Solve each inequality. state the solution set using interval notation
and graph the solution set.  # 1. 4x^2 - 8x >= 0      #2. 3x - x^2 > 0
                         # 3. x^2 + 25 < 10x       A explanation how
you got it, on each process would be very desireable.  Thanx

Dear calliblue,

The first step in each case is to reformulate the equation so that the
right side is 0 while the left side is a quadratic expression. The roots
of the quadratic expression, or the values of x for which it evaluates
to 0, are given by a well-known formula.

A quadratic expression of the form

  ax^2 + bx + c = 0

has roots

  x = (-b + sqrt(b^2 - 4ac)) / 2a

and

  x = (-b - sqrt(b^2 - 4ac)) / 2a .

For more information on quadratic equations, consult the MathWorld
article devoted to this subject.

MathWorld: Quadratic Equation
http://mathworld.wolfram.com/QuadraticEquation.html

After finding the roots of the expression, we can determine by trial and
error or by graphing where they fall in relation to the solution set,
of which we initially know only that it uses the roots as endpoints. The
solution set may fall between the roots or beyond them.


#1. 4x^2 - 8x >= 0

This equation is already in quadratic form. Comparing the expression

  4x^2 - 8x

to the canonical

  ax^2 + bx + c ,

we see that

  a = 4
  b = -8
  c = 0 .

Plugging these into the formulas above, we obtain

  x = (8 + sqrt(64 - 0)) / 8
    = 16 / 8
    = 2

and

  x = (8 - sqrt(64)) / 8
    = 0 .

Since a quadratic equation has a parabolic shape, the solution set of the
inequality must lie either between or beyond the lines x = 0 and x = 2.
To determine which case applies here, we take an x value between 0 and 2,
such as 1, and see what it yields.

  4*1^2 - 8*1 = 4 - 8
              = -4
              < 0 .

Since x = 1 does not satisfy the inequality, we know that the solution
set falls beyond 0 and 2. Furthermore, because the quadratic expression
has 0 and 2 as its roots, the solution set must include 0 and 2. In
conclusion, the solution set is the union of (-inf, 0] and [2, +inf).


#2. 3x - x^2 > 0

We rearrange the left-hand side slightly to obtain

  -x^2 + 3x ,

from which it is evident that

  a = -1
  b = 3
  c = 0 .

These coefficients yield roots

  x = (-3 + sqrt(9 - 0)) / -2
    = (-3 + 3) / 2
    = 0

and

  x = (-3 - sqrt(9 - 0)) / -2
    = -6 / -2
    = 3 .

To see whether the solution set falls between or beyond these lines,
we test the intermediate value x = 2.

  3*2 - 2^2 = 6 - 4
            = 2
            > 0

This satisfies the inequality, so the solution set falls between 0 and
3. The interval is open at both ends, since the quadratic expression is
not greater than 0 for values x = 0 and x = 3. Hence, the solution set is
(0, 3).


#3. x^2 + 25 < 10x

We subtract 10x from each side of the inequality to obtain a quadratic
expression at left.

  x^2 - 10x + 25 < 0

Thus,

  a = 1
  b = -10
  c = 25 .

The roots are therefore

  x = (10 + sqrt(100 - 4*25)) / 2
    = 10 / 2
    = 5

and

  x = (10 - sqrt(100 - 4*25)) / 2
    = 10 / 2
    = 5

which are, of course, the same. We must now determine whether the solution
set falls beyond x = 5. For x = 4, we have

  4^2 - 10*4 + 25 = 16 - 40 + 25
                  = 1
                  > 0 ,

and for x = 6,
  
  6^2 - 10*6 + 25 = 36 - 60 + 25
                  = 1
                  > 0 .
  
Hence, the curve y = x^2 - 10x + 25 just touches the horizontal axis,
y = 0, where x = 5, and is otherwise restricted to the positive range. In
other words, x^2 - 10x + 25 < 0 is never satisfied. The solution set is
therefore the empty set, which you can denote with the symbol ?.

  
I know you want the solution set graphed, and I could do that easily,
except that I don't have a way to put the diagrams online at the moment.
Once I get home, I'll be able to use my digital camera to take snapshots
of the diagrams and upload them to a website. If you wait a few hours,
I'll use an Answer Clarification to post links to the pictures. Hang
tight! You're in good hands!
    
Regards,

leapinglizard

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Clarification of Answer by leapinglizard-ga on 27 Sep 2004 21:18 PDT

As promised, I've drawn and photographed the solution set for each case.


#1.

I drew a curve based on the following points.

  (-3, 60)
  (-2, 32)
  (-1, 12)
  (0, 0)
  (1, -4)
  (2, 0)
  (3, 12)
  (4, 32)
  (5, 60)

http://plg.uwaterloo.ca/~mlaszlo/answers/algebra.1.jpg


#2.

I plotted the following points for this curve.

  (-2, -10)
  (-1, -4)
  (0, 0)
  (1, 2)
  (2, 2)
  (3, 0)
  (4, -4)
  (5, -10)

http://plg.uwaterloo.ca/~mlaszlo/answers/algebra.2.jpg


#3.

This curve is based on the following points.

  (1, 16)
  (2, 9)
  (3, 4)
  (4, 1)
  (5, 0)
  (6, 1)
  (7, 4)
  (8, 9)
  (9, 16)

http://plg.uwaterloo.ca/~mlaszlo/answers/algebra.3.jpg


And that's the lot!

Regards,

leapinglizard

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Request for Answer Clarification by calliblue-ga on 03 Oct 2004 13:30 PDT

I needed them in sigh form, but I didnt clarify. Also not sure on your
technique to get anwser. Im going to put more problems up today. I
just need them more step by step broken down, other then that I like
your work.
                                          Thanx Calliblue

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Clarification of Answer by leapinglizard-ga on 03 Oct 2004 20:57 PDT

I see. Thank you.

leapinglizard

I want them more broken down step by step.