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

**, came up with an explicit solution to the equation \(ax^2+bx=c\).**

*Bramhagupta*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 \(-\frac{-21 – \sqrt{41}}{4}\).

**FAQ**s

**Can the roots ever be imaginary and repeating ?**No. A necessary requirement for repeating roots is \(\Delta = 0\), which is not satisfied by the imaginary root requirement \(\Delta < 0\).

**How to represent the quadratic equation in terms of its roots?**If \(\alpha\) and \(\beta\) are the roots then, the quadratic equation can be written in the form:

\(x^2 -(\alpha + \beta)x + \alpha \cdot \beta = 0\).