GoHabsGo's Thread of Algebra help

[gohabsgo]

I don't want to keep making threads when I come across some puzzling algebra concepts or problems, so I'll just use this thread as a reference point in the future when this may happen.

I'm having trouble understanding the following:
1/x³ = -1 The author of my text says to cross multiply and yield the answer: x³ = -1 What am I missing here? I thought by cross multiplying x³ became -x³. For example, another piece of this larger equation has me cross multiply 2/x³ = 3 which yields 3x³ = 2. What step have I forgotten here in the cross multiplication that isn't changing the x³ negative?

 

[Doc Al]

What does it matter? How is x³ = -1 any different from -x³ = 1?

 

[Mark44]

I don't want to keep making threads when I come across some puzzling algebra concepts or problems, so I'll just use this thread as a reference point in the future when this may happen.

I'm having trouble understanding the following:
1/x³ = -1 The author of my text says to cross multiply and yield the answer: x³ = -1 What am I missing here? I thought by cross multiplying x³ became -x³. For example, another piece of this larger equation has me cross multiply 2/x³ = 3 which yields 3x³ = 2. What step have I forgotten here in the cross multiplication that isn't changing the x³ negative?

What the author is doing here is multiplying both sides of the equation by x³. This yields a new equation, 1 = -x³, or equivalently, x³ = -1.

 

[gohabsgo]

What the author is doing here is multiplying both sides of the equation by x³. This yields a new equation, 1 = -x³, or equivalently, x³ = -1.

The embarrassing thing about this is that I'm not understanding how 1 = -x³ is equivalent to the x³ = -1. Because I can't use the first equation in the Multiplication Property of Zero. I eventually have to take the cubed root of -1, yielding -1. If I take the cubed root of 1, I get 1. To be clear on everything, I'm solving for 2x^-6 -x^-3 -3 = 0 if that helps. The two answers the author leaves in the book are x= the cubed root of 2/3 and x= -1. I believe I'm over-thinking everything.

 

[Mark44]

If you multiply both sides of an equation by a nonzero number, you get a new equation that is equivalent to the first equation. "Equivalent to" means that both equations have exactly the same solutions sets.

 

[eumyang]

The embarrassing thing about this is that I'm not understanding how 1 = -x³ is equivalent to the x³ = -1. Because I can't use the first equation in the Multiplication Property of Zero. I eventually have to take the cubed root of -1, yielding -1. If I take the cubed root of 1, I get 1. To be clear on everything, I'm solving for 2x^-6 -x^-3 -3 = 0 if that helps. The two answers the author leaves in the book are x= the cubed root of 2/3 and x= -1. I believe I'm over-thinking everything.

You multiply both sides of the equation by -1:
[tex]\begin{aligned}
1 &= -x^3 \\
(-1)(1) &= (-1)(-x^3) \\
-1 &= x^3
\end{aligned}[/tex]

 

[gohabsgo]

If you multiply both sides of an equation by a nonzero number, you get a new equation that is equivalent to the first equation. "Equivalent to" means that both equations have exactly the same solutions sets.

You multiply both sides of the equation by -1:
[tex]\begin{aligned}
1 &= -x^3 \\
(-1)(1) &= (-1)(-x^3) \\
-1 &= x^3
\end{aligned}[/tex]

Thanks Mark and Eumyang! I can't believe I forgot about that! I sometimes forget such small concepts when learning new ways to work equations.

 

[gohabsgo]

Ok guys and gals, I'm getting stumped, again, on the following small problem. Hopefully I've written everything out with clarity to explain why I'm confused.

y^2/3 - 8 = 0

y^2/3 = 8

(y^2/3)^3/2 = 8^3/2

I'm getting confused after these steps. My book gives the following:

y=(2³)^3/2 : Firstly, why was the cube root be taken from 8 resulting in 8³ with ^3/2 outside the brackets?

Anyways, from there I understand that it becomes 2^9/2.

However I become discombobulated when the following is given: 2⁴ X 2^1/2. What just happened here?

The result becomes y = 16squareroot2(i don't know how to make the squareroot brackets.) This I understand though. It's those two areas above that are stumping me, as I can't seem to ever recall doing other problems like this.

 

[Mark44]

Ok guys and gals, I'm getting stumped, again, on the following small problem. Hopefully I've written everything out with clarity to explain why I'm confused.

y^2/3 - 8 = 0

y^2/3 = 8

(y^2/3)^3/2 = 8^3/2

I'm getting confused after these steps. My book gives the following:

y=(2³)^3/2 : Firstly, why was the cube root be taken from 8 resulting in 8³ with ^3/2 outside the brackets?

