Quadratic Help

[Smokinoakum]

Re:

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.

When I plug in A=5, B=7, and C= -23 in my quadratic function on my calculatorI get...12953/8324, and -2.956102835 ???

 

[mmm4444bot]

You are correct.

I really messed up that post.

(I'm juggling several sheets of previously-used scratch paper, and my work on your behalf spans over the remaining white space on all of them. :lol: )

So, reverse the y-substitution, as you planned.

You should end up with my incorrect values for y posted above, as your two Real solutions for x.

I think!

Anyways, you'll know when you have the two Real solutions for x because you're going to CHECK YOUR CANDIDATES, yes ?

 

[Smokinoakum]

Re:

You are correct.

I really messed up that post.

(I'm juggling several sheets of previously-used scratch paper, and my work on your behalf spans over the remaining white space on all of them. :lol: )

So, reverse the y-substitution, as you planned.

You should end up with my incorrect values for y posted above, as your two Real solutions for x.

I think!

Anyways, you'll know when you have the two Real solutions for x because you're going to CHECK YOUR CANDIDATES, yes ?

I checked your answer, and yes, you are correct, but I'm still having trouble getting my equations to work out to the same answer...

 

[mmm4444bot]

:idea: If you simply state what you've done so far (versus saying something vague like "still need more help"), you'll be on your way sooner.

I cannot see what you did, so obviously I have no way to discern what types of "trouble" you're having.

Please show your work; somebody here will correct it.

wjm11 just stopped by with a friend, to talk me into running off to some Ben Affleck movie. Gotta go.

Cheers ~ Mark

 

[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.

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

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

substitute:

u = x[sup:2ukgd5rz]2/3[/sup:2ukgd5rz]

2u[sup:2ukgd5rz]2[/sup:2ukgd5rz] + 5u -13 = 0

u [sub:2ukgd5rz]1,2[/sub:2ukgd5rz] = [-5 ±?(25+104)]/4 = 1.589454173 , -4.089454173

Only the positive solution will provide real root.Assuming only real solutions are needed:

x = ±(1.589454173)[sup:2ukgd5rz]3/2[/sup:2ukgd5rz]

x = ± 2.00388141

2)

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

u = (3x-4)[sup:2ukgd5rz]3[/sup:2ukgd5rz]

5u[sup:2ukgd5rz]2[/sup:2ukgd5rz] + 7u - 23 =0

u [sub:2ukgd5rz]1,2[/sub:2ukgd5rz] = [-7 ±?(49+460)]/10 = 1.556102835, -2.956102835

In this case, both the answers will provide real solutions (those will also provide complex solutions - which we will ignore):

(3x-4)[sup:2ukgd5rz]3[/sup:2ukgd5rz] = 1.556102835

3x - 4 = 1.158811415

x = 1.719603805

and

(3x-4)[sup:2ukgd5rz]3[/sup:2ukgd5rz] = -2.956102835

3x - 4 = -1.435180459

x = 0.854939847

I'll leave the "checking" to you.

 

[Smokinoakum]

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

substitute:

u = x[sup:5lgxe5yn]2/3[/sup:5lgxe5yn]

2u[sup:5lgxe5yn]2[/sup:5lgxe5yn] + 5u -13 = 0

u [sub:5lgxe5yn]1,2[/sub:5lgxe5yn] = [-5 ±?(25+104)]/4 = 1.589454173 , -4.089454173

Only the positive solution will provide real root.Assuming only real solutions are needed:

x = ±(1.589454173)[sup:5lgxe5yn]3/2[/sup:5lgxe5yn]

x = ± 2.00388141

2)

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

u = (3x-4)[sup:5lgxe5yn]3[/sup:5lgxe5yn]

5u[sup:5lgxe5yn]2[/sup:5lgxe5yn] + 7u - 23 =0

u [sub:5lgxe5yn]1,2[/sub:5lgxe5yn] = [-7 ±?(49+460)]/10 = 1.556102835, -2.956102835

In this case, both the answers will provide real solutions (those will also provide complex solutions - which we will ignore):

(3x-4)[sup:5lgxe5yn]3[/sup:5lgxe5yn] = 1.556102835

3x - 4 = 1.158811415

x = 1.719603805

and

(3x-4)[sup:5lgxe5yn]3[/sup:5lgxe5yn] = -2.956102835

3x - 4 = -1.435180459

x = 0.854939847

I'll leave the "checking" to you.[/quote]

WOW! This is exactly what I was looking for. I completely understand now.

Thanks for all your help! I think I will be spending a lot of time here since my math class is online anyway. You guys rock!

Neil

 

[Deleted member 4993]

Unless your fundamental UNDERSTANDING of mathematics is very STRONG - do NOT take on-line math courses.

 

[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.

\(\displaystyle >>> \ \\)5x^2/3 + 2x^4/3-13=0 \(\displaystyle \ \ <<<\)

substitute:

u = x[sup:vud6k5eo]2/3[/sup:vud6k5eo]

Subhotosh Khan and Smokinoakum,

please stop passing along the error of what this equation is not equivalent to.

