Question 171658
Let {{{f(x)=Kx^2+(k+3)x+3-4k}}}, 



If one root is 2, then this means that {{{x=2}}} and {{{f(2)=0}}}



{{{f(x)=Kx^2+(k+3)x+3-4k}}} Start with the given equation.



{{{f(2)=K(2)^2+(k+3)(2)+3-4k}}} Plug in {{{x=2}}}.



{{{f(2)=4k+(k+3)(2)+3-4k}}} Square 2 to get 4.



{{{f(2)=4k+2k+6+3-4k}}} Distribute.



{{{4k+2k+6+3-4k=0}}} Set the right side equal to zero.



{{{2k+9=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=(-9)/(2)}}} Divide both sides by {{{2}}} to isolate {{{k}}}.



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Answer:


So the answer is {{{k=-9/2}}} which in decimal form is {{{k=-4.5}}}.