Solving Quadratic Equations
This lesson will act as a summary of all the previous lessons in this unit for how to determine the y-value and the \(x\)-values/roots of quadratic equations using the various different methods.
Quadratic Formula
We can determine the roots of a quadratic equation using the given formula:
This formula is useful for determining the zeroes of a parabola or when it's difficult to factor a quadratic equation. Remember to set the equation to \(0\) before determining the values of \(\textcolor{red}{a},\textcolor{green}{b},\) and \(\textcolor{blue}{c}\).
First, we can shift all terms onto one side:
Next, we can use the quadratic formula to determine the roots. In this instance, \(\textcolor{red}{a = 2}\), \(\textcolor{green}{b = 7}\) and \(\textcolor{blue}{c = -4}\):
\(x = \cfrac{-\textcolor{green}{b} \pm \sqrt{\textcolor{green}{b}^2 - 4\textcolor{red}{a}\textcolor{blue}{c}}}{2\textcolor{red}{a}}\)
\(x = \cfrac{-(\textcolor{green}{7}) \pm \sqrt{(\textcolor{green}{7})^2 - 4(\textcolor{red}{2})(\textcolor{blue}{-4})}}{2(\textcolor{red}{2})}\)
\(x = \cfrac{-7 \pm \sqrt{49 + 32}}{4}\)
\(x = \cfrac{-7 \pm \sqrt{81}}{4}\)
Using \(\pm\), we can determine the \(2\) roots either by adding or subtracting the square root value.
We can determine the first root by adding the square root value:
\(x_1 = \cfrac{-7 + 9}{4}\)
\(x_1 = \cfrac{2}{4}\)
\(x_1 = \cfrac{1}{2}\)
We can determine the second root by adding the square root value:
\(x_2 = \cfrac{-7 - 9}{4}\)
\(x_2 = \cfrac{-16}{4}\)
\(x_2 = -4\)
We can now determine the \(y\)-intercept by setting \(x = 0\):
\(y = 2(0)² + 7(0) - 4\)
\(y = - 4\)
In order to determine the vertex, we first need to identify the Axis of Symmetry. We can do this by dividing the sum of the \(x\)-intercepts by \(2\):
\(\text{AOS} = \cfrac{x_1 + x_2}{2}\)
\(\text{AOS} = \cfrac{1/2 - 4}{2}\)
\(\text{AOS} = \cfrac{-7/2}{2}\)
\(\text{AOS} = \cfrac{-7}{4}\)
We can plug this value into the original equation to determine the Optimal Point:
\(y = 2c² + 7c - 4\)
\(\text{OP} = 2\left(\cfrac{-7}{4}\right)² + 7\left(\cfrac{-7}{4}\right) - 4\)
\(\text{OP} = 2\left(\cfrac{49}{16}\right) - \cfrac{49}{4} - 4\)
\(\text{OP} = \cfrac{98}{16} - \cfrac{196}{16} - \cfrac{64}{16}\)
\(\text{OP} = \cfrac{-162}{16}\)
As we determined that the vertex is \(\boldsymbol{\left(\cfrac{-7}{4}, \cfrac{-162}{16}\right)}\), we can now draw our graph:
Standard Form
The Standard Form of a quadratic equation is expressed as such:
In order to determine the \(x\)-intercepts, we need to fully factor the equation or use the quadratic equation. In order to determine the \(y\)-intercept, we need to set \(x = 0\) and solve for \(y\).
