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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 $.

Expert Answer

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:

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

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

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