Sum and Product of Roots — Without Solving the Equation

Question

For a quadratic equation $2x^2 - 5x + 3 = 0$, find the sum and product of its roots without solving the equation.

Also: if the roots are $\alpha$ and $\beta$, find the value of $\alpha^2 + \beta^2$.

Solution — Step by Step

Why This Works

Every quadratic $ax^2 + bx + c = 0$ can be written as $a(x - \alpha)(x - \beta) = 0$, where $\alpha$ and $\beta$ are the roots. Expanding: $a[x^2 - (\alpha + \beta)x + \alpha\beta] = 0$, which gives $ax^2 - a(\alpha+\beta)x + a\alpha\beta = 0$.

Comparing coefficients with $ax^2 + bx + c = 0$: we get $b = -a(\alpha+\beta)$ and $c = a\alpha\beta$. Rearranging gives Vieta’s formulas. The elegance here is that these relations hold regardless of whether the roots are rational, irrational, or even complex.

This technique has serious weightage — JEE Main 2024 Shift 1 had a question asking for $\frac{1}{\alpha^2} + \frac{1}{\beta^2}$ directly, and students who tried to find roots first wasted 3 minutes. With Vieta’s: $\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{(\alpha\beta)^2}$ — done in 30 seconds.

Alternative Method

You can verify by actually solving $2x^2 - 5x + 3 = 0$ using the quadratic formula or factoring.

Factoring: $2x^2 - 5x + 3 = (2x - 3)(x - 1) = 0$, giving $\alpha = \frac{3}{2}$ and $\beta = 1$.

Check: $\alpha + \beta = \frac{3}{2} + 1 = \frac{5}{2}$ ✓ and $\alpha\beta = \frac{3}{2} \cdot 1 = \frac{3}{2}$ ✓

Then $\alpha^2 + \beta^2 = \frac{9}{4} + 1 = \frac{13}{4}$ ✓

This verification works here because the roots came out rational. For equations with irrational roots like $x^2 - 4x + 1 = 0$ (roots: $2 \pm \sqrt{3}$), computing $\alpha^2 + \beta^2$ by direct substitution is painful. Vieta’s approach costs the same effort regardless.

Common Mistake