Quadratic Help

[Smokinoakum]

Hi...this is my first post so forgive me if I break edicate. I'm having problems with a couple of homework questions.

Here they are...

1. 5x^2/3 + 2x^4/3-13=0 With this one it's the 2/3, and 4/3 powers that are thowing me off.

2. 5(3x-4)^6 + 7(3x-4)^3 -23=0 I don't even know where to start.

Thanks for any help you guys can offer.

Neil

 

[Deleted member 4993]

Hi...this is my first post so forgive me if I break edicate. I'm having problems with a couple of homework questions.

Here they are...

1. 5x^2/3 + 2x^4/3-13=0 With this one it's the 2/3, and 4/3 powers that are thowing me off.

substitute

u = x [sup:3jbg0w1r]2/3[/sup:3jbg0w1r] ? u[sup:3jbg0w1r]2[/sup:3jbg0w1r] = x [sup:3jbg0w1r]4/3[/sup:3jbg0w1r]

2. 5(3x-4)^6 + 7(3x-4)^3 -23=0 I don't even know where to start.

substitute

u = (3x -4) [sup:3jbg0w1r]3[/sup:3jbg0w1r] ? u[sup:3jbg0w1r]2[/sup:3jbg0w1r] = (3x -4) [sup:3jbg0w1r]6[/sup:3jbg0w1r]

Thanks for any help you guys can offer.

Neil

 

[Smokinoakum]

Yes, I got that far...What I'm confused about is what to do with the solution afterwards. Which power do I use to 'unfold' the answer?

 

[Smokinoakum]

Yes, I got that far...What I'm confused about is what to do with the solution afterwards. Which power do I use to 'unfold' the answer?

Do I maybe use the square root of the answers?

 

[Smokinoakum]

Anybody?

 

[Deleted member 4993]

Yes, I got that far...What I'm confused about is what to do with the solution afterwards. Which power do I use to 'unfold' the answer?

Do I maybe use the square root of the answers?

No ....

say, for problem 1, you got

u = 9

then

x[sup:137fudd2]2/3[/sup:137fudd2] = 9

x = 9[sup:137fudd2]3/2[/sup:137fudd2] = ± 27

 

[Smokinoakum]

Yes, I got that far...What I'm confused about is what to do with the solution afterwards. Which power do I use to 'unfold' the answer?

Do I maybe use the square root of the answers?

No ....

say, for problem 1, you got

u = 9

then

x[sup:1t5uqywp]2/3[/sup:1t5uqywp] = 9

x = 9[sup:1t5uqywp]3/2[/sup:1t5uqywp] = ± 27

Why did you change the 2/3 to 3/2?

 

[Smokinoakum]

Also for the problem 5x[sup:2l3nekmj]2/3[/sup:2l3nekmj]+2x[sup:2l3nekmj]4/3[/sup:2l3nekmj]-13=0 shouldn't I change the equation around so it reads 2x[sup:2l3nekmj]4/3[/sup:2l3nekmj]+5x[sup:2l3nekmj]2/3[/sup:2l3nekmj]-13=0 ? This would arrange the powers in the correct form for the quadratic.

 

[mmm4444bot]

Yes, I got that far Why did you not say so, initially ?

:idea: The more information that you withhold from tutors, the more personal time you waste.

.

Why did you change the 2/3 to 3/2?

x^(2/3) = 9

To solve this equation for x, we raise both sides to the power of 3/2.

[x^(2/3)]^(3/2) = 9^(3/2)

This step does NOT "change the 2/3 to 3/2".

It changes the exponent on x^(2/3) to 1.

x^1 = 9^(3/2)

.

shouldn't I change the equation around so it reads 2x[sup:sbi85plg]4/3[/sup:sbi85plg]+5x[sup:sbi85plg]2/3[/sup:sbi85plg]-13=0

That is not necessary, but, if doing so helps clarify things for you, then you may write any terms in any sums in any order you like. The property that allows this is called "The Commutative Property of Addition".

Cheers ~ Mark

 

[Smokinoakum]

Re:

Yes, I got that far Why did you not say so, initially ?

:idea: The more information that you withhold from tutors, the more personal time you waste.

Sorry about that! As I said, this is my first post...I really appreciate this help.

.

Why did you change the 2/3 to 3/2?

x^(2/3) = 9

To solve this equation for x, we raise both sides to the power of 3/2.

[x^(2/3)]^(3/2) = 9^(3/2)

This step does NOT "change the 2/3 to 3/2".

