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

Algebra ->  Polynomials-and-rational-expressions -> 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      Log On


   

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) About Me  (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) About Me  (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

ax%5E2%2Bbx%2Bc=0

the general solution using the quadratic equation is:

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29

So lets solve 2%2Ax%5E2-1=0 (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 2%2Ax%5E2%2B0%2Ax-1=0 notice a=2, b=0, and c=-1)

x+=+%28-0+%2B-+sqrt%28+%280%29%5E2-4%2A2%2A-1+%29%29%2F%282%2A2%29 Plug in a=2, b=0, and c=-1



x+=+%28-0+%2B-+sqrt%28+0-4%2A2%2A-1+%29%29%2F%282%2A2%29 Square 0 to get 0



x+=+%28-0+%2B-+sqrt%28+0%2B8+%29%29%2F%282%2A2%29 Multiply -4%2A-1%2A2 to get 8



x+=+%28-0+%2B-+sqrt%28+8+%29%29%2F%282%2A2%29 Combine like terms in the radicand (everything under the square root)



x+=+%28-0+%2B-+2%2Asqrt%282%29%29%2F%282%2A2%29 Simplify the square root



x+=+%28-0+%2B-+2%2Asqrt%282%29%29%2F4 Multiply 2 and 2 to get 4

So now the expression breaks down into two parts

x+=+%28-0+%2B+2%2Asqrt%282%29%29%2F4 or x+=+%28-0+-+2%2Asqrt%282%29%29%2F4


Which simplifies to

x+=+sqrt%282%29%2F2 or x+=+-+sqrt%282%29%2F2


So the roots approximate to

x=0.707106781186548 or x=-0.707106781186548


So our solutions are:
x=0.707106781186548 or x=-0.707106781186548

Notice when we graph 2%2Ax%5E2-1, we get:



when we use the root finder feature on a calculator, we find that x=0.707106781186548 and x=-0.707106781186548.So this verifies our answer


note:
Since 0%2B-sqrt%282%29%2F2=0%2B-sqrt%281%2F2%29 this means you were really close. You just need to make sure that the negative is placed on the outside of the square root.