Expert: Sherman D. - 7/2/2004

Question
By step#4, why isnt it, (-6)^2 - 4(3)(k) >= 0 instead of just =  to 0,and the answer will then be k<=3 because of the division of the negative which changes the sign?, because they say they want all values that are real and if the discriminant is higher or equal to 0, the roots will be real.Please explain if it was a misprint(like we all do), or I'm incorrect.Thank you for your help.And also, thank you for your quick response.


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Followup To
Question -
Name the values of k for which the roots of 3x^2+k=6x are real.Please explain all the ways to do this.Thank you.
Answer -
3x^2 + k = 6x

3x^2 - 6x + k = 0

b^2 - 4ac = 0

(-6)^2 - 4(3)(k) = 0

36 - 12k = 0

-12k = -36

k = 3

That is the answer i got.

If you are wondering why i used b^2 - 4ac, is because of the formula x = (-b ± sqrt(b^2 - 4ac))/2a, and b^2 - 4ac is what lets you know if you will get 2 real roots, or 2 imaginary roots.

For ex:

if using b^2 - 4ac gets you a 0, then you have one real root, if it gets you a postive value, you get 2 real roots, rather its rational or irrational, but if it gets you a negative value, then you would get 2 imaginary(complex) roots.

Answer
you may be correct, however to me when it asks for the values of "k" for which the roots of your problem is real, i get k = 3, for which by roots it means where your graph curves. "k" being 3 makes the roots of your problem real.

Info found at
www.hyper-ad.com/tutoring/math/algebra/Quadratic%20Formula%20Examples.html

There might be other solutions, not certain, but k = 3 is the best solution.

other numbers that are divisible by 3 would probably give you irrational or rational roots, which ever it is called.