Vieta's formulas for $ax^2$ + bx + c = 0 with roots alpha, beta: alpha + beta = -b/a (sum) and alpha * beta = $\frac{c}{a}$ (product). These extend to higher-degree polynomials: for a cubic with roots alpha, beta, gamma, the elementary symmetric polynomials give sum, sum of pairwise products, and triple product.

The power of Vieta's formulas lies in computing symmetric expressions without finding individual roots. Key identities: $alpha^2$ + $beta^2$ = (alpha+beta)^2 - 2alphabeta. $alpha^3$ + $beta^3$ = (alpha+beta)^3 - 3alphabeta*(alpha+beta). |alpha - beta| = sqrt((alpha+beta)^2 - 4alphabeta) = sqrt(D)/|a|. 1/alpha + 1/beta = $\frac{alpha+beta}{(alpha*beta)}$.

Newton's identity provides a recurrence for power sums $S_k$ = $alpha^k$ + $beta^k$: $S_k$ = (alpha+beta)S_(k-1) - alphabeta*S_(k-2) with $S_0$ = 2, $S_1$ = alpha+beta. This efficiently computes $alpha^n$ + $beta^n$ for large n without finding the roots.

Transformation of equations: to find the equation whose roots are f(alpha), f(beta), use the substitution method. For roots kalpha, kbeta: replace x by x/k. For roots alpha + h, beta + h: replace x by x - h. For roots 1/alpha, 1/beta: replace x by 1/x and clear fractions. For roots $alpha^2$, $beta^2$: sum = (alpha+beta)^2 - 2alphabeta, product = (alpha*beta)^2.

JEE trap: Students often compute $alpha^2$ + $beta^2$ by finding individual roots and squaring. This is both slower and more error-prone than using (alpha+beta)^2 - 2alphabeta directly from coefficients.