SOLUTION: I really need some help for this problems because I have been confuse of this problems since the start and now the Examination is coming near and this is going to be added, I reall

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: I really need some help for this problems because I have been confuse of this problems since the start and now the Examination is coming near and this is going to be added, I reall      Log On


   



Question 206107: I really need some help for this problems because I have been confuse of this problems since the start and now the Examination is coming near and this is going to be added, I really will appreciate some help in this questions:
---find the value of k so that the equation +3x%5E2-7x%2Bk=0+ will have equal roots.
---find the value of k so that the equation +3x%5E2-%285%2Bk%29x-17=0+ will have one roots numerically equal but opposite in sign
--what should be the range of the value of k so that the equation +2x%5E2%2B3kx-9=0+ will have real and unequal roots.
---what should be the value of k so that the equation +x%5E2%2B%28k-2%29x%2B4=0+
will have equal roots?

Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
I really need some help for this problems because I have been confuse of this problems since the start and now the Examination is coming near and this is going to be added, I really will appreciate some help in this questions:
---find the value of k so that the equation +3x%5E2-7x%2Bk=0+ will have equal roots.

Find the discriminant b%5E2-4ac=%28-7%29%5E2-4%283%29%28k%29=49-12k

To have equal roots the discriminant must equal 0, so

49-12k=0
-12k=-49
k=%28-49%29%2F%28-12%29
k=49%2F12

To check, we substitute 49%2F12 for k in

+3x%5E2-7x%2Bk=0+
+3x%5E2-7x%2B49%2F12=0+
Multiply through by 12 to clear of fractions:

12%2A3x%5E2-12%2A7x%2B12%2A%2849%2F12%29=12%2A0

36x%5E2-84x%2B49=0

%286x-7%29%286x-7%29=0

6x-7=0; 6x-7=0
6x=7  ; 6x=7
x=7%2F6 ; x=7%2F6

So the roots are equal. 

----------------------------------- 


---find the value of k so that the equation +3x%5E2-%285%2Bk%29x-17=0+ will have one roots numerically equal but opposite in sign

Then it will have to be true that we also get zero when we 
substitute -x for x:

+3%28-x%29%5E2-%285%2Bk%29%28-x%29-17=0+
3x%5E2%2B%285%2Bk%29x-17=0

And since we started with +3x%5E2-%285%2Bk%29x-17=0+
We can set the left sides equal:

3x%5E2%2B%285%2Bk%29x-17=+3x%5E2-%285%2Bk%29x-17=0+

3x%5E2%2B5x%2Bkx-17=+3x%5E2-5x-kx-17=0+

10x%2B2kx=0+

Divide thru by 2

5x%2Bkx=0

Factor out x

x%285%2Bk%29=0

x=0; 5%2Bk=0 
           k=-5

So the value of k that will cause the
roots to be numerically equal but opposite
in sign.

Check by substituting -5 for k and solve:

+3x%5E2-%285%2Bk%29x-17=0+
+3x%5E2-%285%2B%28-5%29%29x-17=0+
+3x%5E2-17=0+
3x%5E2=17
x%5E2=17%2F3


sqrt%2851%29%2F3 and -sqrt%2851%29%2F3 are numerically 
equal but opposite in sign.


--what should be the range of the value of k so that the equation
+2x%5E2%2B3kx-9=0+ will have real and unequal roots.

That is when the discriminant is positive

Find the discriminant b%5E2-4ac=%283k%29%5E2-4%282%29%28-9%29=9k%5E2%2B72

"Positive" means the same as "greater than 0"

So we set 9k%5E2%2B72%3E0

Divide through by 9

k%5E2%2B8%3E0

Since k%5E2 is never negative the left side will always
positive no matter what value of k we use, so all values of 
k will yield real and unequal roots.

So -infinity%3Ck%3Cinfinity


---what should be the value of k so that the equation +x%5E2%2B%28k-2%29x%2B4=0+
will have equal roots?

Find the discriminant b%5E2-4ac=%28k-2%29%5E2-4%281%29%284%29=4%28k-2%29%5E2-16

To have equal roots the discriminant must equal 0, so

4%28k-2%29%5E2-16=0

Divide thru by 4

%28k-2%29%5E2-4=0

%28k-2%29%5E2=4

k-2=%22%22%2B-sqrt%284%29

k-2=%22%22%2B-2

k=2%2B-2

Using the %22%2B%22, k=2%2B2=4

Using the %22-%22, k=2-2=0

Edwin