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

JEE Main 2024 — 29 January, Morning Shift. Previous Year Question.

Problem

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

Key insight. Because $\alpha$ satisfies $\alpha^2-\alpha+2=0$, every higher power of $\alpha$ can be rewritten using $\alpha^2=\alpha-2$ — repeatedly substituting this reduces any polynomial in $\alpha$ down to just a linear expression $p\alpha+q$. The same trick applies to $\beta$, since it satisfies the identical equation.

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Approach

Since $\alpha,\beta$ are roots of $x^2-x+2=0$, both satisfy $x^2=x-2$. Use this repeatedly to reduce $\alpha^4$, $\alpha^6$, and $\beta^4$ down to linear expressions in $\alpha$ or $\beta$. Substitute everything back into the target expression, collect terms, and use Vieta’s formula for $\alpha+\beta$ to finish.

Solution

Step 1 — Write the reduction rule

Since $\alpha^2-\alpha+2=0$:

$$\alpha^2 = \alpha-2$$

(and identically, $\beta^2=\beta-2$, since $\beta$ satisfies the same equation)

Step 2 — Reduce α⁴

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

Step 3 — Reduce α⁶

$$\alpha^6 = \alpha^2\cdot\alpha^4 = (\alpha-2)(-3\alpha+2) = -3\alpha^2+2\alpha+6\alpha-4 = -3\alpha^2+8\alpha-4$$

$$= -3(\alpha-2)+8\alpha-4 = -3\alpha+6+8\alpha-4 = 5\alpha+2$$

Step 4 — Reduce β⁴ (same formula as α⁴, since β satisfies the same equation)

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

Step 5 — Substitute everything into the target expression

$$\alpha^6+\alpha^4+\beta^4-5\alpha^2 = (5\alpha+2)+(-3\alpha+2)+(-3\beta+2)-5(\alpha-2)$$

Step 6 — Expand and collect like terms

$$= 5\alpha+2-3\alpha+2-3\beta+2-5\alpha+10$$

$$= (5\alpha-3\alpha-5\alpha) + (-3\beta) + (2+2+2+10)$$

$$= -3\alpha-3\beta+16 = -3(\alpha+\beta)+16$$

Step 7 — Find α + β using Vieta’s formula

For $x^2-x+2=0$: sum of roots $=-\dfrac{-1}{1}=1$, so $\alpha+\beta=1$.

Step 8 — Substitute and finish

$$-3(1)+16 = -3+16 = 13$$

Answer

Common mistakes

• Trying to find $\alpha$ and $\beta$ explicitly (as complex numbers) and substituting into the expression directly. This works in principle but involves binomial-expanding complex numbers to the 6th power — far more error-prone than the reduction trick, which never requires knowing $\alpha$ and $\beta$’s actual values at all.
• Forgetting that $\beta$ satisfies the same reduction rule as $\alpha$. Since both are roots of the identical equation $x^2-x+2=0$, the relation $x^2=x-2$ applies to each of them individually — there’s no need for a separate derivation for $\beta$.

Practise next

• Let $\alpha,\beta$ be the roots of $x^2-2x+3=0$. Find $\alpha^6+\alpha^4+\beta^4+5\alpha^2$, using the same power-reduction approach.