# simplify expression

• Sep 2nd 2008, 05:19 PM
fishguts
simplify expression
I am confused over a few problems I solved/ tried to solve for a problem set.
My 1st question is:

Simplify x ^ -2 / x^-2+ y^-2

The simplified version I got was

x^2 + y^2 / x^2

I am not to sure if this is correct if anyone can verify it that would greatly help.

My next question is

Consider a cube with legnth of the side equal to s, then write a function of s whose value is the volume of a cube.

I believe the correct answer is

f(x) = s^3

but I am also not to sure about the answer to this one.

My last question is if I am solving the equation for a circle and I have

4x^2 + 16x + 64 + 4y^2 - 8y + 16 = Root 78

I cannot factor out the terms for x and y to end up with the normal circle equations. Can I use the quadratic formula to solve for the center and radius?

Thanks for any help anyone may provide.
• Sep 2nd 2008, 05:35 PM
i dont feel like checkin ur algebra on #1, #2 looks good tho,

also #3 youre going to want to look into "completing the square"

hope this helps
• Sep 2nd 2008, 05:59 PM
fishguts
Quote:

also #3 youre going to want to look into "completing the square"

hope this helps

Hi for number 3 that is what i calculated after completing the square. The original formula was

4x^2 + 4y^2 - 8y + 16x + 2 = 0

I just cannot factor 4x^2 + 16x + 64 + 4y^2 - 8y + 16 = Root 78 to get the center.
• Sep 2nd 2008, 06:08 PM
ticbol
Quote:

Originally Posted by fishguts
I am confused over a few problems I solved/ tried to solve for a problem set.
My 1st question is:

Simplify x ^ -2 / x^-2+ y^-2

The simplified version I got was

x^2 + y^2 / x^2

I am not to sure if this is correct if anyone can verify it that would greatly help.

My next question is

Consider a cube with legnth of the side equal to s, then write a function of s whose value is the volume of a cube.

I believe the correct answer is

f(x) = s^3

but I am also not to sure about the answer to this one.

My last question is if I am solving the equation for a circle and I have

4x^2 + 16x + 64 + 4y^2 - 8y + 16 = Root 78

I cannot factor out the terms for x and y to end up with the normal circle equations. Can I use the quadratic formula to solve for the center and radius?

Thanks for any help anyone may provide.

Too many.
But since you tried to solve them, let me give a comment.

Simplify x ^ -2 / x^-2+ y^-2
= [1 / x^2] / [(1 / x^2) + (1 / y^2)]
Combine the two fractions of the RHS into one,
= [1 / x^2] / [(y^2 + x^2) /(x^2 y^2)]
= [1 / x^2] * [(x^2 y^2) / (x^2 +y^2)]
= [x^2 y^2] / [(x^2)(x^2 +y^2)]
= y^2 / (x^2 +y^2) --------------------answer.

---------------------------
Consider a cube with legnth of the side equal to s, then write a function of s whose value is the volume of a cube.

function of s is f(s). Not f(x).

V = s^3

So, V = f(s) = s^3 -------answer.

-----------------------
4x^2 + 16x + 64 + 4y^2 - 8y + 16 = Root 78

Are you sure the RHS is sqrt(78)?

4x^2 + 16x + 64 + 4y^2 - 8y + 16 = Root 78
4[x^2 +4x +16] +4[y^2 -2y +4] = sqrt(78)

Umm, I see what you mean. The quadratic inside the brackets are notperfect squares.
No, do not use the quadratuc formula because you will get two roots for each of the quadratics. See first if you can manipulate to get perfect squares in those quadratics. If not, then ue "completing the square" in each of them instead.

To continue,
Divide both sides by 4,
[x^2 +4x +16] +[y^2 -2y +4] = (1/4)sqrt(78)
[x^2 +4x +4 +12] +[y^2 -2y + 1 +3] = (1/4)sqrt(78)
[x^2 +4x +4] +[y^2 -2y +1] = (1/4)sqrt(78) -12 -3
[(x +2)^2] +[(y -1)^2] = (1/4)sqrt(78) -15 -----------the equation of the circle.

Its center is at (-2,1) and its radius is sqrt[(1/4)sqrt(78) -15].
• Sep 2nd 2008, 06:17 PM
ticbol
Quote:

Originally Posted by fishguts
Hi for number 3 that is what i calculated after completing the square. The original formula was

4x^2 + 4y^2 - 8y + 16x + 2 = 0

I just cannot factor 4x^2 + 16x + 64 + 4y^2 - 8y + 16 = Root 78 to get the center.

Oh, the original equation was 4x^2 +4y^2 -8y +16x +2 = 0?

Let us do that then.

[4x^2 +16x] +[4y^2 -8y] = -2
4[x^2 +4x] +4[y^2 -2y] = -2
[x^2 +4x] +[y^2 -2y] = -1/2
[x^2 +4x +4] +[y^2 -2y +1] = -1/2 +4 +1
(x +2)^2 +(y -1)^2 = 9/2 -------------the circle.

Its center is at (-2,1) and its radius is sqrt(9/2) = (1.5)sqrt(2).
• Sep 2nd 2008, 06:20 PM
fishguts
Quote:

Originally Posted by ticbol
Oh, the original equation was 4x^2 +4y^2 -8y +16x +2 = 0?

Let us do that then.

[4x^2 +16x] +[4y^2 -8y] = -2
4[x^2 +4x] +4[y^2 -2y] = -2
[x^2 +4x] +[y^2 -2y] = -1/2
[x^2 +4x +4] +[y^2 -2y +1] = -1/2 +4 +1
(x +2)^2 +(y -1)^2 = 9/2 -------------the circle.

Its center is at (-2,1) and its radius is sqrt(9/2) = (1.5)sqrt(2).

Thanks man, I appreciate it