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March 29, 2024, 12:14:52 am

Author Topic: Δ=b^2-4ac  (Read 1313 times)  Share 

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Lukey

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Δ=b^2-4ac
« on: November 06, 2011, 09:33:27 pm »
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Can someone please explain the concept of Δ=b^2-4ac and how Δ>0 for 2 solutions.  really hope this doesnt come up in exam

b^3

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Re: Δ=b^2-4ac
« Reply #1 on: November 06, 2011, 09:41:18 pm »
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In your quadratic formula you have

Now you can't take the square root of a negative number, so the thing under the root (i.e. b2-4ac which we call the discriminant(Δ)) cannot be less than 0 if solutions are to exsits.

If if equals zero, then the formula simplifes to i.e. there is only one solution.

If it is greater than zero, then the root exsits and as we have a plus or minus in there, there will be two different solutions.

So to sum up
Δ<0 no real solutions
Δ=0 1 real solution
Δ>0 2 real solutions
« Last Edit: November 06, 2011, 10:00:59 pm by b^3 »
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Lukey

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Re: Δ=b^2-4ac
« Reply #2 on: November 06, 2011, 10:00:02 pm »
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ahh thanks alot, i assumed Δ was meant to represent change like Δy/Δx