## tilt 2 years ago 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.

1. 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.

2. tilt

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

3. tilt

|dw:1359306307751:dw|

4. 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

5. 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

6. tilt

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

7. phi

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

8. 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<k<10, instead of a number beteen 2 and 10.

9. tilt

Wait lemme try the other 2 first.

10. 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)

11. tilt

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

12. phi

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

13. tilt

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

14. 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.

15. phi

**25x^2 + 10 x + 1

16. tilt

Then do I put 2 or 10 for the answer?

17. phi

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

18. tilt

Ohhhh I see.

19. tilt

Wait let me try the third and last one.

20. tilt

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

21. 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

22. tilt

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

23. phi

ask again, I don't understand the question.

24. tilt

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?

25. 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

26. tilt

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?

27. tilt

I am still wondering how you chose the inequalities.

28. 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) ?

29. tilt

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?

30. 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) ?

31. tilt

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

32. 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 ?

33. phi

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

34. tilt

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

35. precal

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

36. phi

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

37. precal

|dw:1359309310961:dw|

38. phi

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

39. 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

40. tilt

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

41. phi

are you sure?

42. tilt

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

43. tilt

wait nevermind.

44. tilt

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

45. 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|

46. phi

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

47. tilt

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

48. tilt

Thanks lol. I got it now :D