# Show that a root of x2 – 5x – 1 = 0 is real.

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

Given:

$x^{ 2 } \ – \ 5 x \ – \ 1 \ = \ 0$

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 specific 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:

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

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

$x \ = \ \dfrac{ 5 \pm \sqrt{ 29 } }{ 2 }$

$x \ = \ \dfrac{ 5 \pm 5.38 }{ 2 }$

$x \ = \ \dfrac{ 5 \ + \ 5.38 }{ 2 } , \ \dfrac{ 5 \ – \ 5.38 }{ 2 }$

$x \ = \ \dfrac{ 10.38 }{ 2 } , \ \dfrac{ – 0.38 }{ 2 }$

$x \ = \ 5.19 , \ -0.19$

## Numerical Result

$x \ = \ 5.19 , \ -0.19$

Hence, both the roots are real.

## Example

Calculate the roots of $x^{ 2 } \ – \ 5 x \ + \ 1 \ = \ 0$.

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

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

$\Rightarrow x \ = \ 4.79, \ 0.21$