An Algebraic Equation in one variable with degree 2 is called a Quadratic Equation. We use the Quadratic Formula for finding the solution to these equations. Suppose the \(x\) is an independent variable. The general form of a quadratic equation is:
\(ax^2 + bx + c = 0\)
Where, \(a\), \(b\) and \(c\) are real constant coefficients and \(a\neq 0\)
Some examples of quadratic equations are:
\(4x^2 + 6x + 1 = 0 ; a = 4, b = 6, c = 1\)
\(x^2 + 33x + 7 = 0 ; a = 1, b = 33, c = 7\)
\(x^2 – 9 = 0 ; a = 1, b = 0, c = 9\)
The solution to these equations involve solving for \(x\). In general, 2 solutions are possible for a quadratic equation. These solutions are called “Roots” or “Zeros” of the given quadratic equation. Usually the two roots of the quadratic equation are denoted as \(\bf{\alpha}\) and \(\bf{\beta}\). Depending upon the values of \(a\), \(b\) and \(c\) , the roots can be real or complex in nature.
Index
History
The first references to problems relating to the quadratic formula were found as early as 2000 BC in Babylonian Clay Tablets. The tablets dealt with the solution to an area problem which required the solution to a quadratic equation. In around 300 BC, Euclid, the Greek mathematician , formulated an abstract geometrical method to calculate the positive root for a quadratic equation. In 628 AD , the Indian mathematician Bramhagupta, came up with an explicit solution to the equation \(ax^2+bx=c\).
In 1637, Rene Descartes, the famous French mathematician produced the quadratic solution as we know today.
Quadratic Formula and Determinant
Consider the quadratic equation;
\(ax^2 + bx + c = 0\)
Where \(a\), \(b\) and \(c\) are real numbers and \(a \neq 0\).
The general method to solve this quadratic equation is by use of the quadratic formula:
\(x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}\)
Here, the two roots are denoted as \(\alpha\) and \(\beta\):
\(\alpha = \frac{-b + \sqrt{b^2 -4ac}}{2a}\)
\(\beta = \frac{-b – \sqrt{b^2 -4ac}}{2a}\)
A useful quantity called the ‘Determinant’, denoted as \(\Delta\) is defined as:
\(\Delta = b^2 – 4ac\)
This quantity is helpful in determining the nature of the roots \(\alpha\) and \(\beta\). In terms of the determinant \(\Delta\), the solution to the quadratic equation using quadratic formula becomes:
\(x = \frac{-b \pm \sqrt{\Delta }}{2a}\)
\(\alpha = \frac{-b + \sqrt{\Delta}}{2a}\)
\(\beta = \frac{-b – \sqrt{\Delta}}{2a}\)
Related: Use this Quadratic Formula Calculator to find the solution to your quadratic equations.
Nature of Roots
Depending upon the sign of the determinant \(\Delta\), the roots \(\alpha\) and \(\beta\) can be:
Real and Distinct
When \(\bf{\Delta > 0}\) , the roots are real and distinct or unique.
For example, consider the quadratic equation \(2x^2 + 21x + 50 = 0\) has the roots \(-\frac{21}{4} + \frac{\sqrt{41}}{4}\) and \(-\frac{21}{4}-\frac{\sqrt{41}}{4}\).
Here, \(a = 2, b = 21\) and \(c = 50\). Therefore, we have,
\(\begin{equation}
\begin{split}
\Delta & = b^2 -4ac \\
& = 21^2- 4 \times 2 \times 50 \\
& = 441 – 400 \\
& = 41 \\
\end{split}
\end{equation}
\)
Real and Repeating
When \(\bf{\Delta = 0}\), the roots are real and repeating, i.e., \(\alpha = \beta\).
For example, consider the quadratic equation \(2x^2 + 20x + 50 = 0\) has roots \(\alpha = \beta = -5\).
Here, \(a = 2, b = 20\) and \(c = 50\). Therefore, we have,
\(\begin{equation}
\begin{split}
\Delta & = b^2 -4ac \\
& = 20^2- 4 \times 2 \times 50 \\
& = 400 – 400 \\
& = 0 \\
\end{split}
\end{equation}
\)
Imaginary and Distinct
When \(\Delta < 0\) , the roots are complex or imaginary.
For example, consider the quadratic equation \(2x^2 + 20x + 51 = 0\) has the roots \(-5+\frac{\sqrt{-2}}{2}\) and \(-5-\frac{\sqrt{-2}}{2}\).
