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

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#

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#