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April 16, 2014

April 16, 2014

Posted by **Anonymous** on Sunday, October 3, 2010 at 6:35pm.

which you can't factor

but then you can use the quadratic formula

but than why is the answer no real solutions when I can use the quadratic formula?

- Algebra -
**Reiny**, Sunday, October 3, 2010 at 6:44pmdid 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)

- Algebra -
**Anonymous**, Sunday, October 3, 2010 at 6:47pmthank you

I forgot about having a negative for b^2 - 4ac was no real solutions

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