If α and β are roots of 2x²-5x+3=0 find α²+β²

Question

If $\alpha$ and $\beta$ are the roots of $2x^2 - 5x + 3 = 0$ , find $\alpha^2 + \beta^2$ .

Solution — Step by Step

For a quadratic $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$ , Vieta’s formulas give:

\alpha + \beta = -\frac{b}{a}, \quad \alpha\beta = \frac{c}{a}

For $2x^2 - 5x + 3 = 0$ : $a = 2$ , $b = -5$ , $c = 3$

\alpha + \beta = -\frac{-5}{2} = \frac{5}{2}

\alpha\beta = \frac{3}{2}

We use the key identity:

\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta

This avoids finding the roots themselves — we work entirely with the sum and product.

\alpha^2 + \beta^2 = \left(\frac{5}{2}\right)^2 - 2 \times \frac{3}{2}

= \frac{25}{4} - 3

= \frac{25}{4} - \frac{12}{4}

= \boxed{\frac{13}{4}}

Why This Works

Vieta’s formulas extract the sum and product of roots directly from the coefficients, without having to actually solve for the individual roots. This is powerful because many expressions involving roots (like $\alpha^2 + \beta^2$ , $\alpha^3 + \beta^3$ , $\alpha^2\beta + \alpha\beta^2$ ) can be expressed in terms of $(\alpha + \beta)$ and $\alpha\beta$ using algebraic identities.

We never needed to find $\alpha = 1$ and $\beta = 3/2$ separately. Verify: $\alpha^2 + \beta^2 = 1 + 9/4 = 4/4 + 9/4 = 13/4$ ✓

This approach is faster and less error-prone than factoring or using the quadratic formula.

Alternative Method

Direct calculation (using the actual roots):

$2x^2 - 5x + 3 = 0$

$(2x - 3)(x - 1) = 0$ → $x = 3/2$ or $x = 1$

So $\alpha = 3/2$ , $\beta = 1$ (or vice versa)

$\alpha^2 + \beta^2 = (3/2)^2 + 1^2 = 9/4 + 1 = 9/4 + 4/4 = 13/4$ ✓

Same answer. For this problem, direct factoring works easily. But if the quadratic doesn’t factor nicely (e.g., $2x^2 - 5x + 1 = 0$ ), the Vieta’s formula approach is indispensable.

Build a toolkit of symmetric function identities. Given $\alpha + \beta = s$ and $\alpha\beta = p$ :

$\alpha^2 + \beta^2 = s^2 - 2p$

$\alpha^3 + \beta^3 = s^3 - 3sp = s(s^2 - 3p)$

$\alpha^2\beta + \alpha\beta^2 = sp$

$(\alpha - \beta)^2 = s^2 - 4p$ (discriminant!)

$\alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2(\alpha\beta)^2 = (s^2 - 2p)^2 - 2p^2$

These identities come up repeatedly in JEE and Class 10-12 exams. Practice deriving them from $(\alpha + \beta)^2 = \alpha^2 + 2\alpha\beta + \beta^2$ until they feel automatic.

Common Mistake

Students often compute $(\alpha + \beta)^2 = \alpha^2 + \beta^2$ directly — forgetting the middle term $2\alpha\beta$ . The correct expansion is $(\alpha + \beta)^2 = \alpha^2 + 2\alpha\beta + \beta^2$ , so $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$ . This is the most common single-step error in this type of problem. Double-check: $(5/2)^2 = 25/4$ , then subtract $2 \times 3/2 = 3$ . That gives $25/4 - 12/4 = 13/4$ .

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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