anonymous
  • anonymous
Help me, please?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • 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
  • anonymous
I need help finding the radicand.
anonymous
  • anonymous
@peachpi Do you think you could help me with this one?

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anonymous
  • anonymous
i am sure we can do this
anonymous
  • anonymous
Awesome, thank you!
anonymous
  • anonymous
\[3x^2 − 16x + 2 = 0 \] is the equation right?
anonymous
  • anonymous
'Tis correct~
anonymous
  • anonymous
ok the radicand of \[ax^2+bx+c=0\] is \(b^2-4ac\)
anonymous
  • 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
  • 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
  • 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
  • anonymous
Okay, so we'll have to find those irrational solutions, right? But how?
anonymous
  • anonymous
they do not ask you to find them, but we can
anonymous
  • anonymous
solution is \[\frac{-b\pm\sqrt{b^2-4ac}}{2a}\] lets plug in the numbers and see what we get
anonymous
  • anonymous
we already know \(b^2-4ac=232\)
anonymous
  • anonymous
plugging them in give \[\frac{16\pm\sqrt{232}}{2\times 3}\]
anonymous
  • anonymous
then some more arithmetic is needed
anonymous
  • 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
  • anonymous
it is a short hand for saying there are two different solution one with the plus, and another with the minus
anonymous
  • 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
  • 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
  • anonymous
Right, okay.
anonymous
  • anonymous
first of \[232=4\times 58\] so \[\sqrt{232}=\sqrt{4\times 58}=\sqrt{4}\sqrt{58}=2\sqrt{58}\]\]
anonymous
  • anonymous
therefore \[\frac{16\pm\sqrt{232}}{2\times 3}=\frac{16\pm2\sqrt{58}}{6}\]
anonymous
  • anonymous
now cancel a 2 top and bottom to finish with \[\frac{8\pm\sqrt{58}}{3}\]
anonymous
  • anonymous
since 58 is not a prefect square, \(\sqrt{58}\) is irrational
anonymous
  • anonymous
Oh okay, I see that now.
anonymous
  • anonymous
that is your final answer to the first one second one is easier
anonymous
  • anonymous
Hopefully, because I'm terrible at those cx
anonymous
  • anonymous
the second one factors
anonymous
  • anonymous
Oh so its just a factoring problem?
anonymous
  • anonymous
yeah pretty much
anonymous
  • anonymous
\[ 9x^2 + 3x − 2 = 0 \\ (3x+1)(3x-2)=0\] set each factor equal to zero and solve for \(x\)
anonymous
  • anonymous
Is that the finished product?
anonymous
  • 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
  • anonymous
The first one would be \[-\frac{ 1 }{ 3 }\] Right?
anonymous
  • anonymous
And the second one \[\frac{ 2 }{ 3 }\]
anonymous
  • anonymous
yup
anonymous
  • anonymous
oh wait a second
anonymous
  • anonymous
dang i made a mistake sorry
anonymous
  • anonymous
Its alright ^.^
anonymous
  • anonymous
it is not \[(3x+1)(3x-2)=0\] it is \[(3x-1)(3x+2)=0\]
anonymous
  • anonymous
so solutions are \(\frac{1}{3}\) and \(-\frac{2}{3}\)
anonymous
  • anonymous
Alright cool, so that's it?
anonymous
  • anonymous
yup thats it
anonymous
  • anonymous
Thank you so much! That helped so much!
anonymous
  • anonymous
yw

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