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Phoresis
Oct24-04, 02:00 PM
Ok i have the following question:

Given the functions:

F(x) = 2x^2 + 3x - 2
G(x) = 1 - 2x^2

Find:

a) the zeros of f(x), g(x)

Now ive used the following formula

-b ±√b²-4ac²
2a

and worked out the zeros of f(x) fine, but I'm confused as to how to accomplish the same for g(x) when g(x) only has 2 elements. Any help?

FarazAli
Oct24-04, 02:11 PM
dont u just set g(x) = 0? if so you get x = \frac{\sqrt{2}}{2}

arildno
Oct24-04, 02:14 PM
Ok i have the following question:

Given the functions:

F(x) = 2x^2 + 3x - 2
G(x) = 1 - 2x^2

Find:

a) the zeros of f(x), g(x)

Now ive used the following formula

-b ±√b²-4ac²
2a

and worked out the zeros of f(x) fine, but I'm confused as to how to accomplish the same for g(x) when g(x) only has 2 elements. Any help?
Why do think g has "only two elements"?

Math Is Hard
Oct24-04, 02:15 PM
I hope I remember this correctly, but I believe you can express g(x) as
g(x) = -2x2+ 0x + 1

also, the c should not be squared in your quadratic formula.

Phoresis
Oct24-04, 02:21 PM
yup thanks that was a typo

roger
Oct24-04, 02:22 PM
for g(x) :

x= +/- root2/2

dav2008
Oct24-04, 02:24 PM
You don't need the quadratic formula for g(x)

If you want to find the zeros:

g(x)=1 - 2x2

0=1 - 2x2

2x2=1
x2=1/2

Then, like faraz said x = \frac{\sqrt{2}}{2}

but also

x = -\frac{\sqrt{2}}{2}

Phoresis
Oct24-04, 02:25 PM
oh i see. ok. thanks for your help guys. much appreciated.

Phoresis
Oct24-04, 02:28 PM
how do you plot x = \frac{\sqrt{2}}{2} on a graph though?

dav2008
Oct24-04, 02:35 PM
how do you plot x = \frac{\sqrt{2}}{2} on a graph though?
Why would you want to?

I guess if you really wanted to plot it an a x-y coordinate plane it would just be a vertical line at x = \frac{\sqrt{2}}{2}

FarazAli
Oct24-04, 02:35 PM
you don't need the negative in front of the fraction, because the result of the radical is automatically assumed to be plus and minus.

\sqrt{2} = \pm1.414...

dav2008
Oct24-04, 02:37 PM
you don't need the negative in front of the fraction, because the result of the radical is automatically assumed to be plus and minus.

\sqrt{2} = \pm1.414...
I believe the square root function (being a function and all) only outputs the positive root.

tink
Oct24-04, 02:42 PM
the function we are originally dealing with is not the square root function, it's g(x)=1 - 2x^2. Therefore, you have to include both + and - values of g(0). If you plot the graph, it is a parabola, so will have one, two, or no roots. In this case, it has two, one positive and one negative. Hence the +- of your square root.