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|dw:1359306307751:dw|

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

Wait lemme try the other 2 first.

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.

**25x^2 + 10 x + 1

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

Ohhhh I see.

Wait let me try the third and last one.

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

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

ask again, I don't understand the question.

I am still wondering how you chose the inequalities.

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

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 ( - * - = +)

|dw:1359309310961:dw|

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

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

are you sure?

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

wait nevermind.

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

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

Thanks lol. I got it now :D