Vieta's Relations & Root-Coefficient Identities

If $\alpha, \beta$ are the roots of $ax^2 + bx + c = 0$, then

• Reciprocal roots (one root $= 1/$other) $\iff \alpha\beta = 1 \iff c = a$.
• Roots equal in magnitude, opposite sign $\iff \alpha+\beta = 0 \iff b = 0$.
• Sum of roots $=$ product of roots $\iff -b = c$.
• Sign reading: if $a,b,c$ are all positive then $\alpha+\beta<0$ and $\alpha\beta>0$, so both roots are negative.
• The full sign table for REAL roots ($D\ge0$): product $>0$ means same sign, and then the SUM decides — sum $>0$ both positive, sum $<0$ both negative; product $<0$ means opposite signs, and the sign of the sum tells which root is larger in magnitude; product $=0$ means one root is 0. Read $c/a$ first, then $-b/a$.

Let $s = \alpha+\beta$, $p = \alpha\beta$. The standard identities:

• $\alpha^2 + \beta^2 = s^2 - 2p$
• $(\alpha-\beta)^2 = s^2 - 4p$
• $\alpha^3 + \beta^3 = s^3 - 3ps = s(s^2 - 3p)$
• $\dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{s}{p}$

Forming a new equation: if the new roots have sum $S$ and product $P$, the quadratic is $x^2 - Sx + P = 0$. Compute $S$ and $P$ as symmetric functions of $\alpha,\beta$. Beyond $\alpha^3+\beta^3$ — a root satisfies its own equation. Since $a\alpha^2+b\alpha+c=0$, any power of $\alpha$ reduces: $\alpha^2=-\tfrac{b\alpha+c}{a}$, so e.g. $4\alpha^2=1-2\alpha$ when $4x^2+2x-1=0$. Multiplying by $\alpha^{n-1}$ and adding the $\beta$ copy gives Newton's recurrence for $p_n=\alpha^n+\beta^n$: $a\,p_{n+1}+b\,p_n+c\,p_{n-1}=0$, i.e. $p_{n+1}=s\,p_n-p\,p_{n-1}$ with $p_0=2$, $p_1=s$. So $\alpha^4+\beta^4=s\,p_3-p\,p_2$ and $\alpha^5+\beta^5=s\,p_4-p\,p_3$ follow in two lines each. The other use of a shift: to find an equation whose roots are $\alpha+1,\beta+1$ (or $2\alpha,2\beta$, or $1/\alpha,1/\beta$) substitute $x\to x-1$ (or $x/2$, or $1/x$) into the ORIGINAL equation — the quantity you are asked about becomes a root of the transformed quadratic, and Vieta on it answers the question directly.

For roots $\alpha, \beta$ of $ax^2+bx+c=0$:

• AM $= \dfrac{\alpha+\beta}{2} = -\dfrac{b}{2a}$
• GM $= \sqrt{\alpha\beta} = \sqrt{\dfrac{c}{a}}$
• HM $= \dfrac{2\alpha\beta}{\alpha+\beta} = \dfrac{2c}{-b} = -\dfrac{2c}{b}$
• Roots of equal magnitude, opposite sign: $\alpha+\beta = 0$ (so $b = 0$) and $\alpha\beta < 0$ (so $\frac{c}{a} < 0$). Both conditions are required.
• Denesting the GM's surd: $\sqrt{c/a}$ often comes out as $\sqrt{p+q\sqrt r}$; match it to $(\sqrt m+\sqrt n)^2=m+n+2\sqrt{mn}$ — $3+2\sqrt2=(\sqrt2+1)^2$, $8+4\sqrt3=2(\sqrt3+1)^2$, $7-4\sqrt3=(2-\sqrt3)^2$, $5+2\sqrt6=(\sqrt2+\sqrt3)^2$. The options are written in the denested form.
• Conjugate pair with product 1 lies on the unit circle: if the roots are non-real (so $\beta=\bar\alpha$) and $\alpha\beta=1$, then $|\alpha|^2=\alpha\bar\alpha=1$, i.e. $|\alpha|=|\beta|=1$ — e.g. the roots of $x^2-x+1=0$ or of any $x^2+bx+1=0$ with $|b|<2$.

Two recurring cross-equation setups:

• Swapped roots: "$n$ is a root of $x^2+px+m=0$ and $m$ is a root of $x^2+px+n=0$" — substitute to get $n^2+pn+m=0$ and $m^2+pm+n=0$; subtract them and factor out $(n-m)$ to get $m+n+p = 1$ (when $m \neq n$).
• Same ratio of roots: if $ax^2+bx+c=0$ and $px^2+qx+r=0$ have roots in the same ratio, then $\dfrac{b^2}{ac} = \dfrac{q^2}{pr}$ (equivalently $b^2 pr = q^2 ac$).

For an equation symmetric about $x = m$ (e.g. $(x-a)^4 + (x-b)^4 = k$ with centre $m = \tfrac{a+b}{2}$), substitute $u = x - m$. The odd powers of $u$ cancel, leaving an even equation $u^4 + Au^2 + B = 0$ — a quadratic in $u^2$. Solve for $u^2$, count the real roots (each positive $u^2$ gives two), and for the sum of all roots use Vieta on the degree-$n$ polynomial: sum $= -\dfrac{(\text{coeff of }x^{n-1})}{(\text{coeff of }x^{n})}$ — the complex roots are included.

When the roots feed back into the equation or an extra relation is imposed, set $s = \alpha+\beta$, $p = \alpha\beta$ and translate every condition into $s, p$:

• Coefficients made of the roots: e.g. $x^2 + \alpha x - \beta = 0$ with roots $\alpha, \beta$ gives $\alpha+\beta = -\alpha$ and $\alpha\beta = -\beta$; solve the pair.
• Extra symmetric relation: e.g. $\alpha+\beta = \alpha^2+\beta^2$ becomes $s = s^2 - 2p$; combine with a second given relation to find $s, p$.
• A relation true for ALL such roots is established by checking it against $s$ and $p$, not by finding the roots numerically.

Two structural patterns:

• Integer / rational coefficients force partner roots. A polynomial with integer coefficients that has a root $r$ and is of higher degree often forces a partner (e.g. for a quartic with given roots $-2, 3$, the symmetric construction $(x^2-4)(x^2-9)$ supplies $\pm 2, \pm 3$).
• Equation unchanged under a root transform. "The equation is the same when its roots $\alpha,\beta$ are replaced by $\alpha^2,\beta^2$" means $\{\alpha^2,\beta^2\} = \{\alpha,\beta\}$; enumerate cases ($\alpha^2=\alpha$ or $\alpha^2=\beta$) to count the valid equations. Roots land in $\{0, 1, \omega, \omega^2\}$.