Math Is Fun Forum

Angela27

Guest

Determining the zeros...

Hi, I am stuck with finding zeros for this one... >_<;; It may seem easy, but then I am confused with the answers I looked from the back of the book...

x^(2)-8x+5

The answer is 4 + 11^(1/2) and 4 - 11^(1/2)

May I please have help step by step on this one?

~Angela

ahgua

Member

Registered: 2005-08-24

Posts: 25

Re: Determining the zeros...

Use the formula:

x = -b ± √(b² - 4 ac)
      --------------------- for ax² + bx + c
                 2a
a = 1, b = -8 and c = 5

x = 8 ± √(8² - 4 x 1 x 5)
      -------------------------
                 2(1)

x = 8 ± √(4 x 11)
      -----------------
               2

x = 8 ± 2 √(11)
      ---------------
               2

x = 4 ± √(11)

∴ x = 4 + √(11) or 4 - √(11)

       = 4 + 11^½ or 4 - 11^½

Last edited by ahgua (2005-08-26 13:45:21)

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Life is a passing dream, but the death that follows is eternal...

Angela

Guest

Re: Determining the zeros...

I'm sorry, but what formula was that? I'm not supposed to be knowing the answers before doing the problem... I am wondering how to determine the zeros by starting with the function itself; x^(2) - 8x + 5... Apparently my math skills are not top-notch... Thanks you so much, though...

Angela

Guest

Re: Determining the zeros...

Wow, lol, okay, never mind...

Use the formula:

x = -b ± √(b² - 4 ac)
      --------------------- for ax² + bx + c
                 2a
a = 1, b = -8 and c = 5

x = 8 ± √(8² - 4 x 1 x 5)
      -------------------------
                 2(1)

x = 8 ± √(4 x 11)
      -----------------
               2

x = 8 ± 2 √(11)
      ---------------
               2

x = 4 ± √(11)

∴ x = 4 + √(11) or 4 - √(11)

       = 4 + 11^½ or 4 - 11^½

Angela

Guest

Re: Determining the zeros...

That's awesome... >_<;;;

Angela

Guest

Re: Determining the zeros...

At least I'm learning more... My math skills are pretty stiff... Thanks.

mikau

Member

Registered: 2005-08-22

Posts: 1,504

Re: Determining the zeros...

quadratic formula roxxors t3h big onez!111

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A logarithm is just a misspelled algorithm.

ahgua

Member

Registered: 2005-08-24

Posts: 25

Re: Determining the zeros...

Another way is by the "completing the square".

  x² - 8x + 5 = 0

        x² - 8x = -5

x² - 8x + 4² = -5 + 4² [When LHS = x² - ax, adding (a/2)² to both sides enables it to be factorise to (x - b)² form]

       (x - 4)² = 11

           x - 4 = ± √11

                x = 4 ± √11

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Life is a passing dream, but the death that follows is eternal...

John E. Franklin

Guest

Re: Determining the zeros...

Or another way is to memorize that the center and bottom of an upward facing parabola is at location x=4 because it is one half and negative of the minus eight x.  Then substitute 4 for x and determine how far below the x-axis the bottom of the parabola is.  Hence y=16-32+5 or negative eleven.  Then since the parabola is normal width of one (one x squared term), then take the square root of the eleven height and get the width in either direction on the x-axis.   So the parabola crosses the x-axis at about 0.683 and 7.317.  If the parabola's x squared term was two, then the parabola would probably be skinnier since y gets twice as big.  So in that case, you'd probably take the eleven(distance bottom of parabola is from x-axis) in this example and divide it by two before square rooting it.   I just made this all up, so enjoy it.