x^2+6x+10=0
I need to first find the discriminant, if it's positive there's 2 real solutions, if it's zero there's 1 real solution, and if it's negative there's no real solution
The only problem I have is I'm not sure which formula to use, I know it's either -b+- (sq root) b^2-4ac all over 2a or b^2-4ac
Please help!
I need to first find the discriminant, if it's positive there's 2 real solutions, if it's zero there's 1 real solution, and if it's negative there's no real solution
The only problem I have is I'm not sure which formula to use, I know it's either -b+- (sq root) b^2-4ac all over 2a or b^2-4ac
Please help!
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the discriminate is what is inside the square root which is b^2-4ac
so it becomes
(6)^2-4(1)(10)
36-40=-4
since the answer is -4, it means there are no real solutions
and the quadratic formula -b+- √ b^2-4ac / 2a or b^2-4ac
is used to determine the roots (x intercepts) of the equation while the discriminate b^2-4ac is the find the nature of the roots (how many)
hope this helps
so it becomes
(6)^2-4(1)(10)
36-40=-4
since the answer is -4, it means there are no real solutions
and the quadratic formula -b+- √ b^2-4ac / 2a or b^2-4ac
is used to determine the roots (x intercepts) of the equation while the discriminate b^2-4ac is the find the nature of the roots (how many)
hope this helps
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Use the discriminant: b^2 - 4ac
http://www.mathwarehouse.com/quadratic/d…
http://www.mathwarehouse.com/quadratic/d…