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Mathematics 13 Online
OpenStudy (anonymous):

will someone please teach me how to find the discriminant of the following equation? 9x^2+12x+4=0

OpenStudy (turingtest):

for an equation of the form\[ax^2+bx+c=0\]the discriminant is\[b^2-4ac\]

OpenStudy (anonymous):

so what would this equation look like, in that form?

OpenStudy (anonymous):

Fill in the numbers, he gave you the variables that match up with the equation you gave us. It is all just substitution.

OpenStudy (anonymous):

if a=9 then what is b=? and c=?

OpenStudy (anonymous):

b=12 and c=4?

OpenStudy (anonymous):

correct! now fill in the equation b^2−4ac to find the discriminant

OpenStudy (anonymous):

Once you have the numbers substituted in the correct places..it is just simple math solving and simplifying.

OpenStudy (anonymous):

When you have the answer I can check it for you.

OpenStudy (anonymous):

okay but i still am not to sure. I see where you did that but if i filled out b^2-4ac then would that be 12^2-4?

OpenStudy (anonymous):

you forgot to fill in the (a)(c) part after the 4...

OpenStudy (anonymous):

so 12^2-4(12)(4)? sorry about this... for some reason i am so lost in this.

OpenStudy (anonymous):

close, if a=9 b=12 and c=4 you should have gotten 12^2-4(4)(9).

OpenStudy (anonymous):

okay i see that now thank-you for clearing that up! so the answer would be 12^2-4(4)(9)

OpenStudy (anonymous):

do the math, simplify and you should get your answer

OpenStudy (anonymous):

Did you get an answer?

OpenStudy (anonymous):

12^2-144?

OpenStudy (anonymous):

almost done solving! you need to simplify more. what is 12^2? then subtract 144 from the answer you get.

OpenStudy (anonymous):

so it =0!!

OpenStudy (anonymous):

Correct!! Here is a short explanation. What you are having trouble solving is easier than you think ^.^ I hope you understand!

OpenStudy (anonymous):

i do now thanks to u!

OpenStudy (anonymous):

OH and in my drawing i mixed up the (a) and (c) but when you are solving the equation it usually does not affect your solution.

OpenStudy (anonymous):

ok that is fine

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