Algebra I Demonstration Help GREATLY NEEDED!?

Question

Okay, I am in the 8th grade and have an A in my algebra I class right now and a 4.0 GPA. My Demo is due at the end of the month and I have to pass. There is one problem I cannot seem to figure out. I get to one point where the equation is

25+x^6=1

My parents are both college grads from great schools and they are having a hard time figuring it out as well. I mean, this is only 9th grade math!

The problem is:



7/x+ x^8x^6/4 = (x^4)^5/ 4x^6


the challenge is solve for x

IF ANYONE CAN GIVE ME A TIP ON HOW TO SOLVE THIS OR SOLVE FOR X AND EXPLAIN HOW YOU SOLVED IT I WOULD GREATLY APPRECIATE IT! Thank you!

i'm not kidding. this is how the equation is written! my friends and their parents are also having a hard time figuring this out and i'm not kidding when I say this is Algebra I. (However I seriously think it should be a higher level of math)

Sorry everyone I forgot one part to the equation! I feel so stupid now! Sorry! the part that says x^8x^6 is supposed to be

x^8x^6 -3

Again I am so sorry!

Answer 1

I'm wondering if you're original equation is written properly.

If you simplify, you get:

7/x + x^(8+6)/4 = x^(4*5)/4x^6

which is: 7/x+ x^14/4 = x^20/4x^6 or

7/x+x^14/4 = x^(20-6)/4 or

7/x+x^14/4 =x^14/4

Now, if you subtract x^14/4 from both sides, you are left with 7/x=0 which is meaningless (unless you say that in the limit, x is inifinity, but I don't think that's what you're going for)

So, I think that there is something wrong with the way the equation is written

Answer 2

You seemed to have given two problems

25 + x^6 = 1
x^6 = -24
This would mean that you have complex solutions.

OK the second part looks like
7/x + x^8x^6 - 3 = x^20 / (4x^6)
7/x + x^14 -3 = x^14 / 4
7/x + (3/4)x^14 - 3 = 0

Let's multiply throughout by 4x
28 + 3x^15 - 12x = 0
3*x^15 = 12x - 28

This requires a numerical solution.
If you have matlab, enter the following command:
fzero('3*x^15 - 12*x + 28',1)

You'll get x = -1.1930

Answer 3

7/x +(x^8 x^6) / 4 = (x^4)^5/ (4x^6)
Okay. If I've got the parentheses right, then:
Rewriting x^8 * x^6 as x^14, and (x^4)^5 as x^20, we get:
7/x + x^14/4 = x^20/ (4x^6)

On the right side, we divide out x^6 from numerator and denominator to get:
7/x + x^14/4 = x^14/4

Subtracting x^14/4 from each side, I get the (somewhat distasteful) answer:

7/x = 0

So, I guess x is positive infinity?

Answer 4

First simplify the exponential expressions to:

7/x + (x^14)/4 = (x^14)/4

Multiply by 4:

28/x + x^14 = x^14
28/x = 0

and that has no solution.

Was it perhaps that the 2nd term was x^8 • x^(6/4)? Then you'd have:

7/x + x^(9.5) = (x^14)/4
28 + 4x^(10.5) = x^15,

and solving that's not Alg 1.