Question

If $\alpha, \beta, \gamma, \delta$ are the roots of the equation $x^4 - 4x^3 + 3x^2 + 2x - 2 = 0$ such that $\alpha$ and $\beta$ are integers and $\gamma, \delta$ are irrational numbers, then $\alpha + 2\beta + \gamma^2 + \delta^2 =$

• A. 5
• B. 7
• C. 11
• D. 13

Hint:

When dealing with higher-degree polynomials, always start by checking for simple integer or rational roots using the Rational Root Theorem. Once a root is found, use synthetic division to reduce the degree of the polynomial. This makes it easier to find the remaining roots. For expressions involving sums of powers of roots like $\gamma^2+\delta^2$, it's usually faster to use Vieta's formulas than to calculate the roots explicitly.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
For a polynomial with integer coefficients, any rational roots must be of the form $p/q$, where $p$ is a divisor of the constant term and $q$ is a divisor of the leading coefficient (Rational Root Theorem). Also, irrational roots of polynomials with rational coefficients occur in conjugate pairs.
Step 2: Key Formula or Approach
Let $P(x) = x^4 - 4x^3 + 3x^2 + 2x - 2$.
1. Use the Rational Root Theorem to find the possible integer roots, $\alpha$ and $\beta$.
2. Once the integer roots are found, use polynomial division to find the remaining quadratic factor whose roots are $\gamma$ and $\delta$.
3. Use Vieta's formulas on the quadratic factor to find the value of $\gamma^2 + \delta^2$.
4. Calculate the final expression.
Step 3: Detailed Explanation
1. Find integer roots: According to the Rational Root Theorem, any integer root must be a divisor of the constant term,
-2. The possible integer roots are $\pm 1, \pm 2$.
Let's test these values in $P(x)$:
$P(1) = 1^4 - 4(1)^3 + 3(1)^2 + 2(1) - 2 = 1 - 4 + 3 + 2 - 2 = 0$. So, $x=1$ is a root.
Since there might be another integer root, let's use synthetic division to find the remaining polynomial.
\begin{center} \begin{tabular}{c|ccccc} 1 & 1 & -4 & 3 & 2 & -2
& & 1 & -3 & 0 & 2
\hline & 1 & -3 & 0 & 2 & 0
\end{tabular} \end{center} The reduced polynomial is $Q(x) = x^3 - 3x^2 + 2$. Let's test the possible integer roots in $Q(x)$:
$Q(1) = 1^3 - 3(1)^2 + 2 = 1 - 3 + 2 = 0$. So, $x=1$ is a repeated root.
The two integer roots are $\alpha = 1$ and $\beta = 1$.
2. Find the quadratic factor: We divide $Q(x)$ by $(x-1)$.
\begin{center} \begin{tabular}{c|cccc} 1 & 1 & -3 & 0 & 2
& & 1 & -2 & -2
\hline & 1 & -2 & -2 & 0
\end{tabular} \end{center} The remaining quadratic factor is $x^2 - 2x - 2$. The roots of $x^2 - 2x - 2 = 0$ are the irrational roots $\gamma$ and $\delta$. 3. Find $\gamma^2 + \delta^2$: For the quadratic equation $x^2 - 2x - 2 = 0$, by Vieta's formulas: Sum of roots: $\gamma + \delta = -(-2)/1 = 2$. Product of roots: $\gamma \delta = -2/1 = -2$. We need to find $\gamma^2 + \delta^2$. We use the identity: \[ \gamma^2 + \delta^2 = (\gamma + \delta)^2 - 2\gamma\delta \] \[ \gamma^2 + \delta^2 = (2)^2 - 2(-2) = 4 + 4 = 8 \] 4. Calculate the final expression: We need to calculate $\alpha + 2\beta + \gamma^2 + \delta^2$. Substituting the values we found: $\alpha=1, \beta=1, \gamma^2+\delta^2=8$. \[ 1 + 2(1) + 8 = 1 + 2 + 8 = 11 \] Step 4: Final Answer
The value of the expression is 11.