Solving Quadratic Equations
Solving Quadratic Equations
Solving the quadratic equation ax2 + bx + c = 0 determines the x-intercepts of the parabola (by making y = 0). These are also called the roots (or zeros) of the quadratic function. A quadratic equation may have zero, one, or two real solutions.
There are several ways to find the roots. One method is to look at the graph of the quadratic. If the graph lies above the x-axis, the quadratic equation has zero roots. If the graph of the vertex is on the x-axis, the quadratic has one roots and if the graph crosses the x-axis at two points, the quadratic has two roots.
Another way to find the roots is to factor the quadratic as a product of two binomials, and then use the zero-product property. (If m × n = 0, then either m = 0 or n = 0). This can only be used for quadratic equations that can be factored.
Find the root(s) of z2 – 4z = –4.
Factor as a binomial.
z2 – 4z = –4
z2 – 4z + 4 = 0
(z – 2)(z – 2) = 0
Set each factor equal to zero and solve for z.
(z – 2) = 0
z = 2
All quadratic equations can be solved using the quadratic formula:
\(x = \frac{−b\ \pm\ \sqrt{b^{2}\ −\ 4ac}}{2a}\)
The a, b, and c are from the standard form of quadratic equations. (Note: to use the quadratic equation, the right-hand side of the equation must be equal to zero, ax2 + bx + c = 0.) The part of the formula under the radical (b2 – 4ac) is known as the discriminant. The discriminant tells how many and what type of roots will result without calculating the roots.The Discriminant | ||
|---|---|---|
If b2 – 4ac is | there will be | and the parabola |
zero | only one real root | only one real root
has its vertex on the x-axis (one x-intercept, since the quadratic equation simplifies to
\(x = \frac{−b}{2a}\)). |
positive | two real roots | has two x-intercepts. |
negative | zero real roots (but two complex roots) | has no x-intercepts (never touches the x-axis). |
Solve:
\(3x^{2}\ +\ 16x = −5\)
Write the equation in standard form.
\(3x^{2}\ +\ 16\ +\ 5 = 0\)
Solve using the quadratic formula.
\(a = 3, b = 16, c = 5\)
\(x = \frac{−b\ \pm\ \sqrt{b^{2}\ −\ 4ac}}{2a}\)
\(= \frac{−16\ \pm\ \sqrt{16^{2}\ −\ 4(3)(5)}}{2(3)}\)
\(= \frac{−16\ \pm\ \sqrt{256\ −\ 60}}{6}\)
\(= \frac{16\ \pm\ \sqrt{196}}{6}\)
\(= \frac{16\ \pm\ 14}{6}\)
\(x= \frac{−16\ +\ 14}{6}=−\frac{1}{3}\ or\ x = \frac{−16\ −\ 14}{6} = −5\)
The solutions are –5 and \(−\frac{1}{3}\)