b = 3
• D. a = 0, b = 3

Answer 1

Answer: (A), (B), (C)

A, B, C

Answer 2

Let the given cubic equation be $x^3-ax^2+bx-1=0$.
Let its roots be $\alpha, \beta, \gamma$.
From Vieta's formulas, we have:

1. Sum of roots: $\alpha + \beta + \gamma = a$
2. Sum of products of roots taken two at a time: $\alpha\beta + \beta\gamma + \gamma\alpha = b$
3. Product of roots: $\alpha\beta\gamma = 1$

The new equation has roots $\alpha^2, \beta^2, \gamma^2$.
Let this new equation be $y^3 - A y^2 + B y - C = 0$.
The problem states that this new equation is identical to the given cubic equation.
Therefore, $A=a$, $B=b$, and $C=1$.

Let's express $A, B, C$ in terms of $\alpha, \beta, \gamma$: $A = \alpha^2 + \beta^2 + \gamma^2$
$B = \alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2$
$C = \alpha^2\beta^2\gamma^2$

Now, substitute the relationships from Vieta's formulas:

1. For $C$:
   $C = (\alpha\beta\gamma)^2 = (1)^2 = 1$. This matches the constant term of the given cubic, which is $1$.

2. For $A$:
   $A = \alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)$
   Substitute $a$ and $b$:
   $A = a^2 - 2b$
   Since $A=a$, we have:
   $a = a^2 - 2b$ (Equation 1)

3. For $B$:
   $B = \alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2 = (\alpha\beta + \beta\gamma + \gamma\alpha)^2 - 2\alpha\beta\gamma(\alpha + \beta + \gamma)$
   Substitute $a, b, \alpha\beta\gamma=1$:
   $B = b^2 - 2(1)(a)$
   $B = b^2 - 2a$
   Since $B=b$, we have:
   $b = b^2 - 2a$ (Equation 2)

Now we need to solve the system of equations:
(1) $a = a^2 - 2b$
(2) $b = b^2 - 2a$

Subtract (2) from (1):
$a - b = (a^2 - 2b) - (b^2 - 2a)$
$a - b = a^2 - b^2 + 2a - 2b$
$a - b = (a-b)(a+b) + 2(a-b)$
Rearrange the terms:
$(a-b) - (a-b)(a+b) - 2(a-b) = 0$
$(a-b) [1 - (a+b) - 2] = 0$
$(a-b) [-1 - (a+b)] = 0$
$(a-b) (a+b+1) = 0$

This implies two possible cases:

Case 1: $a-b = 0 \implies a=b$
Substitute $a=b$ into Equation 1:
$a = a^2 - 2a$
$a^2 - 3a = 0$
$a(a-3) = 0$
So, $a=0$ or $a=3$.
If $a=0$, then $b=0$. This gives the solution $(a,b)=(0,0)$.
If $a=3$, then $b=3$. This gives the solution $(a,b)=(3,3)$.

Case 2: $a+b+1 = 0 \implies a+b = -1$
From Equation 1, $2b = a^2 - a$.
Substitute $b = -1-a$ into this expression:
$2(-1-a) = a^2 - a$
$-2 - 2a = a^2 - a$
$a^2 + a + 2 = 0$
The discriminant of this quadratic equation is $D = 1^2 - 4(1)(2) = 1 - 8 = -7$.
Since $D < 0$, the roots for $a$ are complex: $a = \frac{-1 \pm i\sqrt{7}}{2}$.
If $a$ is one root, say $a_1 = \frac{-1 + i\sqrt{7}}{2}$, then $b = -1-a_1 = -1 - \frac{-1 + i\sqrt{7}}{2} = \frac{-2+1-i\sqrt{7}}{2} = \frac{-1 - i\sqrt{7}}{2}$.
Notice that these values of $a$ and $b$ are the two roots of the quadratic equation $x^2+x+2=0$.
So, in this case, $a$ and $b$ are the roots of $x^2+x+2=0$.

We have found three sets of solutions for $(a,b)$:

1. $(0,0)$
2. $(3,3)$
3. $a, b$ are the roots of $x^2+x+2=0$.

Comparing these solutions with the given options: (A) a, b are roots of $x^2+x+2=0$. This matches our third solution. (B) a = b = 0. This matches our first solution. (C) a = b = 3. This matches our second solution. (D) a = 0, b = 3. This is not a solution we found.

Since the question is a single-choice question in a typical competitive exam context, it implies only one option is correct. However, mathematically, all three (A, B, C) are valid scenarios. This suggests a potential ambiguity in the question's format (it might be intended as multiple correct, or there might be an unstated assumption like "a and b are real numbers"). If $a$ and $b$ are restricted to be real numbers, then Case 2 ($a^2+a+2=0$) yields no real solutions, leaving only $(0,0)$ and $(3,3)$ as real solutions. In such a scenario, if only one option is to be chosen, the question is ill-posed.

However, if we strictly follow the derivation without imposing real-number constraints, all three types of solutions are valid. Given the options, and the typical format of such questions in JEE, it's possible that the question expects all valid options to be marked if it's a multiple-correct type, or it's a flawed single-choice question. Assuming it's a single-choice question, there might be a subtle reason to prefer one over the others, but based on pure algebraic derivation, all are valid.

If the question is from a context where multiple options can be correct, then (A), (B), and (C) are all correct. If it's a single correct option question, it's ambiguous. In the absence of further clarification, and since the question does not specify 'real' coefficients, all mathematically derived solutions are valid.

Final solutions are:

1. $a=0, b=0$
2. $a=3, b=3$
3. $a$ and $b$ are the roots of $x^2+x+2=0$ (which are complex conjugates).

Given the options, all three (A), (B), (C) represent valid solutions.