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anonymous
 one year ago
Help me, please?
anonymous
 one year ago
Help me, please?

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0A 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
 one year ago
Best ResponseYou've already chosen the best response.0I need help finding the radicand.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@peachpi Do you think you could help me with this one?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i am sure we can do this

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[3x^2 − 16x + 2 = 0 \] is the equation right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok the radicand of \[ax^2+bx+c=0\] is \(b^24ac\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but that does not give you the actual solution the actual solutions are \[\frac{b\pm\sqrt{b^24ac}}{2a}\]so lets compute it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0actually i see is does not say "find the solutions" it says "Describe the solution(s)" the radicand in this case is \[(6)^24\times 3\times 2=232\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0since 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
 one year ago
Best ResponseYou've already chosen the best response.0Okay, so we'll have to find those irrational solutions, right? But how?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0they do not ask you to find them, but we can

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0solution is \[\frac{b\pm\sqrt{b^24ac}}{2a}\] lets plug in the numbers and see what we get

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we already know \(b^24ac=232\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0plugging them in give \[\frac{16\pm\sqrt{232}}{2\times 3}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0then some more arithmetic is needed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, 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
 one year ago
Best ResponseYou've already chosen the best response.0it is a short hand for saying there are two different solution one with the plus, and another with the minus

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh 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
 one year ago
Best ResponseYou've already chosen the best response.0we 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
 one year ago
Best ResponseYou've already chosen the best response.0first of \[232=4\times 58\] so \[\sqrt{232}=\sqrt{4\times 58}=\sqrt{4}\sqrt{58}=2\sqrt{58}\]\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0therefore \[\frac{16\pm\sqrt{232}}{2\times 3}=\frac{16\pm2\sqrt{58}}{6}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now cancel a 2 top and bottom to finish with \[\frac{8\pm\sqrt{58}}{3}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0since 58 is not a prefect square, \(\sqrt{58}\) is irrational

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh okay, I see that now.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that is your final answer to the first one second one is easier

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hopefully, because I'm terrible at those cx

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the second one factors

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh so its just a factoring problem?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ 9x^2 + 3x − 2 = 0 \\ (3x+1)(3x2)=0\] set each factor equal to zero and solve for \(x\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Is that the finished product?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no that is just what you get when you factor to finish solving for \(x\) solve\[3x+1=0\] and also \[3x2=0\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The first one would be \[\frac{ 1 }{ 3 }\] Right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And the second one \[\frac{ 2 }{ 3 }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dang i made a mistake sorry

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it is not \[(3x+1)(3x2)=0\] it is \[(3x1)(3x+2)=0\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so solutions are \(\frac{1}{3}\) and \(\frac{2}{3}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright cool, so that's it?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you so much! That helped so much!
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