Let α, β Be the Roots of x² − ax − b = 0 With Im(α) < Im(β) and Pₙ = αⁿ − βⁿ. If P₃ = −5√7 i, P₄ = −3√7 i, P₅ = 11√7 i and P₆ = 45√7 i, Then |α⁴ + β⁴| Is Equal To

JEE Main 2025 — 23 January, Evening Shift. Previous Year Question.

Problem

Let $\alpha,\beta$ be the roots of the equation $x^2-ax-b=0$ with $\text{Im}(\alpha)<\text{Im}(\beta)$. Let $P_n=\alpha^n-\beta^n$. If $P_3=-5\sqrt7\,i$, $P_4=-3\sqrt7\,i$, $P_5=11\sqrt7\,i$, and $P_6=45\sqrt7\,i$, then $|\alpha^4+\beta^4|$ is equal to ____.

Key insight. Since $\alpha,\beta$ satisfy $x^2=ax+b$, the sequence $P_n=\alpha^n-\beta^n$ obeys the linear recurrence $P_n=aP_{n-1}+bP_{n-2}$ — the same recurrence structure the quadratic itself generates. Two consecutive applications of this recurrence, using the four given values, give two linear equations that solve directly for $a$ and $b$ — without ever finding $\alpha$ or $\beta$ individually.

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Approach

Derive the recurrence $P_n = aP_{n-1}+bP_{n-2}$ from the fact that $\alpha,\beta$ satisfy $x^2-ax-b=0$. Apply it twice — once with $n=5$ and once with $n=6$ — using the given values of $P_3,P_4,P_5,P_6$, to get two linear equations in $a$ and $b$. Solve for $a$ and $b$, then use $\alpha+\beta=a$ and $\alpha\beta=-b$ to compute $\alpha^4+\beta^4$ via the standard power-sum identities.

Solution

Step 1 — Derive the recurrence for Pn

Since $\alpha,\beta$ satisfy $x^2-ax-b=0$, i.e. $x^2=ax+b$, multiplying by $x^{n-2}$ gives $x^n=ax^{n-1}+bx^{n-2}$. Applying this to both $\alpha$ and $\beta$ and subtracting:

$$P_n = aP_{n-1}+bP_{n-2}$$

Step 2 — Apply the recurrence with n = 5

$$P_5 = aP_4+bP_3$$

$$11\sqrt7\,i = a(-3\sqrt7\,i) + b(-5\sqrt7\,i)$$

Dividing through by $\sqrt7\,i$:

$$11 = -3a-5b \quad \text{…(i)}$$

Step 3 — Apply the recurrence with n = 6

$$P_6 = aP_5+bP_4$$

$$45\sqrt7\,i = a(11\sqrt7\,i)+b(-3\sqrt7\,i)$$

$$45 = 11a-3b \quad \text{…(ii)}$$

Step 4 — Solve the system of equations

From (i): $3a+5b=-11$. From (ii): $11a-3b=45$.

Multiply (i) by 3 and (ii) by 5:

$$9a+15b=-33$$ $$55a-15b=225$$

Adding:

$$64a=192 \implies a=3$$

Substituting into $3a+5b=-11$: $9+5b=-11 \implies 5b=-20 \implies b=-4$

Step 5 — Find α + β and αβ

For $x^2-ax-b=0$ with $a=3$, $b=-4$: the equation is $x^2-3x+4=0$.

Comparing with $x^2-(\text{sum})x+(\text{product})=0$: $\alpha+\beta=3$, $\alpha\beta=4$.

Step 6 — Find α² + β²

$$\alpha^2+\beta^2 = (\alpha+\beta)^2-2\alpha\beta = 9-8=1$$

Step 7 — Find α⁴ + β⁴

$$\alpha^4+\beta^4 = (\alpha^2+\beta^2)^2-2(\alpha\beta)^2 = 1^2-2(16) = 1-32=-31$$

Step 8 — Compute the absolute value

$$|\alpha^4+\beta^4| = |{-31}| = 31$$

Answer

Common mistakes

• Trying to solve for $\alpha$ and $\beta$ individually from the quadratic before using them. The complex roots of $x^2-3x+4=0$ involve $\sqrt7\,i$, and working with them directly (rather than via $\alpha+\beta$ and $\alpha\beta$) makes every subsequent step far messier than necessary.
• Mixing up the sign convention in $x^2-ax-b=0$. Comparing directly with $x^2-(\text{sum})x+(\text{product})=0$ gives $\alpha\beta=-b$, not $+b$ — dropping this sign flip would give the wrong product of roots.

Practise next

• Let $\gamma,\delta$ be roots of $x^2-px-q=0$ with $p>0$, and $Q_n=\gamma^n-\delta^n$. If $Q_2=-\sqrt{19}\,i$, $Q_3=4\sqrt{19}\,i$ and $Q_4=9\sqrt{19}\,i$, find $\left|\gamma^3+\delta^3\right|$, using the same recurrence approach.