Let α, β be the roots of the equation x² - x + 2 = 0 with Im(α) Im(β). Then α⁶ + α⁴ + β⁴ - 5α² is

Q: Wait, why can't we just use the quadratic formula to find $\alpha$ and $\beta$ and plug them in?

You absolutely could, but that is the 'trap' path! If you use the quadratic formula, you will get complex numbers involving $\sqrt{-7}$. Raising those to the power of 6 will lead to massive, messy calculations where a single sign error will ruin your answer. In JEE Advanced, we value elegance over brute force. The recursive reduction method is faster, cleaner, and significantly less prone to error.

Q: I am still confused about the condition $\text{Im}(\alpha) > \text{Im}(\beta)$. Does it change the value of the expression?

Great catch! In this specific problem, the expression is symmetric with respect to $\alpha$ and $\beta$ in a way that the final result remains invariant regardless of which root is which. The condition is purely a distractor. Always check if your final expression depends on the specific identity of the roots or just their symmetric properties (like sum or product).

Q: Is there a faster way to do this using Newton's Sums?

Yes! Newton's Sums are a fantastic tool for higher powers. If $S_n = \alpha^n + \beta^n$, you can derive a recurrence relation from the characteristic equation $x^2 - x + 2 = 0$, which gives $S_n - S_{n-1} + 2S_{n-2} = 0$. While this is very powerful, the recursive reduction method we used is often more intuitive for lower powers like 4 and 6. Both are valid, but choose the one you are most comfortable with under exam pressure.

Question

Let $\alpha, \beta$ be the roots of the equation $x^2 - x + 2 = 0$ with $	ext{Im}(\alpha) > 	ext{Im}(\beta)$. Then $\alpha^6 + \alpha^4 + \beta^4 - 5\alpha^2$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a formidable challenge: evaluating an expression like $\alpha^6 + \alpha^4 + \beta^4 - 5\alpha^2$. At first glance, it looks like a nightmare. You might be tempted to reach for the quadratic formula, find the roots $\alpha$ and $\beta$, and start calculating powers.

But stop! That is exactly the trap the examiner has set for you. In the world of JEE Advanced, brute force is rarely the intended path. Instead, we look for the hidden geometry of the equation.

We are given $x^2 - x + 2 = 0$. This is our universe. Since $\alpha$ is a root, it must obey the laws of this universe: $\alpha^2 - \alpha + 2 = 0$. This is our **Golden Key**. By rearranging it to $\alpha^2 = \alpha - 2$, we have unlocked the ability to reduce any power of $\alpha$ into a simple linear form.

### The Recursive Reduction

Now, let us embark on the journey of reduction. We need $\alpha^4$. Instead of calculating it directly, we square our Golden Key:

$$\alpha^4 = (\alpha - 2)^2 = \alpha^2 - 4\alpha + 4$$

But wait, we still have an $\alpha^2$ term! We simply substitute our Golden Key again: $(\alpha - 2) - 4\alpha + 4$. Simplifying this, we get:

$$\alpha^4 = -3\alpha + 2$$

See how the monster is shrinking? Now for the big one: $\alpha^6$. We know that $\alpha^6 = \alpha^2 \cdot \alpha^4$. Substituting our linear forms, we get $(\alpha - 2)(-3\alpha + 2)$.

Expanding this carefully, we arrive at $-3\alpha^2 + 8\alpha - 4$. Once more, we replace $\alpha^2$ with $\alpha - 2$. The result is:

$$\alpha^6 = 5\alpha + 2$$

We have successfully tamed the high powers.

### The Symmetry Shortcut

Now, what about $\beta^4$? Do we need to repeat the entire process? Absolutely not. This is where we exploit the beauty of symmetry.

Since $\beta$ is also a root of the same equation, it follows the exact same rules. Therefore:

$$\beta^4 = -3\beta + 2$$

We have now reduced every single term in our original expression to a linear form.

### The Grand Finale

Let us assemble our pieces. Our expression $E = \alpha^6 + \alpha^4 + \beta^4 - 5\alpha^2$ becomes:

$$E = (5\alpha + 2) + (-3\alpha + 2) + (-3\beta + 2) - 5(\alpha - 2)$$

Expanding this, we get $5\alpha + 2 - 3\alpha + 2 - 3\beta + 2 - 5\alpha + 10$. Grouping the terms, we find:

$$E = -3\alpha - 3\beta + 16$$

Factoring out the $-3$, we get $E = -3(\alpha + \beta) + 16$. From our original equation $x^2 - x + 2 = 0$, the sum of the roots $\alpha + \beta$ is simply:

$$\alpha + \beta = -\left(\frac{-1}{1}ight) = 1$$

Substituting this in, we get $E = -3(1) + 16 = 13$. The monster is defeated.

The condition $	ext{Im}(\alpha) > 	ext{Im}(\beta)$ was merely a shadow, a distractor meant to make you doubt the path. By trusting the algebraic structure, we found the answer with elegance and speed. Keep this recursive reduction technique in your toolkit; it is a lifesaver in the heat of the exam. The final answer is **13**.

Animated Solution for Mathematics - Quadratic Equations: Let $α, β$ be the roots of the equation $x^{2} - x + 2 = 0$ with $Im (α) > Im (β)$ . Then $α^{6} + α^{4} + β^{4} - 5 α^{2}$ is equal to

Enter Numerical Value:

Visualized Solution

Analyze the Given Equation

Apply the Root Property

Calculate $α^{4}$ Expression

Reduce $α^{4}$ to Linear Form

Setup for $α^{6}$

Expand $α^{6}$ Expression

Reduce $α^{6}$ to Linear Form

Symmetry for $β^{4}$

Substitute into Main Expression

Expand and Group Terms

Factor Out the Constant

Sum of Roots Calculation

Final Substitution and Conclusion

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

The Recursive Reduction

The Symmetry Shortcut

The Grand Finale

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Animated Solution for Mathematics - Quadratic Equations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let α,β be the roots of the equation x2−x+2=0 with Im(α)>Im(β). Then α6+α4+β4−5α2 is equal to