It changes the exponent on x^(2/3) to 1.

x^1 = 9^(3/2)

.

shouldn't I change the equation around so it reads 2x[sup:11z1ktli]4/3[/sup:11z1ktli]+5x[sup:11z1ktli]2/3[/sup:11z1ktli]-13=0

That is not necessary, but, if doing so helps clarify things for you, then you may write any terms in any sums in any order you like. The property that allows this is called "The Commutative Property of Addition".

Cheers ~ Mark

It did help...

I came out with this for the first equation x= 2.00388141 and -8.269857357

Now I'm just stuck on the second question.

5(3x-4)^6 +7(3x-4)^3 -23=0

What I have done so far is substitute. y= (3x-4)^2 giving me... 5y^2 + 7y -23=0

I came out with y= 12953/8324 and -2.56102835

Now I'm a bit confused as what to do with the (y) answer to find (x)

 

[mmm4444bot]

Sorry about that!

Neil, in life there is generally no need to apologize to people offering constructive critisism. You can take it or leave it.

I mean, it's a free world, brother. You can waste as much of your time as you like.



I came out with this for the first equation x= 2.00388141 ? YES (as an approximation)

and -8.269857357 ? NO (not at all)

---------------------------------------------------------------------

just stuck on the second question

What I have done so far is substitute. y = (3x-4)^2 This substitution is not correct.

In the first exercise, your negative approximation for x is not a solution. In fact, I don't think that the second result is even a Real number.

I think that it's supposed to be: -8.269857357 i

After all, you obtained it by taking the square root of a negative number cubed. We cannot take the square root of a negative value, in the Real number system.

My main point is, you would have discovered that the first exercise has only one Real solution, if you had checked your candidates for x.

On the second exercise, your expression for y is not correct. Look at the two given exponents; they are 6 and 3, not 4 and 2.

You're on the right track. Keep trying!

Cheers ~ Mark

PS: Did they instruct you to report all solutions as decimal approximations ?

Did they specifically ask you to find all Real solutions ?



MY EDITS: Changed convoluted sentence structures, and then came back to ask whether or not Complex solutions are also expected.

 

[lookagain]

1. 5x^2/3 + 2x^4/3-13=0 With this one it's the 2/3, and 4/3 powers that are thowing me off.

Smokinoakum,

you must have grouping symbols around the exponents if you type out those expressions horizontally, such as

5^(2/3) + 2x^(4/3) - 13 = 0

------------------------------------------------------------------------------------------------------------------------------------

say, for problem 1, you got

u = 9

then

x[sup:2bcanbjd]2/3[/sup:2bcanbjd] = 9

x = 9[sup:2bcanbjd]3/2[/sup:2bcanbjd] = ± 27

In this hypothetical example, this last line must be

x = ± 9[sup:2bcanbjd]3/2[/sup:2bcanbjd] = ± 27


as \(\displaystyle x^{\frac{2}{3}} = 9 \longrightarrow\)

\(\displaystyle x^2 = 9^3 \longrightarrow\)

\(\displaystyle x = \pm 9^{\frac{3}{2}} = \pm 27\)


Then there are two real solutions, which when rounded to three decimal places, are

\(\displaystyle x = -2.004\) or \(\displaystyle x = 2.004\)

 

[Smokinoakum]

Re:

Sorry about that!

Neil, in life there is generally no need to apologize to people offering constructive critisism. You can take it or leave it.

I mean, it's a free world, brother. You can waste as much of your time as you like.



I came out with this for the first equation x= 2.00388141 ? YES (as an approximation)

and -8.269857357 ? NO (not at all)

---------------------------------------------------------------------

just stuck on the second question

What I have done so far is substitute. y = (3x-4)^2 This substitution is not correct.

In the first exercise, your negative approximation for x is not a solution. In fact, I don't think that the second result is even a Real number.

I think that it's supposed to be: -8.269857357 i

After all, you obtained it by taking the square root of a negative number cubed. We cannot take the square root of a negative value, in the Real number system.

My main point is, you would have discovered that the first exercise has only one Real solution, if you had checked your candidates for x.

WOW! You're right. I pluged 2.00388141 into the original equation, and guess what? It came out to 0

On the second exercise, your expression for y is not correct. Look at the two given exponents; they are 6 and 3, not 4 and 2.

Ooops! That was a typo

You're on the right track. Keep trying!

Cheers ~ Mark

PS: Did they instruct you to report all solutions as decimal approximations ?

Did they specifically ask you to find all Real solutions ?

Yes, and this is an online class with a teacher who does not explain a lot, which is why I am here.


MY EDITS: Changed convoluted sentence structures, and then came back to ask whether or not Complex solutions are also expected.

