Expert: Paul Klarreich - 1/17/2007

Question
given that the equation x^2-(2k-1)x=-k^2 does not touch the x-axis,(i) find the range of values ok
(ii) when k=-2, solve x^2-(2k-1)=k^2

Answer
Questioner:   preety
Category:  Calculus
 
Subject:  add maths
Question:  given that the equation x^2-(2k-1)x=-k^2 does not touch the x-axis,(i) find the range of values of k
(ii) when k=-2, solve x^2-(2k-1)=k^2
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Hi, Preety,

I think you left out some details.  I think your confusion comes from not using the vocabulary correctly.  Words like 'equation', and 'values of k' are algebraic.  Words like 'touch' and 'x-axis' are geometric, and you must not mix them together.

So you cannot say 'the equation x^2-(2k-1)x=-k^2 does not touch the x-axis.'  But you can say:

'the GRAPH OF THE equation x^2-(2k-1)x=-k^2 does not touch the x-axis.'

Now that still does not make sense, because "the GRAPH OF THE equation x^2-(2k-1)x=-k^2" is just a couple of points on a number line, while the x-axis belongs to 2-space.  So you probably mean to write:

For what values of k does the graph of

y = x^2 - (2k-1) + k^2

not touch the x-axis?

Now that is not exactly what you wrote, and it is hard to interpret your equation.  

You cannot write:

y = x^2 - (2k-1)x = -k^2

because it has TWO equal signs, and you MUST have the part that says "y = " in there.  

So I am guessing that the example is what I wrote.

You deal with it this way:

A. The graph of  y = x^2 + (more stuff)  is a parabola.

B. A parabola has a turning point and a direction.

C. Since the direction in this case (the coefficient of x^2 is positive.) is UP, the

turning point is a minimum.

D. If this minimum is already above the x-axis, i.e. positive, then the graph will

never touch the x-axis.

E. In the standard form:  y = ax^2 + bx + c,  there are some rules for finding the

turning point:

   The x-coordinate is given by  x = -b/2a
   The y-coordinate is obtained by substituting that.

Here we go:

y = x^2 - (2k-1)x + k^2   matches up with
y = ax^2  + bx    + c  as follows:

a = 1
b = -(2k-1)
c = k^2

The T.P. has  x = -b/2a = (2k-1)/2, and has:

y = ((2k-1)/2)^2 - (2k - 1)((2k-1)/2) + k^2
   4k^2 - 4k + 1     4k^2 - 4k + 1
y = -------------- -  -------------  + k^2
         4                 2

   4k^2 - 4k + 1     8k^2 - 8k + 2
y = -------------- -  -------------  + k^2
         4                 4

     4k^2 - 4k + 1
y = - -------------  + k^2
           4

     4k^2 - 4k + 1    4k^2
y = - -------------  + -----
           4            4

   4k - 1
y = ------
      4

OK. Now we can get to work.  Since the T.P. must not touch the x-axis, that value must

be positive. So solve the inequality:

4k - 1
------ > 0
  4

4k - 1 > 0
4k > 1
k > 1/4

That's it: the range of values is  1/4 < k < infinity.
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About your second question:

(ii) when k=-2, solve x^2 - (2k-1)  = k^2

Did you mean:

(ii) when k=-2, solve x^2 - (2k-1)x = k^2 ??

Then the equation is

x^2 - (-4-1)x = (-2)^2

x^2 - (-5)x = 4

x^2 + 5x - 4 = 0

which you can solve using the quadratic formula. (I'll leave that to you.)