Question

What are the roots of the equation `x^2-5x+1=0`?

Answer 1

`color(red)(x=4.79128784 or 0.20871215`

Explanation:

You can solve this equation using 2 methods, one being completing the square method, and the other by using the quadratic formula.

1) Commencing completing the square method,
`x^2-5x+1=0`

Subtract 1 on both sides,
`x^2-5x=-1`

Add 6.25 on both sides,
`x^2-5x+6.25=-1+6.25`

Apply perfect quadratic square formula,
`(x-2.5)^2=5.25`

Square root both sides,
`x-2.5=+-sqrt5.25`

Add 2.5 to both sides,
`x=+-sqrt5.25+2.5`

Hence,
`color(red)(x=4.79128784 or 0.20871215`

2) Commencing quadratic formula method,
`x^2-5x+1=0`

Quadratic equation,
`ax^2+bx+c=0`

Substitute `a=1, b=-5, c=1` into the quadratic formula,
`x=(5+-sqrt21)/(2)`

Hence,
`color(red)(x=4.79128784 or 0.20871215`

Answer 2

`x=(5+sqrt(21))/2,` `(5-sqrt(21))/2`

Explanation:

Find the roots:

`x^2-5x+1=0` `lArr` quadratic equation

The standard form for a quadratic equation is `ax^2+bx+c=0`, where `a=1`, `b=-5`, and `c=1`. Note: `a!=0`

Solve this quadratic equation using the quadratic formula:

`x=(-b+-sqrt(b^2-4ac))/(2a)`

Substitute the known values into the formula.

`x=(-(-5)+-sqrt((-5)^2-4*1*1))/(2*1)`

Simplify.

`x=(5+-sqrt(25-4))/2`

Simplify.

`x=(5+-sqrt(21))/2`

Solutions for `x`. (roots)

`x=(5+sqrt(21))/2,` `(5-sqrt(21))/2`