Quadratic Formula
The Quadratic Formula is way of finding the solutions/roots of a quadratic function where \(a ≠ 0\). This works for any quadratic function set in Standard Form, (\(ax^2 + bx + c = 0\)).
This formula can be expressed as:
- \(x\) represents the roots
- \(\textcolor{red}a\) represents the coefficient of \(x^2\)
- \(\textcolor{green}b\) represents the coefficient of \(x\)
- \(\textcolor{blue}c\) represents the constant
The quadratic formula is useful for determining the roots of a quadratic equation without having to graph or factor it the regular way.
Example
Determine the roots for the quadratic function \(2x^2 + 7x - 4 = 0\).
Since we can identify \(\textcolor{red}{a = 2}\), \(\textcolor{green}{b = 7}\), and \(\textcolor{blue}{c= -4}\), we can plug these values into the quadratic formula to solve for the roots:
\(x = \cfrac{-\textcolor{green}b \pm \sqrt{\textcolor{green}b^2 - 4\textcolor{red}a\textcolor{blue}c}}{2\textcolor{red}a} \)
\(x = \cfrac{-\textcolor{green}{7} \pm \sqrt{(\textcolor{green}{7})^2 - 4(\textcolor{red}{2})(\textcolor{blue}{-4})}}{2(\textcolor{red}{2})} \)
\(x = \cfrac{-7 \pm \sqrt{49 + 32}}{4} \)
\(x = \cfrac{-7 \pm \sqrt{81}}{4} \)
\(x = \cfrac{-7 \pm 9}{4} \)
Using \(\pm\), we can determine the 2 roots either by adding or subtracting the 2 values in the numerator:
We can determine the first root by adding the values in the numerator:
\(x₁ = \cfrac{-7 - 9}{4}\)
\(x₁ = \cfrac{-16}{4}\)
\(x₁ = -4\)
We can determine the second root by subtracting the values in the numerator:
\(x₂ = \cfrac{-7 + 9}{4}\)
\(x₂ = \cfrac{2}{4}\)
\(x₂ = \cfrac{1}{2}\)
Therefore, we can determine that the roots are \(\boldsymbol{-4}\) and \(\boldsymbol{\cfrac{1}{2}}\).
Enter in coefficients for the trinomial or click on the button to generate a new question. Then, use the quadratic equation to solve for the roots.
First, we can set the function to Standard Form by using expanding it and moving all terms to one side:
\((x)(2x) + (x)(-3) = 7\)
\(2x^2 - 3x = 7\)
\(2x^2 - 3x - 7 = 0\)
Since we can identify \(\textcolor{red}{a = 2}\), \(\textcolor{green}{b = -3}\) and \(\textcolor{blue}{c = -7}\), we can plug these values into the quadratic formula to solve for the roots:
\(x = \cfrac{-\textcolor{green}b +- \sqrt{\textcolor{green}b^2 - 4\textcolor{red}a\textcolor{blue}c}}{2\textcolor{red}a}\)
\(x = \cfrac{-(\textcolor{green}{-3}) +- \sqrt{(\textcolor{green}{-3})^2 - 4(\textcolor{red}{2})(\textcolor{blue}{-7})}}{2(\textcolor{red}{2})}\)
\(x = \cfrac{3 +- \sqrt{9 + 56}}{4}\)
\(x = \cfrac{3 +- \sqrt{65}}{4}\)
Using \(\pm\), we can determine the roots either by adding or subtracting the \(2\) values in the numerator.
\(x1 = \cfrac{3 + \sqrt{65}}{4}\)
\(x2 = \cfrac{3 - \sqrt{65}}{4}\)
Therefore, we can determine that the roots are \(\boldsymbol{\cfrac{3 + \sqrt{65}}{4}}\) and \(\boldsymbol{\cfrac{3 - \sqrt{65}}{4}}\).
Number of Solutions (Discriminant)
The discriminant is the part of the quadratic formula represented as \(\textcolor{green}{b}^2 - 4\textcolor{red}{a}\textcolor{blue}{c}\). The value of the discriminant is used to determine how many roots/solutions the quadratic function has:
- When \(\textcolor{green}{b}^2 - 4\textcolor{red}a\textcolor{blue}{c} = 0\), there is 1 real solution
- When \(\textcolor{green}{b}^2 - 4\textcolor{red}{a}\textcolor{blue}{c} > 0\), there are 2 real solutions
- When \(\textcolor{green}{b}^2 - 4\textcolor{red}{a}\textcolor{blue}{c} < 0\), there are 0 real solutions or 2 imaginary solutions
If there is only \(1\) real solution, this can be referred to as a double root. This is where \(r\) and \(s\) in \(a(x - r)(x - s)\) are the same value.
If there are no real solutions, this results in \(2\) imaginary solutions that can only be identified by using the quadratic formula instead of factoring.
