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:
\(x = \cfrac{-\textcolor{green}b \pm \sqrt{\textcolor{green}b^2 - 4\textcolor{red}a\textcolor{blue}c}}{2\textcolor{red}a}\)
- \(x\) 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
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 ±, we can determine the 2 roots either by adding the 2 values of subtracting the 2 values:
\(x₁ = \cfrac{-7 - 9}{4}\)
\(x₁ = \cfrac{-16}{4}\)
\(x₁ = -4\)
\(x₂ = \cfrac{-7 + 9}{4}\)
\(x₂ = \cfrac{2}{4}\)
\(x₂ = \cfrac{1}{2}\)
Therefore, we can determine that the roots are \(-4\) and \(\cfrac{1}{2}\).
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.
Determine the roots for the quadratic formula \(x(2x - 3) = 7\).
Show Answer
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 ±, we can determine the 2 roots either by adding the 2 values of subtracting the 2 values:
\(x1 = \cfrac{3 + \sqrt{65}}{4}\)
\(x2 = \cfrac{3 - \sqrt{65}}{4}\)
Therefore, we can determine that the roots are \(\cfrac{3 + \sqrt{65}}{4}\) and \(\cfrac{3 - \sqrt{65}}{4}\).
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.
Determine how many roots the following quadratic functions have:
\(x^2 + 5x + 4\)
Show Answer
We can identify \(\textcolor{red}{a = 1}\), \(\textcolor{green}{b = 5}\), and \(\textcolor{blue}{c = 4}\).
\(\text{Discriminant} = (\textcolor{green}{5})^2 - 4(\textcolor{red}{1})(\textcolor{blue}{4})\)
\(\text{Discriminant} = 25 - 16\)
\(\text{Discriminant} = 9\)
\(9 > 0\)
Therefore, we can determine that \(x^2 + 5x + 4\) has 2 real solutions.
\(3x^2 + -7x + 5\)
Show Answer
We can identify \(\textcolor{red}{a = 3}\), \(\textcolor{green}{b = -7}\), and \(\textcolor{blue}{c = 5}\).
\(\text{Discriminant} = (\textcolor{green}{-7})^2 - 4(\textcolor{red}{3})(\textcolor{blue}{5})\)
\(\text{Discriminant} = 49 - 60\)
\(\text{Discriminant} = -11\)
\(-11 < 0\)
Therefore, we can determine that \(3x^2 + -7x + 5\) has 0 real solutions.
\(-x^2 + 6x - 9\)
Show Answer
We can identify \(\textcolor{red}{a = -1}\), \(\textcolor{green}{b = 6}\), and \(\textcolor{blue}{c = -9}\).
\(\text{Discriminant} = (\textcolor{green}{6})^2 - 4(\textcolor{red}{-1})(\textcolor{blue}{-9})\)
\(\text{Discriminant} = 36 - 36\)
\(\text{Discriminant} = 0\)
\(0 = 0\)
Therefore, we can determine that \(-x^2 + 6x - 9\) has 1 real solution.
The hypoteneuse of a triangle is \(20\;[cm]\). The sum of the lengths of the legs is \(28\;[cm]\). Find the length of each leg.
Show Answer
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 ±, we can determine the 2 roots either by adding the 2 values or subtracting the 2 values:
\(x1 = \cfrac{56 + 8}{4}\)
\(x1 = \cfrac{64}{4}\)
\(x1 = 16\)
\(x2 = \cfrac{56 - 8}{4}\)
\(x2 = \cfrac{48}{4}\)
\(x2 = 12\)
We can plug these numbers into the original equations to verify that our results are correct:
\((16)^2 + (12)^2 = 400\)
\(256 + 144 = 40\)
\(400 = 400\)
\(16 + 12 = 28\)
\(28 = 28\)
Therefore, we can determine that the lengths of the legs are \(16\) and \(12\).