Solve x2 + 6x + 10 = 0.
Apply the same procedure as on the previous page:
This is the original equation. x2 + 6x + 10 = 0
Move the loose number over to the other side. x2 + 6x = – 10
Take half of the coefficient on the x-term (that is, divide it by two, and keeping the sign), and square it. Add this squares value to both sides of the equation. x^2 + 6x + 9 = –10 + 9
Convert the left-hand side to squared form. Simplify the right-hand side.
Note: If you don't know about complex numbers yet, then you have to stop at this step, because a square can't equal a negative number! Otherwise, proceed...
(x + 3)2 = –1
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Square-root both sides. Remember to put the "±" on the right-hand side. x + 3 = ± i
Solve for "x =", and simplify as necessary. x = –3 ± i
If you don't yet know about complex numbers (the numbers with "i" in them), then you would say that the above quadratic has "no solution". If you do know about complexes, then you would say that this quadratic has "no real solution" or that is has a "complex solution".
Since solving "(quadratic) = 0" for x is the same as finding the x-intercepts (assuming the solutions are real numbers), it stands to reason that this quadratic should not intersect the x-axis (since x-intercepts are "real" numbers). As you can see below, the graph does not in fact cross the x-axis.
y = x^2 + 6x + 10
This relationship is always true. If you come up with a real value on the right-hand side of the equation (a zero value is real, by the way; the square root of zero is just zero), then the quadratic will have two x-intercepts (or only one, if you get plus/minus of zero on the right side); if you get a negative on the right-hand side, then the quadratic will not cross the x-axis.
I'll do one last "example". It has become somewhat fashionable to have students derive the