So back to my 2nd problem...

y= (3x-4)^3 = 5y^2 + 7y - 23 = 0
my solutions for my transformed equations are... 12953/8324, and -2.956102835
Now I'm lost as to what to do next.
I'm new at all this, and feel as if I was just thrown in the deap end!

I reaspect all you math wizards out there!

Neil

 

[mmm4444bot]

an online class with a teacher who does not explain a lot

A common lament, when working with machine-teachers.

Do you know ? When taking a class on-line, YOU often end up being the teacher. I mean, you have to teach yourself.

The equation in the first exercise is quadratic "in form".

The substitution shows us this because it leads to a quadratic equation: 2u^2 + 5u - 13 = 0.

The equation in the second exercise is also quadratic "in form".

A carefully-chosen substitution for x will also yield a 2nd-degree polynomial equation, just like it did in the first exercise.

EG:

7x^6 - 3x^3 + 44 = 0

Let y = x^3

Then y^2 = x^6

Substituting leads to 7y^2 - 3y + 44 = 0.

Can you try this type of substitution with (3x - 4)^6 and (3x - 4)^3 versus x^6 and x^3 ?

Finishing is then the same process as it was in the first exercise.

 

[Smokinoakum]

1. 5x^2/3 + 2x^4/3-13=0 With this one it's the 2/3, and 4/3 powers that are thowing me off.

Smokinoakum,

you must have grouping symbols around the exponents if you type out those expressions horizontally, such as

5^(2/3) + 2x^(4/3) - 13 = 0

------------------------------------------------------------------------------------------------------------------------------------

say, for problem 1, you got

u = 9

then

x[sup:8v0esv2k]2/3[/sup:8v0esv2k] = 9

x = 9[sup:8v0esv2k]3/2[/sup:8v0esv2k] = ± 27

In this hypothetical example, this last line must be

x = ± 9[sup:8v0esv2k]3/2[/sup:8v0esv2k] = ± 27


as \(\displaystyle x^{\frac{2}{3}} = 9 \longrightarrow\)

\(\displaystyle x^2 = 9^3 \longrightarrow\)

\(\displaystyle x = \pm 9^{\frac{3}{2}} = \pm 27\)


Then there are two real solutions, which when rounded to three decimal places, are

\(\displaystyle x = -2.004\) or \(\displaystyle x = 2.004\)

x= -2.004 does not work when I plug it in the original equation? Any idea why?
Neil

 

[mmm4444bot]

x= -2.004 does not work when I plug it in the original equation? Any idea why?

Heh, heh, heh, heh, heh.

x = 9^(3/2) is NOT a part of your exercise.

That equation resulted from an EXAMPLE provided by Subhotosh.

What I told you before is true; the first exercise has only ONE Real solution. You already found it. You're done with the first exercise.

 

[Smokinoakum]

Re:

x= -2.004 does not work when I plug it in the original equation? Any idea why?

Heh, heh, heh, heh, heh.

x = 9^(3/2) is NOT a part of your exercise.

That equation resulted from an EXAMPLE provided by Subhotosh.

That's funny because 2.00388141 was the answer to my first equation.

Gotta love irony!
Neil

 

[mmm4444bot]

2.00388141 was the answer It still is, yes ? :?



Gotta love irony Hey !

We all have to let irony bite us in the arse (there's really no avoiding it), but that does not mean that we have to love her, too.

Otherwise, the moral is, "Watch out for devious examples, especially examples coming from Subhotosh." :twisted:

What quadratic equation in y did you come up with for the second exercise ?

 

[Smokinoakum]

Re:

2.00388141 was the answer It still is, yes ? :?



Gotta love irony Hey !

We all have to let irony bite us in the arse (there's really no avoiding it), but that does not mean that we have to love her, too.

What quadratic equation in y did you come up with for the second exercise ?

For the second equation...

5(3x-4)^6 + 7(3x-4)^3 -23 = 0

I substituted this...

y = (3x-4)^3
Thus: 5y^2 + 7y -23 = 0
The answers I got were... 12953/8324, and -2.956102835

I think I should now set up 2 different equations like this...

(3x-4)^3 = 12953/8324

(3x-4)^3 = -2.956102835

Am I on the right track?

Neil

 

[mmm4444bot]

Thus: 5y^2 + 7y -23 = 0

The answers I got were... 12953/8324, and -2.956102835

Ooh. I get the approximations y = 1.719603805 or y = 0.8549398467

We had both better check our work.



I think I should now set up 2 different equations like this ...

Yes, that's correct. You need to "reverse" your substitution and go from the decimal versions of y back to the corresponding values of x.