The direction of opening of a graph and the position of the vertex determines whether the graph has two, one, or no zeroes and indicates whether the corresponding equation has two, one, or no real roots:
Two Real Roots
One Real Root
No Real Roots
\(x^2 + 5x + 4\)
We can identify \(\textcolor{red}{a = 1}\), \(\textcolor{green}{b = 5}\), and \(\textcolor{blue}{c = 4}\). We can plug these values into the appropriate formula and simplify to determine the discriminant:
\(\text{Discriminant} = \textcolor{green}{b}^2 - 4\textcolor{red}{a}\textcolor{blue}{c}\)
\(\text{Discriminant} = (\textcolor{green}{5})^2 - 4(\textcolor{red}{1})(\textcolor{blue}{4})\)
\(\text{Discriminant} = 25 - 16\)
\(\text{Discriminant} = 9\)
\(9 > 0\)
Therefore, we can determine that \(x^2 + 5x + 4\) has \(\boldsymbol{2}\) real solutions.
\(3x^2 + -7x + 5\)
We can identify \(\textcolor{red}{a = 3}\), \(\textcolor{green}{b = -7}\), and \(\textcolor{blue}{c = 5}\). We can plug these values into the appropriate formula and simplify to determine the discriminant:
\(\text{Discriminant} = (\textcolor{green}{5})^2 - 4(\textcolor{red}{1})(\textcolor{blue}{4})\)
\(\text{Discriminant} = (\textcolor{green}{-7})^2 - 4(\textcolor{red}{3})(\textcolor{blue}{5}\)
\(\text{Discriminant} = 49 - 60\)
\(\text{Discriminant} = -11\)
\(-11 < 0\)
Therefore, we can determine that \(3x^2 + -7x + 5\) has \(\boldsymbol{0}\) real solutions.
\(-x^2 + 6x - 9\)
We can identify \(\textcolor{red}{a = -1}\), \(\textcolor{green}{b = 6}\), and \(\textcolor{blue}{c = -9}\). We can plug these values into the appropriate formula and simplify to determine the discriminant:
\(\text{Discriminant} = \textcolor{green}{b}^2 - 4\textcolor{red}{a}\textcolor{blue}{c}\)
\(\text{Discriminant} = (\textcolor{green}{6})^2 - 4(\textcolor{red}{-1})(\textcolor{blue}{-9})\)
\(\text{Discriminant} = 36 - 36\)
\(\text{Discriminant} = 0\)
\(0 = 0\)
Therefore, we can determine that \(-x^2 + 6x - 9\) has \(\boldsymbol{1}\) real solution.
First, we can draw a diagram of the triangle to help visualize the problem:
We can use the given information to determine some equations. For example, we can use the Pythagorean theorem since we are given the value of the hypoteneuse:
\(x^2 + y^2 = 20^2\)
\(x^2 + y^2 = 400\)
In addition, we identify the equation that sums up the \(2\) legs and rearrange it as an input for the first equation:
\(x + y = 28\)
\(x = 28 - y\)
We can now input the second equation into the first equation, expand it using distributive property, and rearrange it accordingly to identify the values for the quadratic formula:
\((28 - y)^2 + y^2 = 400\)
\((28 - y)(28 - y) + y^2 = 400\)
\((28)(28) + (-y)(28) + (-y)(28) + (-y)(-y) + y^2 = 400\)
\(784 - 28y - 28y + y^2 + y^2 = 400\)
\(2y^2 - 56y + 784 - 400 = 0\)
\(2y^2 - 56y + 384 = 0\)
We can now identify \(\textcolor{red}{a = 2}\), \(\textcolor{green}{b = -56}\) and \(\textcolor{blue}{c = 384}\). We can use these values in the quadratic formula to determine the legs:
\(x = \cfrac{-\textcolor{green}{b} +- \sqrt{\textcolor{green}{b}^2 - 4\textcolor{red}{a}\textcolor{blue}{c}}}{2\textcolor{red}{a}}\)
\(x = \cfrac{-(\textcolor{blue}{-56}) +- \sqrt{(\textcolor{green}{-56})^2 - 4(\textcolor{red}{2})(\textcolor{blue}{384})}}{2(\textcolor{red}{2})}\)
\(x = \cfrac{56 +- \sqrt{3136 - 3072}}{4}\)
\(x = \cfrac{56 +- \sqrt{64}}{4}\)
\(x = \cfrac{56 +- 8}{4}\)
Using \(\pm\), we can determine the legs either by adding or subtracting the \(2\) values in the numerator.
We can add the \(2\) values in the numerator to determine leg \(x_1\):
\(x_1 = \cfrac{56 + 8}{4}\)
\(x_1 = \cfrac{64}{4}\)
\(x_1 = 16\)
We can subtract the \(2\) values in the numerator to determine leg \(y\):
\(x_2 = \cfrac{56 - 8}{4}\)
\(x_2 = \cfrac{48}{4}\)
\(x_2 = 12\)
Finally, we can plug these numbers into the original equations to verify that our results are correct.
We can first plug them into \(x^2 + y^2 = 400\):
\(x^2 + y^2 = 400\)
\((16)^2 + (12)^2 = 400\)
\(256 + 144 = 40\)
\(400 = 400\)
We can then plug them into \(x = 28 - y\):
\(x = 28 - y\)
\(16 + 12 = 28\)
\(28 = 28\)
Therefore, we can determine that the lengths of the legs are \(\boldsymbol{16}\) and \(\boldsymbol{12}\).