5x^2/3 + 2x^4/3 - 13 = 0 is equivalent to \(\displaystyle \frac{5x^2}{3} + \frac{2x^4}{3} - 13 = 0.\)

It is not equivalent to \(\displaystyle 5x^{\frac{2}{3}} + 2x^\frac{4}{3} - 13 = 0.\)

 

[Denis]

Subhotosh Khan and Smokinoakum,
please stop passing along the error of what this equation is not equivalent to.
5x^2/3 + 2x^4/3 - 13 = 0 is equivalent to \(\displaystyle \frac{5x^2}{3} + \frac{2x^4}{3} - 13 = 0.\)
It is not equivalent to \(\displaystyle 5x^{\frac{2}{3}} + 2x^\frac{4}{3} - 13 = 0.\)

Take it easy, LA; see 1st post:
> 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.
Even if improperly bracketed, it is clarified with "powers"...

To you, Smokes: 5x^2/3 and 5x^(2/3) are VERY different....HOKAY? :shock:

 

[lookagain]

Subhotosh Khan and Smokinoakum,
please stop passing along the error of what this equation is not equivalent to.
5x^2/3 + 2x^4/3 - 13 = 0 is equivalent to \(\displaystyle \frac{5x^2}{3} + \frac{2x^4}{3} - 13 = 0.\)
It is not equivalent to \(\displaystyle 5x^{\frac{2}{3}} + 2x^\frac{4}{3} - 13 = 0.\)

Take it easy, LA; see 1st post:
> 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.
Even if improperly bracketed, it is clarified with "powers"...

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

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

Take it easy, Denis.

It wasn't "clarified" with "powers" because a clarification would have meant explaining more
about something that was already typed properly. And in the above quote taken from a later
post, I corrected that. So instead of typing something in subsequent posts such as
"when you see/use this equation, these parts of it really mean that," you type out the correct
equation.

 

[BigGlenntheHeavy]

\(\displaystyle 2) \ Given: \ 5(3x-4)^6+7(3x-4)^3-23 \ = \ 0, \ solve \ for \ x.\)

\(\displaystyle Let \ u \ = \ (3x-4)^3, \ then \ u^2 \ = \ [(3x-4)^3]^2 \ = \ (3x-4)^6.\)

\(\displaystyle Hence, \ 5u^2+7u-23 \ = \ 0, \ \implies \ u \ = \ \frac{-7\pm\sqrt{509}}{10}.\)

\(\displaystyle Ergo, \ (3x-4)^3 \ = \ \frac{-7\pm\sqrt{509}}{10}.\)

\(\displaystyle Therefore, \ x \ = \ \frac{\bigg[\bigg(\frac{-7-\sqrt{509}}{10} \bigg)^{1/3}+4\bigg]}{3} \ or \ x \ = \ \frac{\bigg[\bigg(\frac{-7+\sqrt{509}}{10} \bigg)^{1/3}+4\bigg]}{3}\)

\(\displaystyle Note: \ It \ is \ best \ to \ leave \ the \ solutions \ in \ the \ above \ form \ as \ a \ decimal \ approximation\)

\(\displaystyle will \ be \ compounded \ by \ propagation \ errors \ when \ raised \ to \ the \ 6th \ power.\)

 

[Smokinoakum]

\(\displaystyle 2) \ Given: \ 5(3x-4)^6+7(3x-4)^3-23 \ = \ 0, \ solve \ for \ x.\)

\(\displaystyle Let \ u \ = \ (3x-4)^3, \ then \ u^2 \ = \ [(3x-4)^3]^2 \ = \ (3x-4)^6.\)

\(\displaystyle Hence, \ 5u^2+7u-23 \ = \ 0, \ \implies \ u \ = \ \frac{-7\pm\sqrt{509}}{10}.\)

\(\displaystyle Ergo, \ (3x-4)^3 \ = \ \frac{-7\pm\sqrt{509}}{10}.\)

\(\displaystyle Therefore, \ x \ = \ \frac{\bigg[\bigg(\frac{-7-\sqrt{509}}{10} \bigg)^{1/3}+4\bigg]}{3} \ or \ x \ = \ \frac{\bigg[\bigg(\frac{-7+\sqrt{509}}{10} \bigg)^{1/3}+4\bigg]}{3}\)

\(\displaystyle Note: \ It \ is \ best \ to \ leave \ the \ solutions \ in \ the \ above \ form \ as \ a \ decimal \ approximation\)

\(\displaystyle will \ be \ compounded \ by \ propagation \ errors \ when \ raised \ to \ the \ 6th \ power.\)

Yes, I agree, but my teacher wants the answer in decimal form. I wish I had one of you guys as an online teacher. I have gotten more out of this thread, than this whole semester working with him. I'm not sure if that's because of his skills, or yours?

Neil

 

[mmm4444bot]

Hi lizzygirl223:

Copy-and-paste the keywords below (in blue) into the search field at google.com

three points parabola solve system equations

If you cannot understand those lessons, then please use the [NEWTOPIC] button on the board's index page, here, to start your own thread, and ask specific questions about how to set up the equations or solve the system.

Cheers ~ Mark