SOLUTION: use i notation to express any nonreal complex numbers. How do I go about getting this answer? I don't understand this stuff! Thanks in advance 2x^2 – 6x + 1 = 0

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Question 1309: use i notation to express any nonreal complex numbers. How do I go about getting this answer? I don't understand this stuff! Thanks in advance 2x^2 – 6x + 1 = 0
Answer by JamieL38(6) About Me  (Show Source):
You can put this solution on YOUR website!
Remember that the quadratic formula is x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29.
The part inside the square root, b%5E2-4%2Aa%2Ac is called the discriminant. The value of the discriminant determines what kinds of roots the quadratic has. If the discriminant is less than 0, then the quadratic has two nonreal roots. If the discriminant is greater than 0, then the quadratics has two real roots. If the discriminant is equal to 0, then the quadratic has a real double root.
So for your problem the discriminant is b%5E2-4%2Aa%2Ac=%28-6%29%5E2-4%282%29%281%29=36-8=28This value is greater than 0, so your quadratic has two real roots.
They are: x=%286+%2B-+sqrt%2828%29%29%2F%282%2A2%29
Thus, x=%286+%2B-+2sqrt%287%29%29%2F4=%283+%2B-+sqrt%287%29%29%2F2
If you had a negative discriminant then your answer would include an i. Take for example this quadratic that's similar to yours: 2x%5E2-6x%2B5=0. The discriminant is equal to %28-6%29%5E2-4%282%29%285%29=-4. So your solution would be: x=%286+%2B-+sqrt%28-4%29%29%2F%282%2A2%29=%286+%2B-+2i%29%2F4=%283+%2B-+i%29%2F2.