SOLUTION: given the equation Kx^2+(k+3)x+3-4k=0 find the value of k given that one root is 2

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Question 171658: given the equation Kx^2+(k+3)x+3-4k=0 find the value of k given that one root is 2
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Let f%28x%29=Kx%5E2%2B%28k%2B3%29x%2B3-4k,


If one root is 2, then this means that x=2 and f%282%29=0


f%28x%29=Kx%5E2%2B%28k%2B3%29x%2B3-4k Start with the given equation.


f%282%29=K%282%29%5E2%2B%28k%2B3%29%282%29%2B3-4k Plug in x=2.


f%282%29=4k%2B%28k%2B3%29%282%29%2B3-4k Square 2 to get 4.


f%282%29=4k%2B2k%2B6%2B3-4k Distribute.


4k%2B2k%2B6%2B3-4k=0 Set the right side equal to zero.


2k%2B9=0 Combine like terms on the left side.


2k=0-9 Subtract 9 from both sides.


2k=-9 Combine like terms on the right side.


k=%28-9%29%2F%282%29 Divide both sides by 2 to isolate k.


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Answer:

So the answer is k=-9%2F2 which in decimal form is k=-4.5.