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anonymous
 3 years ago
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.
anonymous
 3 years ago
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.

This Question is Closed

ZeHanz
 3 years ago
Best ResponseYou've already chosen the best response.1This 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That's as far as I went. As i was solving for no real roots i got stuck at this: dw:1359306176821:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1359306307751:dw

phi
 3 years ago
Best ResponseYou've already chosen the best response.1what 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

phi
 3 years ago
Best ResponseYou've already chosen the best response.1so 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0whoa whoa whoa where did you (k2)(k10) from?

phi
 3 years ago
Best ResponseYou've already chosen the best response.1k^2 12k+20 < 0 factor into (k2)(k10) < 0

ZeHanz
 3 years ago
Best ResponseYou've already chosen the best response.1First, 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Wait lemme try the other 2 first.

phi
 3 years ago
Best ResponseYou've already chosen the best response.1You 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)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0NOW I am confused. dw:1359306861746:dw

phi
 3 years ago
Best ResponseYou've already chosen the best response.1there are two k values that result in 1 real root (a "repeated" root)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Then they are both answers? Shouldn't be happening though.

phi
 3 years ago
Best ResponseYou've already chosen the best response.1Don'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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Then do I put 2 or 10 for the answer?

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Wait let me try the third and last one.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1359307404608:dw dw:1359307505574:dw

phi
 3 years ago
Best ResponseYou've already chosen the best response.1we 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So you take the bigger one of the two and forge an equation from the two?

phi
 3 years ago
Best ResponseYou've already chosen the best response.1ask again, I don't understand the question.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So 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?

phi
 3 years ago
Best ResponseYou've already chosen the best response.1I 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ohhh. 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I am still wondering how you chose the inequalities.

phi
 3 years ago
Best ResponseYou've already chosen the best response.1so 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) ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No, 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?

phi
 3 years ago
Best ResponseYou've already chosen the best response.1Yes, 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) ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah. The discriminant must be positive for 2 real roots.

phi
 3 years ago
Best 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 ?

phi
 3 years ago
Best ResponseYou've already chosen the best response.1Similarly, both terms could be negative, resulting in their product being positive.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I got it, but where did you get the other one from. I got k>10. How about the k<2?

precal
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1359309274367:dw no solution is where your quadratic function does not cross the x axis

phi
 3 years ago
Best ResponseYou've already chosen the best response.1try doing the analysis with both terms being negative (  *  = +)

phi
 3 years ago
Best ResponseYou've already chosen the best response.1(k2)(k10) > 0 this could happen if (k2) is negative and (k10) is negative

precal
 3 years ago
Best ResponseYou've already chosen the best response.0remember 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yes, so it can also be k<2 and k<10. k<10 works for both too though.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Surely it can't be k<10 and k>10

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Getting mixed up with negatives... I got it now. So k<2 also works for k<10

phi
 3 years ago
Best ResponseYou've already chosen the best response.1yes. 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

phi
 3 years ago
Best ResponseYou've already chosen the best response.1or do lots of problems, in which case you memorize the different cases.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I know how to solve these questions easily... it's just that the teacher decided to stump the class... :(

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thanks lol. I got it now :D
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