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anonymous
 3 years ago
Would it be easier to factor 12t^217t^15=0 or to use the quadratic formula?
anonymous
 3 years ago
Would it be easier to factor 12t^217t^15=0 or to use the quadratic formula?

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ash2326
 3 years ago
Best ResponseYou've already chosen the best response.0Is this your equation? \[12t^217t^{1}5=0\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i dont think you can use the quadratic formula in this case, as we have ve power in this case.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[12t^217t^{1}5=0\]This one won't factor that well. The problem is probably\[12t^217t5=0\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0If it is, you can use the acmethod to factor it. Do you know how to do that?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yes, but that's not the method my teacher wants me to use. Thank you.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what is the problem exactly. is it quadratic or not?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No. it's not quadratic.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The quadratic formula only works on quadratic equations.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0there must have been a mistake with this problem. anyways, the equation can be reduced to cubic form.the equation has a sign change between 1 and 2. so it must have a root between these two numbers. i found it to be 1.246. the other two roots can be found by reducing this cubic equation into quadratic equation and they come out to be imaginary.
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