Express g(x) as g(x) = -2x2+ 0x + 1

[Phoresis]

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 I've used the following formula

[U]-b ±√b²-4ac²[/U]
        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]

dont u just set g(x) = 0? if so you get $$x = \frac{\sqrt{2}}{2}$$

 

[arildno]

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 I've used the following formula

[U]-b ±√b²-4ac²[/U]
        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]

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]

yup thanks that was a typo

 

[roger]

for g(x) :

x= +/- root2/2

 

[dav2008]

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]

oh i see. ok. thanks for your help guys. much appreciated.

 

[Phoresis]

how do you plot $$x = \frac{\sqrt{2}}{2}$$ on a graph though?

 

[dav2008]

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]

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]

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]

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.