Question

Solve each equation, if possible. Write irrational numbers in simplest radical form. Describe the strategy you used to get your solution and tell why you chose that strategy.
〖3x〗^2+27=0
x^2-8x+1=0

Answer 1

Problem 1
Do you know what a complex number is? If you do not, you can get an answer but not one you will like much.

(3x)^2 + 27 = 0               remove the brackets. Remember to square what's inside the brackets.

9x^2 + 27 = 0                Divide both terms by 9
9x^2/9 + 27/9 = 0

x^2 + 3 =  0                  Subtract 3 from both sides.
x^2 = -3                        Take the square root from both sides.
x = sqrt(-3)                    but the square root of - 3 = 3i
x = i*sqrt(3)

Problem 2
x^ - 8x + 1 = 0
a = 1
b = - 8
c = 1

x = [- -8 +/- sqrt(b^2 - 4*a*c) ]/(2*a)     Quadratic formula
x = [ 8 +/- sqrt( (-8)^2 - 4(1*1)]/2          Substitute Givens and combine
x = [ 8 +/- sqrt( 64 - 4 )] /2                    Subtract 4
x = [ 8 +/- sqrt (60)]/2                            Break 60 into 4 * 15 
x = [ 8 +/- sqrt (4*15)]/2                         Notice 4 is a perfect square. sqrt4 = 2
x = [ 8 +/- 2*sqrt(15)] / 2                        Divide through by 2
x = 8/2 +/- sqrt (15)
x = 4 +/- sqrt(15) the twos were gone