Complex Roots of x²-(3-2i)x-(2i-2)=0, Find αγ+βδ

JEE Main 2025 — 28 January, Evening Shift. Previous Year Question.

Problem

If $\alpha+i\beta$ and $\gamma+i\delta$ are roots of the equation $x^2 – (3-2i)x – (2i-2) = 0$, where $i=\sqrt{-1}$ and $\alpha,\beta,\gamma,\delta$ are real numbers, then find the value of $\alpha\gamma+\beta\delta$.

Key insight. With complex coefficients, summing and multiplying the roots via Vieta’s formulas quickly turns into messy simultaneous equations. It’s much faster to notice that the constant term can be rewritten to match part of the linear term, which lets the quadratic factorise directly — handing over the roots immediately.

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Approach

The natural first instinct — sum the roots to get $-\dfrac{b}{a}$ and multiply them to get $\dfrac{c}{a}$ — technically works, but leads to a pair of simultaneous equations in four real unknowns that’s slow to untangle. Many JEE quadratics with complex coefficients are built so that rearranging the equation itself reveals a common factor, letting it split into two linear pieces directly. Spotting that is the whole problem here.

Solution

Step 1 — Look for a way to split the constant term

The equation is:

$$x^2 – (3-2i)x – (2i-2) = 0$$

The constant term is $-(2i-2)$. Notice that if $x=1$, then $1 \times (2i-2) = 2i-2$ — meaning the constant term is exactly $(2i-2)$ multiplied by $1$. This suggests rewriting the coefficient $3-2i$ so that a factor of $(2i-2)$ appears explicitly.

Step 2 — Rewrite $3-2i$ to expose the shared factor

Write $3$ as $1+2$, so that:

$$x^2 – \big[1+2-2i\big]x – (2i-2) = 0$$

Group the $-2+2i$ part together with the constant term’s sign flipped to match:

$$x^2 – x – (2-2i)x – (2i-2) = 0$$

$$x^2 – x + (2i-2)x – (2i-2) = 0$$

Step 3 — Factor by grouping

$$x(x-1) + (2i-2)(x-1) = 0$$

$$(x-1)\big(x + (2i-2)\big) = 0$$

So the two roots are:

$$x = 1 \qquad \text{and} \qquad x = 2-2i$$

Step 4 — Match the roots to $\alpha+i\beta$ and $\gamma+i\delta$

Writing $x=1$ as $1+0i$ and comparing with $\alpha+i\beta$:

$$\alpha = 1, \quad \beta = 0$$

Comparing $x=2-2i$ with $\gamma+i\delta$:

$$\gamma = 2, \quad \delta = -2$$

Step 5 — Compute $\alpha\gamma+\beta\delta$

$$\alpha\gamma+\beta\delta = (1)(2) + (0)(-2) = 2$$

Answer

Common mistakes

• Reaching for Vieta’s formulas by default. They aren’t wrong here, but multiplying $(\alpha+i\beta)(\gamma+i\delta)$ and separating real and imaginary parts to match $\dfrac{c}{a}$ produces extra equations that this factorisation approach avoids entirely.
• Missing the hint in the constant term. The key move — noticing that $2i-2$ appears in both the coefficient (after splitting $3$ into $1+2$) and the constant term — is easy to miss if the equation isn’t inspected for shared factors before jumping into a formula.

Practise next

• Try the same trick on $x^2-(5-3i)x+(4-3i)=0$: look for a value of $x$ that makes the constant term collapse. Then, writing its roots as $\alpha+i\beta$ and $\gamma+i\delta$, find $\alpha\gamma+\beta\delta$.