How do you solve #a^2-sqrt(3)a+1 = 0# ?

2 Answers
May 12, 2015

#(a-sqrt(3)/2)^2 = (a-sqrt(3)/2)(a-sqrt(3)/2)#

#= a^2-(sqrt(3)/2+sqrt(3)/2)a+(sqrt(3)/2)(sqrt(3)/2)#

#= a^2-sqrt(3)a+3/4#

So we have:

#0 = a^2-sqrt(3)a+1 = a^2-sqrt(3)a+3/4+1/4#

#=(a-sqrt(3)/2)^2+1/4#

Subtracting 1/4 from both sides, we get:

#(a-sqrt(3)/2)^2 = -1/4#

This has no real number solutions since the square of any real number is non-negative.

If you want complex solutions,

#a-sqrt(3)/2 = +-sqrt(-1/4) = +-i/2#

Adding #sqrt(3/2)# to both sides, we get

#a = sqrt(3)/2 +- i/2#.

May 12, 2015

I would start applying the formula to solve quadratic equations (in fact, this is a quadratic equation in "a"):

#a=(-b+-sqrt(b^2-4ac))/(2a) => a=(sqrt3+-sqrt((sqrt3)^2-4·1·1))/(2·1) => a=(sqrt3+-sqrt(3-4))/2 => a=(sqrt3+-sqrt(-1))/2#

As you can see, the equation has no real solution, since it has a square root of a negative number (#sqrt(-1)#).

  • So, if you are working with real numbers, the answer is that there is no #a in RR# which makes #a^2-sqrt3a+1 = 0#.

  • But if you are working with complex numbers, then there are two solutions:
    #a_1=(sqrt3+i)/2# and #a_2=(sqrt3-i)/2#.