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

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:

\(x = \cfrac{-\textcolor{green}{7} \pm \sqrt{(\textcolor{green}{7})^2 - 4(\textcolor{red}{2})(\textcolor{blue}{-4})}}{2(\textcolor{red}{2})} \)

\(x = \cfrac{-7 \pm \sqrt{49 + 32}}{4} \)

\(x = \cfrac{-7 \pm \sqrt{81}}{4} \)

\(x = \cfrac{-7 \pm 9}{4} \)

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 \(x(2x - 3) = 7\).

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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:

\(x^2 + 5x + 4\)

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\(3x^2 + -7x + 5\)

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\(-x^2 + 6x - 9\)

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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.

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