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tilt
Group Title
I'm having trouble doing this question...
(3K5)x^2+Kx+1=0
For what value of K will this quadratic equation have no real roots, one real root, or two real roots?
Answer as an inequality if necessary.
 one year ago
 one year ago
tilt Group Title
I'm having trouble doing this question... (3K5)x^2+Kx+1=0 For what value of K will this quadratic equation have no real roots, one real root, or two real roots? Answer as an inequality if necessary.
 one year ago
 one year ago

This Question is Closed

ZeHanz Group TitleBest ResponseYou've already chosen the best response.1
This is a quadratic equation: ax²+bx+c=0. The discriminant, D=b²4ac tells you about the number of real roots: D > 0: 2 real roots, D = 0: 1 real root, D < 0, no real roots. You probably know all this already ;) You just have to see that in your own equation a=3K4, b=K and c=1. Put these values for a, b and c in the formula for D and solve the above three cases for K.
 one year ago

tilt Group TitleBest ResponseYou've already chosen the best response.1
That's as far as I went. As i was solving for no real roots i got stuck at this: dw:1359306176821:dw
 one year ago

tilt Group TitleBest ResponseYou've already chosen the best response.1
dw:1359306307751:dw
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
what you get is (k2)(k10) < 0 this expression is less than 0 if it is negative. it is negative if one of the factors is + and the other is negative
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
so you try (k2)< 0 and k10>0 > k<2 and k>10 can not happen or k2 > 0 and k10 < 0 > k>2 and k<10 or 2 < k < 10
 one year ago

tilt Group TitleBest ResponseYou've already chosen the best response.1
whoa whoa whoa where did you (k2)(k10) from?
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
k^2 12k+20 < 0 factor into (k2)(k10) < 0
 one year ago

ZeHanz Group TitleBest ResponseYou've already chosen the best response.1
First, you get: K²12K+20<0 This is the same as (K2)(K10)<0. It equals 0 if K = 2 or K=10. For other values of K it will be positive or negative. If you try K=0 (very simple to do), you get a positive number. To get a negative outcome, you therefore have to have 2<k<10, instead of a number beteen 2 and 10.
 one year ago

tilt Group TitleBest ResponseYou've already chosen the best response.1
Wait lemme try the other 2 first.
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
You used "completing the square" to find the roots of K²12K+20=0 you got k=2 and k= 10 which implies k2=0 and k10= 0 this means you could write the quadratic as (k2)(k10) (in case you could not factor the quadratic)
 one year ago

tilt Group TitleBest ResponseYou've already chosen the best response.1
NOW I am confused. dw:1359306861746:dw
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
there are two k values that result in 1 real root (a "repeated" root)
 one year ago

tilt Group TitleBest ResponseYou've already chosen the best response.1
Then they are both answers? Shouldn't be happening though.
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
Don't get confused. There are two k values that result in a quadratic: x^2+2x+1 =0 5x^2 +10x +1 =0 each has one 1 (repeated) root.
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
**25x^2 + 10 x + 1
 one year ago

tilt Group TitleBest ResponseYou've already chosen the best response.1
Then do I put 2 or 10 for the answer?
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
For what value of K will this quadratic equation have one real root? K=2 or K=10
 one year ago

tilt Group TitleBest ResponseYou've already chosen the best response.1
Wait let me try the third and last one.
 one year ago

tilt Group TitleBest ResponseYou've already chosen the best response.1
dw:1359307404608:dw dw:1359307505574:dw
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
we use the same idea (k2)(k10) > 0 > 0 means the expression is positive (things that are positive are > 0) to be +, both terms must be + or both terms must be  let's first do both positive: k2>0 and k10>0 > k>2 and k>10 both conditions have to be true, so k>10 is the answer now both terms are  k2<0 and k10<0 > k<2 and k<10 Both must be true, so k<2 works
 one year ago

tilt Group TitleBest ResponseYou've already chosen the best response.1
So you take the bigger one of the two and forge an equation from the two?
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
ask again, I don't understand the question.
 one year ago

