The aim of this question is to understand the **solution of a quadratic equation** using the **standard form** of its roots.

A **quadratic equation** is a polynomial** equation with a degree equal to 2**. A standard quadratic equation can be written **mathematically** as the following formula:

\[ a x^{ 2 } \ + \ b x \ + \ c \ = \ 0 \]

Where $ a $, $ b $, $ c $ are **some constants** and $ x $ is the **independent variable**. The **roots of the quadratic equation** can be written **mathematically** as the following formula:

\[ x \ = \ \dfrac{ – \ b \pm \sqrt{ b^{ 2 } \ – \ 4 a c } }{ 2 a } \]

The specificÂ **roots of a quadratic equation** may be **real or complex** depending upon the values of the constants $ a $, $ b $, $ c $.

## Expert Answer

**Given:**

\[ x^{ 2 } \ â€“ \ 5 x \ â€“ \ 1 \ = \ 0 \]

**Comparing** the above equation with the following **standard equation**:

\[ a x^{ 2 } \ + \ b x \ + \ c \ = \ 0 \]

We can see that:

\[ a \ = \ 1 , \ b \ = \ – 5 , \text{ and } c \ = \ – 1 \]

The specificÂ **roots of the quadratic equation** can be calculated using the following formula:

\[ x \ = \ \dfrac{ – \ b \pm \sqrt{ b^{ 2 } \ – \ 4 a c } }{ 2 a } \]

**Substituting values:**

\[ x \ = \ \dfrac{ – \ ( – 5 ) \pm \sqrt{ ( – 5 )^{ 2 } \ – \ 4 ( 1 ) ( – 1 ) } }{ 2 ( 1 ) } \]

\[ x \ = \ \dfrac{ 5 \pm \sqrt{ 25 \ + \ 4 } }{ 2 } \]

\[ x \ = \ \dfrac{ 5 \pm \sqrt{ 29 } }{ 2 } \]

\[ x \ = \ \dfrac{ 5 \pm 5.38 }{ 2 } \]

\[ x \ = \ \dfrac{ 5 \ + \Â 5.38 }{ 2 } , \ \dfrac{ 5 \ – \ 5.38 }{ 2 } \]

\[ x \ = \ \dfrac{ 10.38 }{ 2 } , \ \dfrac{ – 0.38 }{ 2 } \]

\[ x \ = \ 5.19 , \ -0.19 \]

## Numerical Result

\[ x \ = \ 5.19 , \ -0.19 \]

Hence,Â **both the roots are real.**

## Example

**Calculate the roots** of $Â x^{ 2 } \ â€“ \ 5 x \ + \ 1 \ = \ 0 $.

The specificÂ **roots of the quadratic equation** can be calculated using the following formula:

\[ x \ = \ \dfrac{ – \ ( – 5 ) \pm \sqrt{ ( – 5 )^{ 2 } \ – \ 4 ( 1 ) ( 1 ) } }{ 2 ( 1 ) } \]

\[ \Rightarrow x \ = \ 4.79, \ 0.21 \]