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I'm having trouble doing this question... (3K-5)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 is a quadratic equation: ax²+bx+c=0. The discriminant, D=b²-4ac 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=3K-4, 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.
That's as far as I went. As i was solving for no real roots i got stuck at this: |dw:1359306176821:dw|
what you get is (k-2)(k-10) < 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
so you try (k-2)< 0 and k-10>0 --> k<2 and k>10 can not happen or k-2 > 0 and k-10 < 0 ---> k>2 and k<10 or 2 < k < 10
whoa whoa whoa where did you (k-2)(k-10) from?
k^2 -12k+20 < 0 factor into (k-2)(k-10) < 0
First, you get: K²-12K+20<0 This is the same as (K-2)(K-10)<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.
Wait lemme try the other 2 first.
You used "completing the square" to find the roots of K²-12K+20=0 you got k=2 and k= 10 which implies k-2=0 and k-10= 0 this means you could write the quadratic as (k-2)(k-10) (in case you could not factor the quadratic)
NOW I am confused. |dw:1359306861746:dw|
there are two k values that result in 1 real root (a "repeated" root)
Then they are both answers? Shouldn't be happening though.
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.
Then do I put 2 or 10 for the answer?
For what value of K will this quadratic equation have one real root? K=2 or K=10
Wait let me try the third and last one.
|dw:1359307404608:dw| |dw:1359307505574:dw|
we use the same idea (k-2)(k-10) > 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: k-2>0 and k-10>0 --> k>2 and k>10 both conditions have to be true, so k>10 is the answer now both terms are - k-2<0 and k-10<0 --> k<2 and k<10 Both must be true, so k<2 works
So you take the bigger one of the two and forge an equation from the two?
ask again, I don't understand the question.
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?
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
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?
I am still wondering how you chose the inequalities.
so I take the bigger inequalities ? 1 step at a time: (k-2)(k-10) > 0 this is the condition for 2 real roots You agree that this means (k-2)(k-10) is positive (almost by definition) ?
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?
Yes, I understand the question, and it will take a few simple sentences to explain. But the first question is (k-2)(k-10) > 0 this is the condition for 2 real roots You agree that this means (k-2)(k-10) is positive (almost by definition) ?
Yeah. The discriminant must be positive for 2 real roots.
(k-2)(k-10) > 0 this means that both (k-2) is positive and (k-10) is positive or k-2 > 0 and at the same time k-10>0 or k>2 and at the same time k>10 obviously k=3 would make the (k-2) positive, but the (k-10) 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. k-2 will be positive and k-10 will be positive. Does that answer the question ?
Similarly, both terms could be negative, resulting in their product being positive.
I got it, but where did you get the other one from. I got k>10. How about the k<2?
|dw:1359309274367:dw| no solution is where your quadratic function does not cross the x axis
try doing the analysis with both terms being negative ( - * - = +)
(k-2)(k-10) > 0 this could happen if (k-2) is negative and (k-10) is negative
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
Yes, so it can also be k<2 and k<10. k<10 works for both too though.
Surely it can't be k<10 and k>10
Getting mixed up with negatives... I got it now. So k<2 also works for k<10
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|
or do lots of problems, in which case you memorize the different cases.
I know how to solve these questions easily... it's just that the teacher decided to stump the class... :(
Thanks lol. I got it now :D