How do you solve #x^2-6x+25=0#?

1 Answer
Jul 4, 2015

Answer:

This has no real solution. It does have imaginary solutions however.

Explanation:

You can see this in two ways.

1) Without the discriminant
#x^2-6x+9+16=0 \color(white)(.............)#(Completing the square)
#(x-3)^2+16=0#
#(x-3)^2=-16#
#(x-3)=+-sqrt(-16)#

Since #sqrt(-16)# doesn't exist (at least not with real numbers), this equation has no real solutions.

If you want the imaginary solutions:
#(x-3)=+-sqrt(-16)#
#x-3=+-sqrt(16)*sqrt(-1)#
#x-3=+-4*i#
#x=+-4i+3#

2) With the discriminant
The discriminant is: #D = b^2-4ac# if you have a quadratic equation of the form #ax^2+bx+c=0#.
It can be proven that if this discriminant is less than 0, the equation has no solutions:
#D = (-6)^2-4*1*25=36-100=-64 < 0#