First, we can easily factor this equation in order to determine the x-intercepts:
\(y = 3x² - 3x - 9x + 9\)
\(y = 3x(x - 1)-9(x - 1)\)
\(y = (x - 1)(3x - 9)\)
\(y = 3(x - 1)(x - 3)\)
We can determine the roots as \(1\) and \(3\). We can now determine the \(y\)-intercept:
\(y = (3(0)² - 12(0) + 9\)
\(y = 9\)
In order to determine the vertex, we first need to identify the Axis of Symmetry. We can do so by dividing the sum of the \(x\)-intercepts by \(2\):
\(\text{AOS} = \cfrac{x_1 + x_2}{2}\)
\(\text{AOS} = \cfrac{3 + 1}{2}\)
\(\text{AOS} = \cfrac{4}{2}\)
\(\text{AOS} = 2\)
We can substitute the Axis of Symmetry for \(x\) in the original equation to determine the Optimal Point:
\(y = 3x² - 12x + 9\)
\(\text{OP} = 3(2)² + -12(2) + 9\)
\(\text{OP} = (3)(4) - 24 + 9\)
\(\text{OP} = 12 - 24 + 9\)
\(\text{OP} = -3\)
As we determined that the vertex is \(\boldsymbol{(2, -3)}\), we can now draw our graph:
Factored Form
The Factored Form of a quadratic equation is expressed as such:
The \(x\)-intercepts are \(p\) and \(q\) which will make the equation equal \(0\).
The \(y\)-intercept can be determined by setting \(x = 0\), then solving for \(y\).
Based on the given equation, we can easily determine the \(x\)-intercepts:
\(x₂ = 7\)
We can now determine the \(y\)-intercept:
\(y = -\cfrac{1}{2}(0-3)(0-7)\)
\(y = -\cfrac{1}{2}(-3)(-7)\)
\(y = -\cfrac{1}{2}(21)\)
\(y = -\cfrac{21}{2}\)
In order to determine the vertex, we first need to identify the Axis of Symmetry. We can do so by dividing the sum of the \(x\)-intercepts by \(2\)::
\(\text{AOS} = \cfrac{x_1 + x_2}{2}\)
\(\text{AOS} = \cfrac{3 + 7}{2}\)
\(\text{AOS} = \cfrac{10}{2}\)
\(\text{AOS} = 5\)
We can substitute the Axis of Symmetry for \(x\) in the original equation to determine the Optimal Point:
\(\text{OP} = -\cfrac{1}{2}(x-3)(x-7)\)
\(\text{OP} = -\cfrac{1}{2}(5-3)(5-7)\)
\(\text{OP} = -\cfrac{1}{2}(2)(-2)\)
\(\text{OP} = -\cfrac{1}{2}(-4)\)
\(\text{OP} = 2\)
As we determined that the vertex is \(\boldsymbol{(5, 2)}\), we can now draw our graph:
Vertex Form
The Vertex Form of a quadratic equation is expressed as such:
The vertex is already provided with \(h\) representing the \(x\)-coordinate and \(k\) representing the \(y\)-coordinate. The \(y\)-intercept can be determined by setting \(x = 0\). Depending on the equation, there may be \(0\), \(1\), or \(2\) \(x\)-intercepts.
Based on the given equation, we can determine the \(y\)-intercept:
\(y = -2(0 + 4)² + 3\)
\(y = -2(4)² + 3\)
\(y = -2(16) + 3\)
\(y = -32 + 3\)
\(y = 29\)
Since the equation is shifted upwards and has a negative sign we can determine that this equation has \(x\)-intercept(s):
\(-2(x + 4)² = -3\)
\(\cfrac{-2(x + 4)²}{-2} = \cfrac{-3}{-2}\)
\(\sqrt{(x + 4)²} = \sqrt{\cfrac{3}{2}}\)
\(x + 4 = \sqrt{\cfrac{3}{2}}\)
\(x = -4 \pm \sqrt{\cfrac{3}{2}}\)
We can determine the \(2\) \(x\)-intercepts either by adding or subtracting the square root value from \(-4\):
We can determine the first root by adding the square root value:
\(x_1 = -4 + \sqrt{\cfrac{3}{2}}\)
\(x_1 = -2.78\)
We can determine the second root by subtracting the square root value:
\(x_2 = -4 - \sqrt{\cfrac{3}{2}}\)
\(x_2 = -5.22\)
Therefore, we can determine that the \(x\)-intercepts are \(\boldsymbol{-2.78}\) and \(\boldsymbol{-5.22}\), and the \(y\)-intercept is \(\boldsymbol{29}\).
From looking at the equation, we can identify that \(h = -4\) and \(k = 3\). Therefore, we can determine the vertex is \(\boldsymbol{(-4, 3)}\).
With all values identified, we can now draw our graph:
