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

Question

Eva Abramson · Accepted Answer

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

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))# After the values have been changed in the equation, we must first solve the square root's interior. #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)

Jace Anderson · Answer

undefined Sure, to solve the equation $x^2 - 6x + 6 = 0$ using the quadratic formula, you would follow these steps:

1. Identify the coefficients $a$, $b$, and $c$ in the equation $ax^2 + bx + c = 0$.
   - In this case, $a = 1$, $b = -6$, and $c = 6$.

2. Substitute these values into the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

3. Plug in the values for $a$, $b$, and $c$:
   - $x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(6)}}{2(1)}$.

4. Simplify the equation:
   - $x = \frac{6 \pm \sqrt{36 - 24}}{2}$,
   - $x = \frac{6 \pm \sqrt{12}}{2}$,
   - $x = \frac{6 \pm 2\sqrt{3}}{2}$.

5. Split into two solutions:
   - $x_1 = \frac{6 + 2\sqrt{3}}{2} = 3 + \sqrt{3}$,
   - $x_2 = \frac{6 - 2\sqrt{3}}{2} = 3 - \sqrt{3}$.

So, the solutions to the equation $x^2 - 6x + 6 = 0$ are $x = 3 + \sqrt{3}$ and $x = 3 - \sqrt{3}$.