eddy2017
Elite Member
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- Oct 27, 2017
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Easy(2x+ *)^2 = (2x)^2 + 2(2x)(*) + (*)^2 = 4x^2 + 4x* + *^2 = 4x^2 + 12xy + 9y^2
You need 4x* = 12xy. Divide by sides by 4x to get * = ...
Who is 'we', and what post are you talking about? Or, were you just thinking out loud about something that doesn't really matter in this thread.We were talking about quadratic equation in my post.
Oh, the tutors were discussing that I should learn about quadratic equations in a post I made. I saw this post and I want to know if any of the equations here qualify as quadratic. That is what I said we( referring to the tutors and me ). Take a look at my post here #10. I hope I am not breaking any rules.Who is 'we', and what post are you talking about? Or, were you just thinking out loud about something that doesn't really matter in this thread.
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Very well. Please do that in your own threads. Thank you!tutors were discussing that I sohuld learn about quadratic equations in a post I made.
One question, mmm. Just to be sure I don't mess up.Very well. Please do that in your own threads. Thank you!
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You're free to comment, but not for the purposes of starting your own discussions. If you're looking for tutoring on your own questions (like, what is or isn't a quadratic polynomial), then please start your own threads.Am I allowed to reply to other threads?
That is one specific form, Eddy, but there are many other quadratic forms.A quadratic equation takes this form: ax^2+bx+c=0
They aren't asking you to solve equations for any variable(s), Eddy. They're asking you to find a value for each unknown parameter (i.e., the numbers represented by asterisks).Just wondering in case these can be solved as a Q.E
That is fairly easy. At least I was able to do number 1.They aren't asking you to solve equations for any variable(s), Eddy. They're asking you to find a value for each unknown parameter (i.e., the numbers represented by asterisks).
Prof Steven, seconding your reply that a QE has two solutions I have found this statement:Agree. A quadratic equation has exactly two solutions.
When is product ab = 0?Prof Steven, seconding your reply that a QE has two solutions I have found this statement:
"A quadratic expression can be written as the product of two linear factors and each factor can be equated to zero, So there exist two solutions.
Can you provide an example where this can be seen?.
Why two solutions?
when one of the factors is equal to zero.When is product ab = 0?
So when we have a product of 2 linear factors equal to 0, either one can be 0 -> 2 solutions.when one of the factors is equal to zero.
Either a or b
Thanks, prof.So when we have a product of 2 linear factors equal to 0, either one can be 0 -> 2 solutions.