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John E. Franklin
2005-08-30 07:44:11

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.

ahgua
2005-08-27 13:54:31

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

mikau
2005-08-27 13:30:42

quadratic formula roxxors t3h big onez!111

Angela
2005-08-27 12:45:20

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

kylekatarn
2005-08-27 12:31:03

I think there is a simple way to do this
you can check if the polinomial can be reduced to the form (x-a)(x-b)

lets find the reduced form

(x-a)(x-b)=x²-8x+5
x²-(a+b)x+ab=x²-8x+5
-(a+b)x+ab=-8x+5

now solve the system
a+b=8
ab=5
wich gives us two (a;b) pairs. One one of them is a solution of P(x)=0. By substituition we can find the rigth one.
I find this method very fast and easy but this isn't always that simple. Some quadratics can't be reduced to the (x-a)(x-b) form.

Just use the quadratic formula solver because it works on all cases. I posted this just for fun:)

Angela
2005-08-27 11:53:23

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

Angela
2005-08-27 11:52:26

Wow, lol, okay, never mind...

#### ahgua wrote:

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
2005-08-27 11:51:10

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...

ahgua
2005-08-27 11:35:02

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

Angela27
2005-08-27 11:25:42

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