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.

- anonymous

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- ZeHanz

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.

- anonymous

That's as far as I went.
As i was solving for no real roots i got stuck at this:
|dw:1359306176821:dw|

- anonymous

|dw:1359306307751:dw|

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## More answers

- phi

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

- phi

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

- anonymous

whoa whoa whoa where did you (k-2)(k-10) from?

- phi

k^2 -12k+20 < 0
factor into
(k-2)(k-10) < 0

- ZeHanz

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

- anonymous

Wait lemme try the other 2 first.

- phi

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)

- anonymous

NOW I am confused.
|dw:1359306861746:dw|

- phi

there are two k values that result in 1 real root (a "repeated" root)

- anonymous

Then they are both answers? Shouldn't be happening though.

- phi

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.

- phi

**25x^2 + 10 x + 1

- anonymous

Then do I put 2 or 10 for the answer?

- phi

For what value of K will this quadratic equation have one real root? K=2 or K=10

- anonymous

Ohhhh I see.

- anonymous

Wait let me try the third and last one.

- anonymous

|dw:1359307404608:dw|
|dw:1359307505574:dw|

- phi

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

- anonymous

So you take the bigger one of the two and forge an equation from the two?

- phi

ask again, I don't understand the question.

- anonymous

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?

- phi

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

- anonymous

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?

- anonymous

I am still wondering how you chose the inequalities.

- phi

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) ?

- anonymous

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?

- phi

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) ?

- anonymous

Yeah. The discriminant must be positive for 2 real roots.

- phi

(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 ?

- phi

Similarly, both terms could be negative, resulting in their product being positive.

- anonymous

I got it, but where did you get the other one from. I got k>10. How about the k<2?

- precal

|dw:1359309274367:dw|
no solution is where your quadratic function does not cross the x axis

- phi

try doing the analysis with both terms being negative ( - * - = +)

- precal

|dw:1359309310961:dw|

- phi

(k-2)(k-10) > 0
this could happen if (k-2) is negative and (k-10) is negative

- precal

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

- anonymous

Yes, so it can also be k<2 and k<10. k<10 works for both too though.

- phi

are you sure?

- anonymous

Surely it can't be k<10 and k>10

- anonymous

wait nevermind.

- anonymous

Getting mixed up with negatives... I got it now.
So k<2 also works for k<10

- phi

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|

- phi

or do lots of problems, in which case you memorize the different cases.

- anonymous

I know how to solve these questions easily... it's just that the teacher decided to stump the class... :(

- anonymous

Thanks lol. I got it now :D

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