Every quadratic equation ax² + bx + c = 0 is solved by the quadratic formula x = − b ± b 2 − 4 a c 2 a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} x = 2 a − b ± b 2 − 4 a c . The expression under the root, b² - 4ac, is the discriminant, and its sign decides how many real solutions there are: positive gives two, zero gives one repeated solution, and negative gives none, which on the graph means the parabola never touches the x-axis. For 2x² - 3x - 4 = 0 the discriminant is 9 + 32 = 41, so there are two solutions, x = 3 ± 41 4 x = \frac{3 \pm \sqrt{41}}{4} x = 4 3 ± 41 . When the discriminant is a perfect square the quadratic factors over the integers and factoring is faster; when it is not, the formula is the tool, and the answer choices will be written with a radical.
The vertex form of a quadratic, y = a(x - h)² + k, names its vertex (h, k) directly, with a still telling you whether the parabola opens up (a positive, k a minimum) or down (a negative, k a maximum). To convert from standard form y = x² + bx + c, complete the square: take half of b, square it, and add and subtract it so that the first three terms form a perfect square. For y = x² + 6x + 2, half of 6 is 3 and 3² = 9, so y = (x² + 6x + 9) - 9 + 2 = (x + 3)² - 7, with vertex (-3, -7). When a is not 1, factor it out of the x-terms first. The vertex's x-coordinate is also -b/(2a), which is quicker when only the location is needed. In context, the vertex is the maximum height, the minimum cost, or the time of the peak, and the word "maximum" or "minimum" in a question is the signal to find it.
An exponential function f(t) = a times b^t has starting value a and growth factor b per unit of t. A quantity growing r percent per period has b = 1 + r/100; one shrinking r percent has b = 1 - r/100. When the period of the rate and the unit of t differ, the exponent is adjusted: a population that doubles every 3 hours is P(t) = P₀ times 2^(t/3) with t in hours, and money earning 6 percent a year compounded monthly is A = P(1 + 0.06/12)^(12t). A table whose outputs have a constant ratio is exponential and the ratio is b; one with a constant difference is linear. Questions ask for the model, for a value at a given t, or for what a or b means: a is the amount at time zero, and b - 1 is the percent change per period as a decimal.
Rational exponents are radicals: x^(1/n) is the nth root of x, and x^(m/n) is the nth root of x^m, which equals the nth root of x raised to the m. So 8^(2/3) = (8^(1/3))² = 2² = 4, and x^(3/2) = x times √x. The exponent rules from Foundations apply unchanged. Radical expressions simplify by pulling out perfect-square factors, √50 = √(25 times 2) = 5 √2, and radical equations are solved by isolating the radical and raising both sides to the matching power. That step can introduce extraneous solutions, values that satisfy the squared equation but not the original: √(2x + 3) = x squares to 2x + 3 = x², whose solutions are x = 3 and x = -1, but only x = 3 works in the original because a square root is never negative. Always check.
A polynomial's zeros and factors are the same information. If (x - a) is a factor of p(x) then p(a) = 0 and the graph has an x-intercept at a; conversely, if p(a) = 0 then (x - a) divides p(x) exactly. A zero with an even multiplicity, such as (x - 2)², touches the axis without crossing; an odd multiplicity crosses. The degree of a polynomial bounds the number of zeros and sets its end behavior: even degree with a positive leading coefficient rises on both ends, odd degree rises on the right and falls on the left. Given a table of values with p(4) = 0, the test expects you to know that x - 4 is a factor, and given a factored form, to read the zeros with their signs reversed.
A rational expression is a fraction of polynomials. Simplify by factoring numerator and denominator completely and cancelling common factors; the values that make the original denominator zero are excluded from the domain even if the factor cancels. For (x² - 9)/(x² - x - 6) = (x - 3)(x + 3)/((x - 3)(x + 2)) = (x + 3)/(x + 2), with x not equal to 3 or -2. To add or subtract rational expressions, use a common denominator exactly as with numeric fractions. An equation with rational expressions is cleared by multiplying through by the common denominator, and any solution that makes a denominator zero is rejected.
A system of one linear and one quadratic equation is solved by substitution: replace y in the quadratic with the linear expression, collect to one side, and solve the resulting quadratic. The solutions are the x-coordinates where the line meets the parabola, and there are two, one or none according to that quadratic's discriminant. For y = x² - 4x + 3 and y = x - 1, substitution gives x² - 5x + 4 = 0, so x = 1 or x = 4, and the intersection points are (1, 0) and (4, 3). A question may ask for the points, for the number of solutions, or for the sum of the x-coordinates, which is available from the collected quadratic without solving it.