How do you solve #x^2 - 6x + 6 = 0# using the quadratic formula?

1 Answer
Jul 6, 2016

Answer:

#x_1 = 4.73# (3sf) #x_2 = 1.27# (3sf)

Explanation:

The quadratic formula is: #(-b +- sqrt(b^2 - 4ac))/(2a)#, Therefore the first step is to identify #a,b# & #c#. We need to ensure that the equation is #=0# and in this case that is true.

Then we are able to obtain the values to substitute into the formula; the general quadratic formula layout is #ax^2 + bx +c# therefore we look at your equation: #x^2 - 6x + 6#. The value of #a# has to be #1# since #x^2# is multiplied by #1#. Then #b = -6#, maintaining the negative is essential to the calculation (the sign will also stay with the value that is next in the equation). Then #c = 6#.

We now substitute #a= 1#, #b=-6# and #c= 6# into the quadratic formula.

#x = (-b +- sqrt(b^2 - 4ac))/(2a)# #= (-(-6) +- sqrt((-6)^2 - 4(1)(6)))/(2(1))#

Now that we have the values substituted into the equation we first solve the inside of the square root. So,

#x = (-(-6) +- sqrt(12))/(2(1))#

Since the value under the root (discriminant) is positive we know that there will be two real solutions for #x#.

Separate the two #x# values so that we solve for the positive root first.

#x_1 = (-(-6) + sqrt(12))/(2(1))# = #4.73# (3 sf)

Now solve the second #x# value by using the negative root.

#x_2 = (-(-6) - sqrt(12))/(2(1))# = #1.27# (3 sf)