Vieta's Formulas: Answering Without Solving

An interviewer gives you $x^2 - 7x + 10 = 0$ and asks for the sum of the squares of the roots. The instinct is to solve the quadratic, get $x = 2$ and $x = 5$, then compute $2^2 + 5^2 = 29$. That works fine here because the roots are clean integers — but change the coefficients slightly and the roots become ugly irrational numbers, and "just solve it" turns into a mess of nested radicals. Vieta's formulas let you skip solving entirely: the coefficients themselves already encode the sum and product of the roots.

The formulas, one symbol at a time

For a quadratic $ax^2 + bx + c = 0$ with roots $r_1, r_2$:

$$r_1 + r_2 = -\frac{b}{a}, \qquad r_1 r_2 = \frac{c}{a} .$$

In plain English: the sum of the roots is minus the linear coefficient over the leading coefficient, and the product of the roots is the constant term over the leading coefficient — no need to know what the individual roots are. For a cubic $ax^3+bx^2+cx+d=0$ with roots $r_1,r_2,r_3$, the pattern extends: $r_1+r_2+r_3 = -b/a$, the sum of pairwise products $r_1r_2+r_1r_3+r_2r_3 = c/a$, and $r_1r_2r_3 = -d/a$. Each coefficient, in order, gives you the next symmetric combination of the roots.

Worked example 1: sum of squares without solving

For $x^2 - 7x + 10 = 0$: $r_1+r_2 = 7$ and $r_1 r_2 = 10$. The sum of squares can be built from these using the identity $r_1^2+r_2^2 = (r_1+r_2)^2 - 2r_1r_2$ — a rearrangement of $(r_1+r_2)^2 = r_1^2+2r_1r_2+r_2^2$. Plugging in: $7^2 - 2(10) = 49 - 20 = 29$, matching the direct calculation above but generalizing instantly to coefficients that don't factor nicely, like $x^2 - 7x + 11 = 0$, where the roots involve $\sqrt{5}$ but the sum of squares is still $7^2 - 2(11) = 27$ exactly.

Worked example 2: reconstructing a cubic from its roots

An interviewer asks for a cubic with roots $2, -1, 3$. Rather than expanding $(x-2)(x+1)(x-3)$ term by term, use Vieta's in reverse: sum $= 2+(-1)+3=4$, pairwise sum $= (2)(-1)+(2)(3)+(-1)(3) = -2+6-3=1$, product $= (2)(-1)(3) = -6$. For a monic cubic $x^3 - (\text{sum})x^2 + (\text{pairwise sum})x - (\text{product})$, that gives $x^3 - 4x^2 + x + 6$. Multiplying out the original factored form to check confirms the same polynomial — but Vieta's got there by reading off three sums instead of multiplying three binomials.