SOLUTION: Hello, I am having diffuclties understanding this problem. Find all real and imaginary zeros for each polynomial factor. f(x) = 2x^2-1=0 Here is the work that I have starte

Question 88435: Hello, I am having diffuclties understanding this problem.
Find all real and imaginary zeros for each polynomial factor.
f(x) = 2x^2-1=0
Here is the work that I have started with but I am unsure if I am doing this properly
2x^2-1=0
+1 = +1
2x^2=1
------
2
x^2=1/2

would my answer be x= square root of +1/2 and square root of -1/2?
Thank you for your help
Found 2 solutions by stanbon, jim_thompson5910:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Find all real and imaginary zeros for each polynomial factor.
f(x) = 2x^2-1=0
2x^2-1=0
2x^2 = 1
x^2 = 1/2
x = sqrt(1/2) or x = -sqrt(1/2)
x = (sqrt2/20 or x = (-sqrt2)/2
===========
Cheers,
Stan H.

Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Now let's use the quadratic formula to solve for x:

Starting with the general quadratic

the general solution using the quadratic equation is:

So lets solve (note: since the polynomial does not have an "x" term, the 2nd coefficient is zero. In other words, b=0. So that means the polynomial really looks like notice , , and )

Plug in a=2, b=0, and c=-1

Square 0 to get 0

Multiply to get

Combine like terms in the radicand (everything under the square root)

Simplify the square root

Multiply 2 and 2 to get 4

So now the expression breaks down into two parts

or

Which simplifies to

or

So the roots approximate to

or

So our solutions are:
or

Notice when we graph , we get:

when we use the root finder feature on a calculator, we find that and .So this verifies our answer

note:
Since this means you were really close. You just need to make sure that the negative is placed on the outside of the square root.
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