# using the quadratic formula

• Aug 25th 2009, 05:33 AM
Just4
using the quadratic formula
Use the quadratic formula to solve the equation

x^2 - x = -7

x^2 - x + 7 = -7 + 7

x^2 - x + 7 = 0

a= 1, b= -1, c= 7

- (-1) + or - sqrt (-1)^2-4(1)(7)/2(1)

1 + or - sqrt 1-28/ 2
1 + or - sqrt -27 /2

am I on the right track here?
• Aug 25th 2009, 05:35 AM
Quote:

Originally Posted by Just4
Use the quadratic formula to solve the equation

x^2 - x = -7

x^2 - x + 7 = -7 + 7

x^2 - x + 7 = 0

a= 1, b= -1, c= 7

- (-1) + or - sqrt (-1)^2-4(1)(7)/2(1)

1 + or - sqrt 1-28/ 2
1 + or - sqrt -27 /2

am I on the right track here?

You are correct . You can continue from here to find its complex roots .
• Aug 25th 2009, 05:59 AM
Brama Bull
I thought this was the finally answer

1 + or - sqrt -27/2

could you explain to me how I would find the complex roots?
• Aug 25th 2009, 07:17 AM
HallsofIvy
Quote:

Originally Posted by Brama Bull
I thought this was the finally answer

1 + or - sqrt -27/2

could you explain to me how I would find the complex roots?

First, be careful with your parentheses! Most people would read "1 + or - sqrt -27/2" as $1\pm\sqrt{\frac{-27}{2}}$ but what you mean is $\frac{1\pm\sqrt{27}}{2}$ which, without LaTex, should be written as (1 + or - sqrt(27))/2.

Now, for the square root: $\sqrt{-27}= \sqrt{(-1)(27)}= \sqrt{-1}\sqrt{27}$. The square root of -1 is, by definition, "i". So $\sqrt{-27}= i\sqrt{27}$. It is also true that 27= 9(3) so $\sqrt{27}= \sqrt{9}\sqrt{3}= 3\sqrt{3}$. Your teacher might insist that you write the answer as $\frac{1\pm 3i\sqrt{3}}{2}$ and, in any case, it is better to do so.