SOLUTION: Please help me with determining the number of real solutions based on the discriminate. {{{ x^2-4x-5=0 }}} I know that the discriminate is b^2-4ac, I just don't understand how t

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Please help me with determining the number of real solutions based on the discriminate. {{{ x^2-4x-5=0 }}} I know that the discriminate is b^2-4ac, I just don't understand how t      Log On


   

Question 412670: Please help me with determining the number of real solutions based on the discriminate.
+x%5E2-4x-5=0+
I know that the discriminate is b^2-4ac, I just don't understand how to apply it when figuring out if it is equal to zero, less than zero, or greater than zero. If you could help that would be greatly appreciated, thank you so much!
Found 2 solutions by rfer, lwsshak3:
Answer by rfer(16322) About Me  (Show Source):
You can put this solution on YOUR website!
a=1, b=-4, c=-5
-4^2-4*1*-5
16+20=36
It is greater than zero so there are two real solutions

Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Please help me with determining the number of real solutions based on the discriminate.
..
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
..
If the discriminant:
b^2-4ac>0. then the equation has two distinct real roots. You can see this from the figure above . If the discriminate turns out to be a perfect square, you probably could factor the equation to determine the roots
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b^2-4ac=0, then the equation has one real root, called a double root. You can see this from the figure above, the root would be equal to -b/2a
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b^2-4ac>0, Then the equation has no real roots. You would still, however, have two nonreal or imaginary complex number roots, because you are taking the sqrt of a negative number.