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

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To find the zeros of the function g(x) = 1 - 2x^2, set g(x) to zero, resulting in the equation 2x^2 = 1, which simplifies to x^2 = 1/2. The solutions are x = ±√(2)/2, indicating that g(x) has two zeros, one positive and one negative. The quadratic formula is not necessary for this function, as it can be solved directly. The discussion also clarifies that when plotting the zeros on a graph, they represent points on the x-axis where the function intersects. Understanding the nature of the function as a parabola is crucial for interpreting its roots.
Phoresis
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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
Code:
[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?
 
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dont u just set g(x) = 0? if so you get x = \frac{\sqrt{2}}{2}
 
Phoresis said:
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
Code:
[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"?
 
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.
 
yup thanks that was a typo
 
for g(x) :

x= +/- root2/2
 
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}
 
oh i see. ok. thanks for your help guys. much appreciated.
 
how do you plot x = \frac{\sqrt{2}}{2} on a graph though?
 
  • #10
Phoresis said:
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}
 
  • #11
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...
 
  • #12
FarazAli said:
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.
 
  • #13
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.
 
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