Advanced: Advanced Math

The hardest Advanced Math questions replace a number with a constant and ask what the constant must be, or ask about the solutions of an equation without letting you find them, or hand you two expressions and ask when they are the same. These questions cannot be brute-forced with a calculator; they reward knowing three structural facts about polynomials.

The first is that the discriminant controls the number of solutions, so a condition like "exactly one solution" or "the line is tangent to the parabola" is an equation in the constant. The second is that the coefficients of a quadratic encode the sum and product of its roots, so questions about the roots often need no roots. The third is that a polynomial's value at a equals its remainder on division by x - a, so a table of values is a table of factors.

This lesson works through each, then finishes with the equation types the Medium lesson introduced, now with their traps: rational equations that lose their only candidate, exponential equations that need a common base, and rates that must be rewritten per month instead of per year.

A quadratic with a constant in it, such as x² + kx + 16 = 0, has a number of real solutions that depends on k through the discriminant k² - 64. Exactly one real solution requires k² - 64 = 0, so k = 8 or k = -8; no real solutions requires k² < 64, so k is strictly between -8 and 8; two distinct real solutions requires k² > 64. When the constant sits in the leading coefficient or the constant term, the same idea applies, with the caution that a leading coefficient of zero would make the equation linear, not quadratic. A line y = mx + c is tangent to a parabola exactly when substituting the line into the parabola gives a quadratic with discriminant zero: y = 2x + c and y = x² meet where x² - 2x - c = 0, whose discriminant 4 + 4c is zero at c = -1, and the tangent point is then (1, 1). For c greater than -1 the line cuts the parabola twice, for c less than -1 it misses.

If r and s are the roots of ax² + bx + c = 0, then ax² + bx + c = a(x - r)(x - s) = a x² - a(r + s)x + a rs, and matching coefficients gives r + s = -b/a and rs = c/a. These two facts answer questions about the roots without finding them. For 2x² - 7x + 3 = 0 the roots sum to 7/2 and multiply to 3/2, so the sum of their reciprocals, (r + s)/(rs), is 7/3, and the sum of their squares, (r + s)² - 2rs, is 49/4 - 3 = 37/4. For 3x² - 12x + 5 = 0 the sum of the solutions is 12/3 = 4, a fact the test asks about directly and one that also locates the axis of symmetry at x = 2. Given one root of a quadratic with a constant, the other root and the constant both follow from the sum and product.

Two polynomial expressions are equivalent, equal for every value of x, exactly when the coefficients of each power of x agree. To find constants that make (2x + a)(x - 3) equivalent to 2x² + bx - 15, expand the left side: 2x² - 6x + ax - 3a = 2x² + (a - 6)x - 3a. Matching constants gives -3a = -15, so a = 5; matching x-terms then gives b = a - 6 = -1. When a question asks for one constant and the expansion produces several equations, at least one of them involves only that constant. Substituting a specific value of x gives one equation relating the constants and is a good check, but it is rarely enough on its own. Equivalent-expression questions with rational expressions work the same way after multiplying through by the common denominator, and questions with a radical or a rational exponent usually want the exponent rules from the Medium lesson applied to expressions with letters in place of numbers.

Dividing a polynomial p(x) by x - a leaves a remainder equal to p(a): p(x) = (x - a)q(x) + p(a) for some polynomial q. So x - a is a factor exactly when p(a) = 0, and a table of values with p(2) = 0 announces the factor x - 2 without any division. For p(x) = x³ - 4x² + x + 6, p(2) = 8 - 16 + 2 + 6 = 0, so x - 2 is a factor, while p(1) = 1 - 4 + 1 + 6 = 4, so division by x - 1 leaves remainder 4 and x - 1 is not a factor. Given a statement in the form p(x) = (x - 3)q(x) + 5, you can read p(3) = 5 directly. Multiplicity matters for graphs: a factor (x - a)² makes the graph touch the axis at a without crossing, and the number of distinct real zeros is at most the degree.

A rational equation is cleared by multiplying every term by the common denominator, and every candidate solution must be checked against the original denominators. Sometimes the only candidate is excluded and the equation has no solution: 3/(x - 2) + 1 = x/(x - 2) becomes 3 + (x - 2) = x after multiplying by x - 2, which simplifies to 1 = 0, a contradiction, so there is no solution. A rational function is undefined where its denominator is zero, and a question that asks for what value the expression is undefined is asking for the zero of the denominator, even when that factor cancels with the numerator.

Exponential equations with the variable in the exponent are solved by rewriting both sides as powers of one base and equating the exponents: 2^(x + 1) = 8^(x - 1) becomes 2^(x + 1) = 2^(3x - 3), so x + 1 = 3x - 3 and x = 2. Equivalent-rate questions rewrite an exponential model so its exponent counts a different period. Since (b^(1/n))^(nt) = b^t, a yearly growth of 6 percent, factor 1.06, is the same as a monthly factor of 1.06^(1/12), about 1.0049, a monthly rate near 0.49 percent, and a quantity that halves every 10 hours, factor (1/2)^(t/10), loses about 6.7 percent per hour, not 5 percent. The test's standard wrong answer is the yearly rate divided by 12.

Translating a graph changes the equation in a predictable way. Replacing x by x - h shifts the graph h units right; adding k to the output shifts it k units up; multiplying the output by -1 reflects across the x-axis. So if f has vertex (1, -4), then g(x) = f(x - 3) + 2 has vertex (4, -2), and the same rule moves the zeros, the intercepts and the maximum. A parabola read from its graph is best written in factored form when its x-intercepts are visible, y = a(x - p)(x - q), with a fixed by one more point, or in vertex form when the vertex is visible. Questions that show two graphs and ask how one was obtained from the other are asking which of these substitutions connects them.

Solving the quadratic for x when the question asks for the constant that produces a given number of solutions. Write the discriminant condition instead.

Forgetting the sign in the sum of roots: r + s is -b/a, not b/a.

Matching only one coefficient and stopping, or substituting one value of x and treating a single equation as a full match.

Dividing polynomials by long division when a single evaluation gives the remainder.

Accepting x = 2 as a solution of an equation whose denominator is x - 2, or failing to notice that a rational equation collapsed to a contradiction.

Equating exponents before the bases are the same: 2^(x + 1) = 8^(x - 1) does not give x + 1 = x - 1.

Converting a yearly rate to a monthly one by dividing by 12. Take the twelfth root of the yearly factor.

Shifting a graph left for f(x - 3): a subtraction inside moves it right.