tilt Group TitleBest ResponseYou've already chosen the best response.1
So I take the bigger inequalities from both the positive and the negative trials and put them together? Like take the bigger one from k>2 / k>10 and k<2 / k<10? In which case 2>k>10?
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
I would not write it 2>k>10 (you only use this notation if k is between 2 and 10) in this case k<2 results in quadratics with 2 real roots and k>10 results in quadratics with 2 real roots I would say For what value of K will this quadratic equation have 2 real roots? for K<2 or k>10
 one year ago

tilt Group TitleBest ResponseYou've already chosen the best response.1
Ohhh. So I don't clump them together. The only time when I do is if I am solving for when there aren't any real roots. Anyways, so I take the bigger inequalities from the above question for my answer?
 one year ago

tilt Group TitleBest ResponseYou've already chosen the best response.1
I am still wondering how you chose the inequalities.
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
so I take the bigger inequalities ? 1 step at a time: (k2)(k10) > 0 this is the condition for 2 real roots You agree that this means (k2)(k10) is positive (almost by definition) ?
 one year ago

tilt Group TitleBest ResponseYou've already chosen the best response.1
No, I mean like for K>2 K>10 and K<2 and K<10. How did you choose 2 of them? Can't you also choose k<10 and k>2?
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
Yes, I understand the question, and it will take a few simple sentences to explain. But the first question is (k2)(k10) > 0 this is the condition for 2 real roots You agree that this means (k2)(k10) is positive (almost by definition) ?
 one year ago

tilt Group TitleBest ResponseYou've already chosen the best response.1
Yeah. The discriminant must be positive for 2 real roots.
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
(k2)(k10) > 0 this means that both (k2) is positive and (k10) is positive or k2 > 0 and at the same time k10>0 or k>2 and at the same time k>10 obviously k=3 would make the (k2) positive, but the (k10) term would be negative. We need both to be positive. so if you need k>2 and k>10, k>10 works for both terms i.e. k2 will be positive and k10 will be positive. Does that answer the question ?
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
Similarly, both terms could be negative, resulting in their product being positive.
 one year ago

tilt Group TitleBest ResponseYou've already chosen the best response.1
I got it, but where did you get the other one from. I got k>10. How about the k<2?
 one year ago

precal Group TitleBest ResponseYou've already chosen the best response.0
dw:1359309274367:dw no solution is where your quadratic function does not cross the x axis
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
try doing the analysis with both terms being negative (  *  = +)
 one year ago

precal Group TitleBest ResponseYou've already chosen the best response.0
dw:1359309310961:dw
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
(k2)(k10) > 0 this could happen if (k2) is negative and (k10) is negative
 one year ago

precal Group TitleBest ResponseYou've already chosen the best response.0
remember the number of times the function crosses the x axis is the number of solutions otherwise you are on the correct path, just reminding you about you are looking for Listen to phi
 one year ago

tilt Group TitleBest ResponseYou've already chosen the best response.1
Yes, so it can also be k<2 and k<10. k<10 works for both too though.
 one year ago

tilt Group TitleBest ResponseYou've already chosen the best response.1
Surely it can't be k<10 and k>10
 one year ago

tilt Group TitleBest ResponseYou've already chosen the best response.1
wait nevermind.
 one year ago

tilt Group TitleBest ResponseYou've already chosen the best response.1
Getting mixed up with negatives... I got it now. So k<2 also works for k<10
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
yes. I draw a number line, and arrows. the place where I get 2 lines overlapping is the answer so for k<2 and k<10 I would draw dw:1359309685962:dw
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
or do lots of problems, in which case you memorize the different cases.
 one year ago

tilt Group TitleBest ResponseYou've already chosen the best response.1
I know how to solve these questions easily... it's just that the teacher decided to stump the class... :(
 one year ago

tilt Group TitleBest ResponseYou've already chosen the best response.1
Thanks lol. I got it now :D
 one year ago
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