Your question doesn't make sense. Starting with
y^2/3 = 8
they raised each side to the 3/2 power. This resulted in
(y^2/3)^3/2 = 8^3/2

Anyways, from there I understand that it becomes 2^9/2.

However I become discombobulated when the following is given: 2⁴ X 2^1/2. What just happened here?

2^9/2 = 2^{4 + 1/2} = 2⁴ * 2^1/2

The result becomes y = 16squareroot2(i don't know how to make the squareroot brackets.) This I understand though. It's those two areas above that are stumping me, as I can't seem to ever recall doing other problems like this.

There is something missing here, since there should be two solutions, unless they explicitly said that y > 0.

The other solution is y = -16[itex]\sqrt{2}[/itex]
You can click the square root expression to see how I did it.

As a check, I will square this expression, and then take the cube root.
[(-16[itex]\sqrt{2}[/itex])²]^1/3 = [256*2]^1/3 = 512^1/3 = 8
This shows that -16[itex]\sqrt{2}[/itex] is a solution of the equation y^3/2 = 8.

 

[gohabsgo]

Your question doesn't make sense. Starting with
y^2/3 = 8
they raised each side to the 3/2 power. This resulted in
(y^2/3)^3/2 = 8^3/2

2^9/2 = 2^{4 + 1/2} = 2⁴ * 2^1/2



There is something missing here, since there should be two solutions, unless they explicitly said that y > 0.

The other solution is y = -16[itex]\sqrt{2}[/itex]
You can click the square root expression to see how I did it.

As a check, I will square this expression, and then take the cube root.
[(-16[itex]\sqrt{2}[/itex])²]^1/3 = [256*2]^1/3 = 512^1/3 = 8
This shows that -16[itex]\sqrt{2}[/itex] is a solution of the equation y^3/2 = 8.

Mark, The bold problem is the problem from the beginning. It's a factor of a larger quadratic-like problem. The other factor I was able to easily solve for when set to 0, but this one is confusing me regarding the exponents. I don't know what to do, or why the author took those steps to achieve the end result.

Why did 8 result in (2³)^3/2 when both sides were powered by ^3/2? That's one thing confusing me. The other is how 2^9/2 became 2⁴ X 2^1/2

 

[BloodyFrozen]

Here's a little hint

y^2/3 - 8 = 0
y^2/3=8

Next, what you do to one side of the equation, you must do to the other side. To cancel the ^2/3, raise y to the ^3/2 power.

This will yield:

(y^2/3)^3/2= 8^3/2

y= 8^3/2

now.. if we use prime factorization,

y= (2³)^3/2

Which will...

y= 2^9/2

y= 2^4.5 if you prefer..

--->
sqrt(2^9)

sqrt(512)

Now do the rest...------> Simplify exact or appropriate. Whatever neccessary.

Atleast that's how I do it

 

[Mark44]

Why did 8 result in (2³)^3/2 when both sides were powered by ^3/2? That's one thing confusing me. The other is how 2^9/2 became 2⁴ X 2^1/2

8 did NOT result in (2³)^3/2. These two numbers are not equal.

What happened was that both sides were raised to the 3/2 power. After this, the right side was 8^3/2, and they rewrote 8 as 2³, so 8^3/2 = (2³)^3/2.

 

[gohabsgo]

8 did NOT result in (2³)^3/2. These two numbers are not equal.

What happened was that both sides were raised to the 3/2 power. After this, the right side was 8^3/2, and they rewrote 8 as 2³, so 8^3/2 = (2³)^3/2.

Thanks Mark! I realize now that prime factorization was involved in that step. But now I'm struggling to grasp the latter part. How 2^9/2 = 2⁴ X 2^1/2

EDIT: I do understand it now, as I totally missed what I was supposed to do. Simplifying squareroot of 512. I just don't get why the above notation was used for that step.

 

[BloodyFrozen]

Thanks Mark! I realize now that prime factorization was involved in that step. But now I'm struggling to grasp the latter part. How 2^9/2 = 2⁴ X 2^1/2

I just PMed you about it

 

[gohabsgo]

I just PMed you about it

Thanks again BloodyFrozen for helping me see the light!

 

[BloodyFrozen]

Sure, no problem