How do you solve using the quadratic formula #x^2-4x+10=0#?

2 Answers
May 3, 2015

Given a quadratic equation of the form
#ax^2+bx+c=0#
the quadratic roots can be evaluated using the formula
#x=(-b+-sqrt(b^2-4ac))/(2a)#

For #x^2-4x+10 = 0#

#x= (4+-sqrt(16-40))/2#

#x= 2+-sqrt(-6)#

This equation has no Real solutions
but if we are allowed Complex solutions:
#x=2+-sqrt(6)i#

May 4, 2015

You notice the 1st degree coefficient is even, so you can use the direct formula
#P(x)=ax^2+bx+c, β=b2#
#x1,x2={−β±sqrt(β^2−ac)}/a#
so #x_1,x_2=2±sqrt(4−10)=2±sqrt(−6)#
So you know there are no solutions in ℝ, and you have the complex conjugated solutions #z=2+sqrt(6)i# and#barz=2−sqrt(6)i#