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how do you find out how many solutions there are for x^2+2x+8
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try to solve or use \(b^2-4ac\) are you allowed complex solutions?
yeah
Fundamental theorem of algebra implies that a polynomial of degree n always has n roots, counting repeated roots. So since it's of degree 2, it has 2 roots.
?
Since it's a second degree polynomial (highest term is a power of 2, the x^2), it automatically must have 2 solutions.
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oh ok
the maximum no. of solutions is 2.............
It has 2 solutions, \[ x=-1±i\sqrt{7}. \]
it has no real solution........
thankx
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