Quadratics

Standard form

A quadratic equation in standard form is \(ax^2 + bx + c = 0\), with \(a \ne 0\). Get this form before factoring or using the formula.

Example: \(2x^2 + 3x = 2\) becomes \(2x^2 + 3x - 2 = 0\): \(a = 2,\ b = 3,\ c = -2\).

Square root property

Square root property: if \(x^2 = k\), then \(x = \pm\sqrt{k}\). Isolate the square first. If \(k \lt 0\), the solutions are complex.

Example: \(x^2 = 18\): \(x = \pm 3\sqrt{2}\). \(\;2(x - 1)^2 = 50\): \((x - 1)^2 = 25\), \(x - 1 = \pm 5\), so \(x = 6\) or \(x = -4\).

Example: \(x^2 = -16\): \(x = \pm 4i\). \(\;(x + 3)^2 = -8\): \(x = -3 \pm 2i\sqrt{2}\).

Watch out: do not forget the \(\pm\): there are usually two solutions.

Completing the square

Completing the square: 1. If \(a \ne 1\), divide every term by \(a\). 2. Move \(c\) to the right. 3. Add \(\left(\dfrac{b}{2}\right)^2\) to both sides. 4. Factor the left as a square. 5. Use the square root property.

Example: \(x^2 + 6x - 7 = 0\): \(x^2 + 6x = 7\); add 9: \((x + 3)^2 = 16\), so \(x = 1\) or \(x = -7\).

Example: \(2x^2 - 8x + 3 = 0\): \(x^2 - 4x = -\dfrac{3}{2}\); add 4: \((x - 2)^2 = \dfrac{5}{2}\), so \(x = 2 \pm \dfrac{\sqrt{10}}{2}\).

Using the quadratic formula

Standard form first. Write down \(a\), \(b\), \(c\) with their signs. Simplify the radical, then divide every term by \(2a\) (or simplify the fraction).

Example: \(x^2 - 4x + 1 = 0\): \(x = \dfrac{4 \pm \sqrt{16 - 4}}{2} = \dfrac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3}\).

Example: \(2x^2 + 3x - 2 = 0\): \(x = \dfrac{-3 \pm \sqrt{25}}{4}\), so \(x = \dfrac{1}{2}\) or \(x = -2\).

Watch out: if \(b\) is negative, \(-b\) is positive; and \(b^2\) is never negative.

The discriminant

Example: \(x^2 + 2x + 5 = 0\): \(4 - 20 = -16 \lt 0\), two complex solutions, \(x = -1 \pm 2i\).

Which method?

Only \(x^2\) or \((x - h)^2\) and a number: square root property. Factors easily: factor. Otherwise: the quadratic formula (it always works).

Graphing a parabola

The graph of \(f(x) = ax^2 + bx + c\) is a parabola. \(a \gt 0\): opens up (has a minimum). \(a \lt 0\): opens down (has a maximum). Vertex at \(x = -\dfrac{b}{2a}\); find \(y\) by evaluating. Axis of symmetry: \(x = -\dfrac{b}{2a}\). y-intercept: \((0, c)\). x-intercepts: solve \(f(x) = 0\).

Example: \(f(x) = x^2 - 4x - 5\): \(x = \dfrac{4}{2} = 2\), \(f(2) = -9\), vertex \((2, -9)\). Zeros: \((x - 5)(x + 1) = 0\).

Vertex form

Vertex form: \(f(x) = a(x - h)^2 + k\) has vertex \((h, k)\).

Example: \(f(x) = -2(x + 1)^2 + 3\): vertex \((-1, 3)\), opens down, maximum value 3.

Watch out: \((x + 1)\) means \(h = -1\): the sign inside flips.

Maximum and minimum word problems

The vertex gives the highest or lowest value. Find \(t\) (or \(x\)) with \(-\dfrac{b}{2a}\), then evaluate to get the max or min.

Example: \(h(t) = -16t^2 + 64t + 5\): \(t = \dfrac{-64}{-32} = 2\) s, \(h(2) = 69\) ft is the highest point.

Example: 40 ft of fence around a rectangle: area \(A = x(20 - x)\), largest at \(x = 10\), \(A = 100\) ft².

Quadratic inequalities

1. Move everything to one side (0 on the other). 2. Find the zeros. 3. Test one number in each interval (or picture the parabola). 4. Write the solutions in interval notation.

Example: \(x^2 - x - 6 \lt 0\): zeros \(-2\) and \(3\). Test \(0\): \(-6 \lt 0\) is true, so \((-2, 3)\).

Example: \(x^2 - x - 6 \ge 0\): \((-\infty, -2] \cup [3, \infty)\).