SOLUTION: 2x^2-2x+7=0 I have to solve this by completing the square. I solved it but I'm unsure of my answer, I got x= 1/2 + or

SOLUTION: 2x^2-2x+7=0 I have to solve this by completing the square. I solved it but I'm unsure of my answer, I got x= 1/2 + or - i times radical 13 over 2.

Question 171112: 2x^2-2x+7=0
I have to solve this by completing the square. I solved it but I'm unsure of my answer, I got x= 1/2 + or - i times radical 13 over 2.
Found 2 solutions by jim_thompson5910, solver91311:
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
You are correct. There are a number of ways to verify your answer, but you can plug in the answer back into the equation and simplify.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Complete the square on

Step 1, starting from a quadratic in standard form like you have is to put the constant term on the right:

Step 2, if the coefficient on the term is other than 1, divide by that coefficient.

Step 3, divide the resulting coefficient on the term by 2 and square the result

Step 4, add this result to both sides of your equation and collect terms

Step 5, the above result has a perfect square on the left (hence the term "completing the square"), so factor it:

Step 6, take the square root of both sides:

or

Which leads us to a great big oops! because you can't take the square root of a negative number. The solution is to use the imaginary number which is defined as , leaving us with:

or
or
or

Step 7, isolate and simplify in each equation

If you want the exact representation of the roots of the given equation, you are done. If you need a numerical approximation of the imaginary parts of your complex numbers, get out your calculator.

Multiplying to verify that the product is, in fact, , is left as an exercise for the student. Alternatively, you could just trust me.

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