Help me, please?

- anonymous

Help me, please?

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- anonymous

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)

- anonymous

I need help finding the radicand.

- anonymous

@peachpi
Do you think you could help me with this one?

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## More answers

- anonymous

i am sure we can do this

- anonymous

Awesome, thank you!

- anonymous

\[3x^2 − 16x + 2 = 0 \] is the equation right?

- anonymous

'Tis correct~

- anonymous

ok
the radicand of \[ax^2+bx+c=0\] is \(b^2-4ac\)

- anonymous

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

- anonymous

actually i see is does not say "find the solutions" it says "Describe the solution(s)"
the radicand in this case is
\[(-6)^2-4\times 3\times 2=232\]

- anonymous

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

- anonymous

Okay, so we'll have to find those irrational solutions, right? But how?

- anonymous

they do not ask you to find them, but we can

- anonymous

solution is
\[\frac{-b\pm\sqrt{b^2-4ac}}{2a}\] lets plug in the numbers and see what we get

- anonymous

we already know \(b^2-4ac=232\)

- anonymous

plugging them in give
\[\frac{16\pm\sqrt{232}}{2\times 3}\]

- anonymous

then some more arithmetic is needed

- anonymous

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?

- anonymous

it is a short hand for saying there are two different solution
one with the plus, and another with the minus

- anonymous

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?

- anonymous

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

- anonymous

Right, okay.

- anonymous

first of \[232=4\times 58\] so
\[\sqrt{232}=\sqrt{4\times 58}=\sqrt{4}\sqrt{58}=2\sqrt{58}\]\]

- anonymous

therefore \[\frac{16\pm\sqrt{232}}{2\times 3}=\frac{16\pm2\sqrt{58}}{6}\]

- anonymous

now cancel a 2 top and bottom to finish with
\[\frac{8\pm\sqrt{58}}{3}\]

- anonymous

since 58 is not a prefect square, \(\sqrt{58}\) is irrational

- anonymous

Oh okay, I see that now.

- anonymous

that is your final answer to the first one
second one is easier

- anonymous

Hopefully, because I'm terrible at those cx

- anonymous

the second one factors

- anonymous

Oh so its just a factoring problem?

- anonymous

yeah pretty much

- anonymous

\[ 9x^2 + 3x − 2 = 0 \\
(3x+1)(3x-2)=0\] set each factor equal to zero and solve for \(x\)

- anonymous

Is that the finished product?

- anonymous

no that is just what you get when you factor
to finish solving for \(x\) solve\[3x+1=0\] and also \[3x-2=0\]

- anonymous

The first one would be
\[-\frac{ 1 }{ 3 }\]
Right?

- anonymous

And the second one
\[\frac{ 2 }{ 3 }\]

- anonymous

yup

- anonymous

oh wait a second

- anonymous

dang i made a mistake sorry

- anonymous

Its alright ^.^

- anonymous

it is not \[(3x+1)(3x-2)=0\] it is \[(3x-1)(3x+2)=0\]

- anonymous

so solutions are \(\frac{1}{3}\) and \(-\frac{2}{3}\)

- anonymous

Alright cool, so that's it?

- anonymous

yup thats it

- anonymous

Thank you so much! That helped so much!

- anonymous

yw

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