Decide how many solutions the equation has
Favorites|Homepage
Subscriptions | sitemap
HOME > > Decide how many solutions the equation has

Decide how many solutions the equation has

[From: ] [author: ] [Date: 12-04-17] [Hit: ]
if its zero theres 1 real solution,The only problem I have is Im not sure which formula to use,Please help!since the answer is -4,http://www.mathwarehouse.......
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!

-
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

-
Use the discriminant: b^2 - 4ac

http://www.mathwarehouse.com/quadratic/d…
1
keywords: equation,the,how,has,solutions,many,Decide,Decide how many solutions the equation has
New
Hot
© 2008-2010 http://www.science-mathematics.com . Program by zplan cms. Theme by wukong .