Quadratics #
Solving by Plotting/Graphing #
To solve the equation \( x^2 + 6x + 9 = 0 \) by graphing, create an xy-table with points. It’s advisable to plot at least 5 points, though 3 can suffice. The axis of symmetry, determined by \( \frac{-b}{2a} \), marks the midpoint.
| x | y |
|---|---|
| -5 | 4 |
| -4 | 1 |
| -3 | 0 |
| -2 | 1 |
| -1 | 4 |
The zero is located at \( x = -3 \), which happens to be our axis of symmetry in this case. If the table had not included the solution, we would plot our 5 points, connect them all, and see where the x-intercepts would be.
Solving by Grouping #
Take the quadratic \( 5x^2 - 3x - 2 = 0 \). We need to find what multiplies to \( ac \) and adds to \( b \). The two numbers that satisfy these properties in this case are \( -5 \) and \( 2 \). Replace the \( b \) term with these terms that add to \( b \). Imagine a bar dividing the first pair of terms and second pair of terms and factor. After factoring, set both factors equal to zero to find your roots.
\[ \begin{aligned} 5x^2 - 3x - 2 &= 0 \\ 5x^2 - 5x + 2x - 2 &= 0 \\ 5x(x - 1) + 2(x - 1) &= 0 \\ (5x + 2)(x - 1) &= 0 \\ x &= -\frac{2}{5},\ 1 \end{aligned} \]Solving by Completing the Square #
To complete the square, your \( a \) coefficient must be 1 and your \( c \) term must be on the right side.
\[ \begin{aligned} 8x^2 - 32x - 64 &= 0 \\ x^2 - 4x - 8 &= 0 \\ x^2 - 4x &= 8 \end{aligned} \]After this, you must add \( \left(\frac{b}{2}\right)^2 \) to both sides. Then, factor the left-hand side to \( \left(x + \frac{b}{2}\right)^2 \), which is \( (x - 2)^2 \) in this case.
\[ \begin{aligned} x^2 - 4x + (-2)^2 &= 8 + (-2)^2 \\ x^2 - 4x + 4 &= 12 \\ (x - 2)^2 &= 12 \end{aligned} \]The rest is algebra. Don’t forget the \( \pm \) when taking the square root to account for the possibility of a second solution.
\[ \begin{aligned} \sqrt{(x - 2)^2} &= \pm\sqrt{12} \\ x - 2 &= \pm\sqrt{12} \\ x &= \pm\sqrt{12} + 2 \end{aligned} \]Solving by the Quadratic Formula #
The quadratic formula is written below. Substitute your coefficients. We will use the quadratic \( x^2 + 2x + 1 = 0 \) as example.
\[ \begin{aligned} x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\ x &= \frac{-2 \pm \sqrt{4 - 4}}{2} \\ x &= \frac{-2 \pm \sqrt{0}}{2} \\ x &= \frac{-2 \pm 0}{2} \\ x &= \frac{-2}{2} \\ x &= -1 \end{aligned} \]What is a complex number? #
Try solving the equation \( x^2 = -1 \). No matter which method you use, your answer will be undefined. To circumvent this, mathematicians defined a new, special number: \( i = \sqrt{-1} \). To the previous equation, we can now say: \( x = \pm i \). We can scale this new “imaginary unit” by normal numbers as well: numbers such as \( 4i \) are equally valid.
We can combine real numbers and imaginary numbers to form a new set of numbers: the complex numbers. These are numbers of the form \( a + bi \), such as \( 2 + 3i \). Try solving \( x^2 + 2x + 1 = 0 \) using the Quadratic Formula.
Writing Vertex Form from a Table #
| x | y |
|---|---|
| -3 | 6 |
| -2 | 3 |
| -1 | 2 |
| 0 | 3 |
| 1 | 6 |
The vertex has the highest or lowest \( y \) value, so our vertex is \( (-1, 2) \). The vertex is located at \( (h, k) \) so we can substitute these numbers into the vertex form formula.
\[ \begin{aligned} y &= a(x - h)^2 + k \\ y &= a(x - (-1))^2 + 2 \\ y &= a(x + 1)^2 + 2 \end{aligned} \]We are still left with an \( a \) coefficient. We can solve for this coefficient by plugging in a non-vertex point for \( x \) and \( y \). Let’s choose \( (-3, 6) \).
\[ \begin{aligned} 6 &= a(-3 + 1)^2 + 2 \\ 6 &= a(-2)^2 + 2 \\ 6 &= 4a + 2 \\ a &= 1 \end{aligned} \]We found our \( a \) coefficient which we can plug in and get our final answer:
\[ \begin{aligned} y &= 1(x + 1)^2 + 2 \\ y &= (x + 1)^2 + 2 \end{aligned} \]