+0  
 
+1
666
6
avatar+17 

x^2-kx+9 = 0 then when the roots are real? k>6, k<6 k = 6 or none?

 Jan 14, 2016

Best Answer 

 #5
avatar+128089 
+5

The roots are real when

 

b^2 - 4ac >= 0    

 

So......the roots are real when

 

k^2 - 4(1)(9) >= 0

 

k^2 - 36 >= 0

 

k^2  >= 36

 

And this  happens when k=6,  k > 6   or when k < -6

 

 

 

cool cool cool

 Jan 14, 2016
 #1
avatar+8581 
0

I would say none. My answer does not show up as any of the multiple choices.

 Jan 14, 2016
 #2
avatar+17 
0

How did you approach it hailey?

 Jan 14, 2016
 #3
avatar+8581 
0

I plugged in the numbers for K. But the answer did not look right. I could be wrong though, I might have did it the wrong way.

 Jan 14, 2016
 #4
avatar
+5

With the usual notation, for real roots you need b^2 - 4ac >= 0.

Plug in the values for a and c, and (-k) for b and see what that tells you about k.

 Jan 14, 2016
 #5
avatar+128089 
+5
Best Answer

The roots are real when

 

b^2 - 4ac >= 0    

 

So......the roots are real when

 

k^2 - 4(1)(9) >= 0

 

k^2 - 36 >= 0

 

k^2  >= 36

 

And this  happens when k=6,  k > 6   or when k < -6

 

 

 

cool cool cool

CPhill Jan 14, 2016
 #6
avatar+118587 
0

Thanks Chris and guest,

 Hi PurplePotato, you have a fabulous username!  Welcome to the forum :)

 Jan 15, 2016

3 Online Users

avatar