Question 1162180
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They want you find the value of "k" in a way that the straight line f(x) = 2x-1 be a tangent line 
to the parabola  g(x) = 3x^2 + 8x + k.


It means that the equation

    3x^2 + 8x + k = 2x - 1 

has only one real root (the two roots merge into one root).


The equation is equivalent to

    3x^2 + 6x + (k+1) = 0.     (*)


Its discriminant is


    b^2 - 4ac = {{{6*2 - 4*3(k+1)}}} = 36 - 12(k+1)


We want the discriminant be zero

    36 - 12(k+1) = 0,   or   36 = 12(k+1),   k+1 = 36/12 = 3,  k = 3-1 = 2.


<U>ANSWER</U>.  k = 2.


<U>CHECK</U>.  At k= 2, the equation (*)  becomes  3x^3 +6x + 3 = 0.

        It is equivalent to  x^2 + 2x + 1 = 0,  which  factors into  {{{(x+1)^2}}} = 0  and has only one real solution x= -1.


        The point  (x,f(x)) = (-1,f(-1)) = (-1,-3)  is the same as  the point  (x,g(x)) = (-1,g(-1)) = (-1,-3).


        It is the unique common point of the two given lines.
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Solved. &nbsp;&nbsp;// &nbsp;&nbsp;The problem was solved using &nbsp;Algebra only, &nbsp;without using &nbsp;Calculus.