Step 1: Find the roots of the given cubic equation.
The given equation is \( x^3 - 2x^2 + 3x - 4 = 0 \). Let the roots be \( \alpha, \beta, \gamma \). Using Vieta's formulas for a cubic equation \( x^3 + ax^2 + bx + c = 0 \):
\( \alpha + \beta + \gamma = -a = -(-2) = 2 \)
\( \alpha\beta + \beta\gamma + \gamma\alpha = b = 3 \)
\( \alpha\beta\gamma = -c = -(-4) = 4 \)
We need to find the cubic equation whose roots are \( \alpha^2, \beta^2, \gamma^2 \).
Step 2: Form the new cubic equation.
The new cubic equation with roots \( \alpha^2, \beta^2, \gamma^2 \) can be written as:
\[ (x - \alpha^2)(x - \beta^2)(x - \gamma^2) = 0 \] This expands to:
\[ x^3 - (\alpha^2 + \beta^2 + \gamma^2)x^2 + (\alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2)x - \alpha^2\beta^2\gamma^2 = 0 \] We need to compute the coefficients using the relationships from Vieta's formulas.
Step 3: Compute \( \alpha^2 + \beta^2 + \gamma^2 \).
Using the identity:
\[ \alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) \] Substitute the known values:
\[ \alpha^2 + \beta^2 + \gamma^2 = (2)^2 - 2(3) = 4 - 6 = -2 \]
Step 4: Compute \( \alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2 \).
First, find \( (\alpha\beta)^2 + (\beta\gamma)^2 + (\gamma\alpha)^2 \): \[ (\alpha\beta)^2 + (\beta\gamma)^2 + (\gamma\alpha)^2 = (\alpha\beta + \beta\gamma + \gamma\alpha)^2 - 2\alpha\beta\gamma(\alpha + \beta + \gamma) \] Substitute: \[ (\alpha\beta)^2 + (\beta\gamma)^2 + (\gamma\alpha)^2 = (3)^2 - 2(4)(2) = 9 - 16 = -7 \] So, \( \alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2 = -7 \).
Step 5: Compute \( \alpha^2\beta^2\gamma^2 \).
Since \( \alpha\beta\gamma = 4 \), we have:
\[ \alpha^2\beta^2\gamma^2 = (\alpha\beta\gamma)^2 = (4)^2 = 16 \] Thus, the constant term in the new equation is \( -\alpha^2\beta^2\gamma^2 = -16 \).
Step 6: Form the new cubic equation.
Using the computed sums:
Coefficient of \( x^2 \): \( -(\alpha^2 + \beta^2 + \gamma^2) = -(-2) = 2 \)
Coefficient of \( x \): \( \alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2 = -7 \)
Constant term: \( -\alpha^2\beta^2\gamma^2 = -16 \) The new cubic equation is: \[ x^3 + 2x^2 - 7x - 16 = 0 \]
Step 7: Match with the options.
The equation \( x^3 + 2x^2 - 7x - 16 = 0 \) matches option (2).
Final Answer:
\[ \boxed{x^3 + 2x^2 - 7x - 16 = 0} \]
| List-I | List-II | ||
|---|---|---|---|
| (A) | $f(x) = \frac{|x+2|}{x+2} , x \ne -2 $ | (I) | $[\frac{1}{3} , 1 ]$ |
| (B) | $(x)=|[x]|,x \in [R$ | (II) | Z |
| (C) | $h(x) = |x - [x]| , x \in [R$ | (III) | W |
| (D) | $f(x) = \frac{1}{2 - \sin 3x} , x \in [R$ | (IV) | [0, 1) |
| (V) | { -1, 1} | ||
| List I | List II | ||
|---|---|---|---|
| (A) | $\lambda=8, \mu \neq 15$ | 1. | Infinitely many solutions |
| (B) | $\lambda \neq 8, \mu \in R$ | 2. | No solution |
| (C) | $\lambda=8, \mu=15$ | 3. | Unique solution |
If the sum of two roots of \( x^3 + px^2 + qx - 5 = 0 \) is equal to its third root, then \( p(q^2 - 4q) = \) ?