Help me, please?

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

Help me, please?

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

A quadratic equation is shown below: 3x2 − 16x + 2 = 0 Part A: Describe the solution(s) to the equation by just determining the radicand. Show your work. (5 points) Part B: Solve 9x2 + 3x − 2 = 0 by using an appropriate method. Show the steps of your work, and explain why you chose the method used. (5 points)
I need help finding the radicand.
@peachpi Do you think you could help me with this one?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

i am sure we can do this
Awesome, thank you!
\[3x^2 − 16x + 2 = 0 \] is the equation right?
'Tis correct~
ok the radicand of \[ax^2+bx+c=0\] is \(b^2-4ac\)
but that does not give you the actual solution the actual solutions are \[\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]so lets compute it
actually i see is does not say "find the solutions" it says "Describe the solution(s)" the radicand in this case is \[(-6)^2-4\times 3\times 2=232\]
since that is a positive number, but not a perfect square (like say 25 or 16) that means there are two real but irrational solutions
Okay, so we'll have to find those irrational solutions, right? But how?
they do not ask you to find them, but we can
solution is \[\frac{-b\pm\sqrt{b^2-4ac}}{2a}\] lets plug in the numbers and see what we get
we already know \(b^2-4ac=232\)
plugging them in give \[\frac{16\pm\sqrt{232}}{2\times 3}\]
then some more arithmetic is needed
Okay, so for the top part does that mean you could add or subtract or is there a specific one you're supposed to use?
it is a short hand for saying there are two different solution one with the plus, and another with the minus
Oh alright, that makes sense. But how would you solve the top part? We've already established that the 232 ends up two real irrational numbers, if we're trying to figure those out now, what do we do on the top?
we can to that but i want to stress that the question does not ask for the actual solutions, only to "describe " them the description is two real irrational solutions now lets find them
Right, okay.
first of \[232=4\times 58\] so \[\sqrt{232}=\sqrt{4\times 58}=\sqrt{4}\sqrt{58}=2\sqrt{58}\]\]
therefore \[\frac{16\pm\sqrt{232}}{2\times 3}=\frac{16\pm2\sqrt{58}}{6}\]
now cancel a 2 top and bottom to finish with \[\frac{8\pm\sqrt{58}}{3}\]
since 58 is not a prefect square, \(\sqrt{58}\) is irrational
Oh okay, I see that now.
that is your final answer to the first one second one is easier
Hopefully, because I'm terrible at those cx
the second one factors
Oh so its just a factoring problem?
yeah pretty much
\[ 9x^2 + 3x − 2 = 0 \\ (3x+1)(3x-2)=0\] set each factor equal to zero and solve for \(x\)
Is that the finished product?
no that is just what you get when you factor to finish solving for \(x\) solve\[3x+1=0\] and also \[3x-2=0\]
The first one would be \[-\frac{ 1 }{ 3 }\] Right?
And the second one \[\frac{ 2 }{ 3 }\]
yup
oh wait a second
dang i made a mistake sorry
Its alright ^.^
it is not \[(3x+1)(3x-2)=0\] it is \[(3x-1)(3x+2)=0\]
so solutions are \(\frac{1}{3}\) and \(-\frac{2}{3}\)
Alright cool, so that's it?
yup thats it
Thank you so much! That helped so much!
yw

Not the answer you are looking for?

Search for more explanations.

Ask your own question