Equations and Inequalities

1Sum and product of roots

For ax^2+bx+c=0 with roots \alpha and \beta, the sum is \alpha+\beta=-\frac{b}{a} and the product is \alpha\beta=\frac{c}{a}. These two identities are usually faster than the quadratic formula when a CAT question asks only for \alpha+\beta, \alpha\beta, \alpha^2+\beta^2, or a symmetric expression in the roots — read sum and product off the coefficients, do not solve first.

To rebuild the monic quadratic from known roots, write x^2-(\alpha+\beta)x+\alpha\beta=0. Scaling by a recovers the general form ax^2+bx+c=0 with the same sum and product.

Figure. Roots \alpha=2 and \beta=3 on y=(x-2)(x-3) (drawn opening up on screen). Read \alpha+\beta=-b/a=5 and \alpha\beta=c/a=6 from the coefficients — do not solve first when the question only wants a symmetric expression.

Reading Vieta off the equation

1. Standard formWrite ax^2+bx+c=0 with a\ne0; rearrange first if the equation is not already quadratic-standard.
2. Read sum and productSum =-b/a, product =c/a. Do not solve for \alpha and \beta unless the question asks for the roots themselves.
3. Rebuild if neededFrom known roots, form x^2-(\text{sum})x+(\text{product})=0, then multiply by a if a leading coefficient is required.

2Nature of roots from the discriminant

For ax^2+bx+c=0, the discriminant \Delta=b^2-4ac decides whether the roots are real and whether they coincide. A positive \Delta gives two distinct real roots; zero gives a repeated real root — one root, not two; a negative \Delta means no real roots at all. On CAT, equal roots always mean \Delta=0, and confusing that condition with the sum or product formulas is a common slip.

Figure. Solid curve: \Delta>0 crosses the axis twice. Dashed curve: \Delta=0 touches once (repeated root). \Delta<0 never meets the real axis — no real roots.

Reading the discriminant

1. Identify coefficientsRead a, b, c from the equation in standard form — rearrange first if needed.
2. Compute \DeltaEvaluate b^2-4ac exactly; keep the sign.
3. Match the caseUse the table to name the root type before solving, unless the question only asks for a parameter from equal roots.

3The quadratic formula

When factoring fails or a parameter sits inside the coefficients, the quadratic formula gives every real root explicitly: x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}. Compute \Delta=b^2-4ac first — if it is negative and only real roots are wanted, stop; if it is zero the \pm collapses to one value.

Figure. Compute Δ = b²−4ac first. If it is negative and only real roots are wanted, stop; if it is zero the ± collapses to one value; otherwise keep both signs. For 2x²+3x−2=0, Δ = 9+16 = 25, so x = (−3±5)/4 gives 1/2 or −2.

Applying the formula

1. Standard formWrite ax^2+bx+c=0 with a\ne0; divide through if the leading coefficient is not 1.
2. Compute \DeltaEvaluate b^2-4ac and check its sign before taking the square root.
3. Substitute both branchesPlug into \frac{-b\pm\sqrt{\Delta}}{2a} and simplify each branch separately.

4Common root of two quadratics

When two quadratics share a common root, subtract their left-hand sides (after aligning both to =0) to cancel the x^2 terms and obtain a linear equation that pins that root — subtract first, do not guess factors. Substitute the candidate back into either original equation to confirm, and to recover the second roots if needed.

Figure. Two quadratics share the root x=2. Subtracting the monic forms cancels x^2 and leaves a linear equation that pins that common root — then substitute back to confirm.

Pinning a shared root

1. Align both equationsWrite each quadratic as =0 with the same variable powers lined up.
2. SubtractSubtract one from the other so the x^2 coefficients cancel (scale first if the leading coefficients differ).
3. Solve and verifySolve the resulting linear equation for the shared root, then substitute into either original quadratic to confirm and find the other root of each.

