+0  
 
0
63
4
avatar+802 

Hi my good friends....please help me with this one..I have tried and tried...:

 

Determine the nature of the roots without solving the equation.

 

\(kx^2+(4k-1)x=2-4k\)

 

I did this....

 

\(kx^2+4kx-x+4k-2=0 \)

\(kx^2+x(4k-1)+2(2k-1)=0\)

 

Not sure..what I would not be suprised of is if I had made a mistake....smiley..

 

Thank you for all your help..always appreciated...

 Mar 3, 2021
 #1
avatar+802 
+1

But then I did this: leave it as 

 

\(kx^2+x(4k-1)+4k-2\)

 

\(d=b^2-4ac\)

\(=(4k-1)^2-4(k)(4k-2)\)

\(=16k^2-8k+1-16k^2+8k\)

\(=1\)

 

So the roots are Real, un-equal and rational....is this right perhaps?

 Mar 3, 2021
 #2
avatar+112972 
+1

Yes, that is correct.   laugh

The symbol for the determinant is a triangle.

 

Backslash triangle.  wink

 \(\triangle \)

Melody  Mar 3, 2021
 #3
avatar+802 
+1

Hello dear Melody.....

 

Thank you for confirming....yes I know its the triangle, just do not know where to get the symbol...I searched but could not find it...smiley

juriemagic  Mar 4, 2021
 #4
avatar+112972 
+1

Your welcome.

And now you know wink

 

If you had search "LaTex triangle symbol" it would have come straight up - in picture form.   :)

Melody  Mar 4, 2021

79 Online Users

avatar
avatar
avatar