How do you find all the real and complex roots of #x^2-2x-5=0#?

1 Answer
Burglar
Jun 10, 2016

You find them applying the formula for the second order equations.

Explanation:

Every second order equation in the form

#ax^2+bx+c=0# can be solved applying the formula

#x=(-b\pmsqrt(b^2-4ac))/(2a)#.

The roots are two, one when you use the sign #+# and one with the sign #-#.
So the only thing that you have to do is to identify #a, b, c# in your equation and substitute in the formula.

your equation is

#x^2-2x-5=0# and if you confront it with
#ax^2+bx+c=0# you see that

#a=1#, #b=-2# and #c=-5#.

Then the solution is

#x=(-(-2)\pmsqrt((-2)^2-4*1*(-2)))/(2*1)#

#=(2\pm sqrt((4+8)))/2#

#=(2\pmsqrt(12))/2#, I use the fact that #sqrt(12)=sqrt(4*3)=sqrt(4)sqrt(3)=2sqrt(3)#

#=(2\pm 2sqrt(3))/2#

#=1\pm sqrt(3)#

The two roots are

#x_1=1+sqrt(3)\approx2.73#
#x_2=1-sqrt(3)\approx-0.73#.

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