5Minimizing with AM-GM

For non-negative x and y, the AM-GM inequality says \frac{x+y}{2}\ge\sqrt{xy}, with equality exactly when x=y. On CAT this is the standard one-line minimum for expressions like x+\frac{k}{x} with x>0: the minimum value is 2\sqrt{k}, reached at x=\sqrt{k} — no calculus required.

Figure. For x>0, (x+1/x)/2 \ge 1, with equality only at x=1. The solid AM curve never dips below the dashed GM line — that contact point is the AM-GM minimum.

AM-GM minimum in three moves

1. Check positivityConfirm every term you feed AM-GM is non-negative (usually x>0 and k>0).
2. Apply AM-GMWrite \frac{x+(k/x)}{2}\ge\sqrt{x\cdot(k/x)}=\sqrt{k}, so x+k/x\ge 2\sqrt{k}.
3. Read equalitySet the two terms equal: x=k/x, hence x=\sqrt{k} (positive branch).

6Modulus equations and inequalities

The inequality |x-a|<r means x lies in the open interval (a-r,a+r), while |x-a|>r means x lies outside the closed interval [a-r,a+r]. For equations that sum several absolute values, split the number line at each critical point (where an inside expression changes sign) and solve a linear equation on each piece, checking the middle interval separately — it often collapses to a constant.

A related CAT shortcut: for |x-a|+|x-b| with a\le b, the minimum value is |a-b|, achieved for every x in the closed interval between a and b.

Figure. The open interval (a-r, a+r) is the solution of |x-a|<r; outside the closed interval is |x-a|>r. Endpoints are excluded for the strict inequalities.

Solving a multi-piece modulus equation

1. Mark critical pointsList every point where an expression inside a modulus is zero; these split the real line into intervals.
2. Solve on each pieceOn each open interval (and at the endpoints if needed), drop the moduli with the correct signs and solve the resulting linear equation.
3. Check the middleOn an interval between two critical points the sum of moduli is often a constant — compare it to the right-hand side before solving.

7Sign-scheme for polynomial inequalities

For a factored polynomial inequality, mark every real root on a number line and test the sign of the product on each open interval those roots create. A standard CAT habit: start from the rightmost interval (where every linear factor is positive for a monic product) and alternate signs as you cross each simple root. Intervals where the product is positive or negative — according to the inequality — are the solution; exclude roots when the inequality is strict.

Figure. Sign chart for (x-1)(x-3): positive left of 1, negative between 1 and 3, positive right of 3. The strict inequality keeps only the middle open interval.

Building the sign chart

1. Factor and mark rootsWrite the polynomial as a product of linear factors over the reals and place each root on the number line.
2. Sign from the rightOn the rightmost interval every factor (x-r) is positive for a monic product; record +, then flip sign each time you cross a simple root.
3. Read the inequalityKeep the intervals whose sign matches the inequality; drop or keep the roots themselves according to whether the inequality is strict.

8Integer solutions by stars and bars

For a linear equation x_1+x_2+\cdots+x_k=n in non-negative integers, the number of solutions is \binom{n+k-1}{k-1} (equivalently \binom{n+k-1}{n}) by stars and bars. The same count covers x+y+z=n with x,y,z\ge 0. If a variable is required to be at least 1, substitute x_i'=x_i-1 and reduce to the non-negative case before applying the formula.

Figure. One placement of k-1 bars among n stars is one non-negative solution of x_1+\cdots+x_k=n; count the placements, not the values.

Counting non-negative solutions

1. State n and kIdentify the target sum n and the number of variables k, and confirm each variable is a non-negative integer (or reduce to that case).
2. Apply the binomialThe count is \binom{n+k-1}{k-1}. Prefer whichever of \binom{\cdot}{k-1} or \binom{\cdot}{n} is smaller to compute.
3. Adjust boundsIf a lower bound is positive, shift variables first; if an upper bound appears, subtract the complementary count.