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I need help finding the radicand.

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 \(b^2-4ac\)

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

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?

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