Posted by Anonymous on Sunday, October 3, 2010 at 6:35pm.
did you mean the equation
x^2 + x + 2 = 0 ?
When you solve an equation of the form
ax^2 + bx + c =0 you are really just asking,
"where does the graph of the corresponding function
y = ax^2 + bx + c cross the x-axis"?
In your case the graph, which is a parabola, lies totally above the x-axis, thus no x-intercepts , and thus no real solution.
If the b^2 - 4ac part of the formula is zero, there will be only one solution.
If the b^2 - 4ac part of the formula is positive, there will be 2 different real solutions.
If the b^2 - 4ac part of the formula is negative, like in your case, there will be no real solution at all.
(there will be two imaginary solutions)
thank you
I forgot about having a negative for b^2 - 4ac was no real solutions
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