Here, \(a = 2, b = 20\) and \(c = 51\). Therefore, we have,
\(\begin{equation}
\begin{split}
\Delta & = b^2 -4ac \\
& = 20^2- 4 \times 2 \times 51\\
& = 400 – 408 \\
& = -8 \\
\end{split}
\end{equation}
\)
Methods to Solve Quadratic Equation
There are various methods to solve a quadratic equation. But mainly two methods are used, first is completing square method and other is by using quadratic formula. Let’s have a look at both of them.
By Completing Squares
In this method, you try to express the given quadratic equation as a sum of a real number and a square of a binomial expression. For example , Consider the quadratic equation \(x^2 + 7x +10 = 0\). Here \(a = 1 , b = 7\) and \(c = 10\).
The given equation could be also written as:
\(\begin{equation}
\begin{split}
x^2 + 2\times \frac{7}{2}x +10 & = 0\\
x^2 + 2 \times \frac{7}{2}x +\frac{49}{4} -\frac{49}{4} +10 & = 0\\
\left( x+ \frac{7}{2} \right) ^2-\frac{9}{4} & = 0 \\
\left( x+ \frac{7}{2} \right) ^2 & = \frac{9}{4} \\
x+ \frac{7}{2} & = \pm \frac{3}{2} \\
x & = -\frac{7 \pm 3}{2} \\
x & = -5, -2
\end{split}
\end{equation}
\)
By Using the Quadratic Formula
Consider the quadratic equation \(ax^2 + bx+ c = 0\).
We have,
\(\begin{equation}
\begin{split}
x^2 + \frac{b}{a}x +\frac{c}{a} & = 0\\
x^2 + 2 \times \frac{b}{2a}x +\frac{b^2}{4a^2} -\frac{b^2}{4a^2} +\frac{c}{a} & = 0\\
\left( x+ \frac{b}{2a} \right) ^2-\frac{4ac-b^2}{4a^2} & = 0 \\
\left( x+ \frac{b}{2a} \right) ^2 & = \frac{b^2-4ac}{4a^2} \\
x+ \frac{b}{2a} & = \pm \frac{\sqrt{b^2-4ac}}{2a} \\
x & = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \\
\end{split}
\end{equation}
\)
Using this quadratic function formula, the solutions can be found out.
Quadratic Formula Examples
Question 1. Find the roots of \(2x^2 + 20x + 51 = 0\).
Solution. Here, \(a = 2, b = 20\) and \(c = 51\). We have,
\(\begin{equation}
\begin{split}
x & = -\frac{-b \pm \sqrt{b^2-4ac}}{2a} \\
& = -\frac{-20 \pm \sqrt{20^2-4\times 2 \times 51}}{2\times 2} \\
& = -\frac{-20 \pm \sqrt{400-408}}{4} \\
& = -\frac{-20 \pm 2\sqrt{2}i}{4} \\
& = -\frac{-10 \pm \sqrt{2}i}{2} \\
\end{split}
\end{equation}
\)
Therefore, by using quadratic equation formula, the roots are \(-\frac{-10 + \sqrt{2}i}{2}\) and \(-\frac{-10 – \sqrt{2}i}{2}\).
Question 2. Find the roots of \(2x^2 + 21x + 50 = 0\).
Solution. Here, \(a = 2, b = 21\) and \(c = 50\). We have,
\(\begin{equation}
\begin{split}
x & = -\frac{-b \pm \sqrt{b^2-4ac}}{2a} \\
& = -\frac{-21 \pm \sqrt{21^2-4\times 2 \times 50}}{2\times 2} \\
& = -\frac{-21 \pm \sqrt{441-400}}{4} \\
& = -\frac{-21 \pm \sqrt{41}}{4} \\
\end{split}
\end{equation}
\)
Therefore, by using quadratic equation formula, the roots are \(-\frac{-21 + \sqrt{41}}{4}\) and \(-\frac{-21 – \sqrt{41}}{4}\).
FAQs
No. A necessary requirement for repeating roots is \(\Delta = 0\), which is not satisfied by the imaginary root requirement \(\Delta < 0\).
If \(\alpha\) and \(\beta\) are the roots then, the quadratic equation can be written in the form:
\(x^2 -(\alpha + \beta)x + \alpha \cdot \beta = 0\).
