How do you solve a^2-sqrt(3)a+1 = 0a23a+1=0 ?

2 Answers
May 12, 2015

(a-sqrt(3)/2)^2 = (a-sqrt(3)/2)(a-sqrt(3)/2)(a32)2=(a32)(a32)

= a^2-(sqrt(3)/2+sqrt(3)/2)a+(sqrt(3)/2)(sqrt(3)/2)=a2(32+32)a+(32)(32)

= a^2-sqrt(3)a+3/4=a23a+34

So we have:

0 = a^2-sqrt(3)a+1 = a^2-sqrt(3)a+3/4+1/40=a23a+1=a23a+34+14

=(a-sqrt(3)/2)^2+1/4=(a32)2+14

Subtracting 1/4 from both sides, we get:

(a-sqrt(3)/2)^2 = -1/4(a32)2=14

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/2a32=±14=±i2

Adding sqrt(3/2)32 to both sides, we get

a = sqrt(3)/2 +- i/2a=32±i2.

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))/2a=b±b24ac2aa=3±(3)241121a=3±342a=3±12

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