Question

Evaluate: 3√8x^3 - 1 = 2x-1

Answer 1

Is that \[\sqrt[3]{8x^3} -1 = 2x -1\]
or
\[\sqrt[3]{8x^3-1} = 2x - 1\]

Answer 2

Ah, must be the latter.

Try cubing both sides...

Answer 3

it's actually \[4\sqrt{3-8x ^{2}} = 2x-1\]

Answer 4

sorry, i had the wrong equation!

Answer 5

Same approach :-)

Answer 6

i multiplied both sides by 4 and ended up with 3-8x^2 = 16x^4

Answer 7

how should i factor this to find x?

Answer 8

How did multiplying both sides by 4 get you that? And is that 4 * the square root, or the 4th root?

Answer 9

it's the 4th root

Answer 10

multiplying by 4 cancels out the 4th root

Answer 11

Okay, to write nth root, you type \ ( \sqrt[n]{stuff} \ )

Answer 12

\[\sqrt[4]{3-8x ^{2}} = 2x\]

Answer 13

then i got the 3-8x\[3-8x ^{2} = 16x^{4}\]

Answer 14

or should i have ended up with 2x instead of 16x^{4}

Answer 15

multiplying by 4 cancels the 4th root? huh?

Answer 16

yeah... o.o

Answer 17

okay nevermind... i'll still give you the "best response" medal though. thanks for your time

Answer 18

Sorry, having some browser problems. You didn't multiply by 4, you raised each side to the 4th power, big difference!

\[(\sqrt[4]{3-8x^2})^4 = (2x)^4\]\[3-8x^2=2^4x^4 = 16x^4\]\[16x^4+8x^2+3 = 0\]
Now how do we factor that?

Answer 19

yes! that's exactly where i'm stuck. i have no idea how to factor this kind of trinomial

Answer 20

by the way shouldn't it be \[16x ^{4} + 8x ^{2} - 3 = 0\], not +3?

Answer 21

Well, this looks like a quadratic, except instead of \(ax^2+bx+c = 0\) we've got \(ax^4 + bx^2 + c = 0\)
Let's try a substitution to make it look like a quadratic. We'll do \(u = x^2\)

\[16u^2 +8u -3 = 0\] (yes, good catch on the sign error!)
That we can solve by any of the usual methods for quadratics. Then we plug \(x^2\) in wherever we see \(u\) in the solution.

Answer 22

Yes, much nicer roots with the -3 vs. the +3 :-)

Answer 23

but how do i solve this once i've plug in x ^{2} ?

Answer 24

\[16x(x ^{2})^{2} + 8(x ^{2}) - 3 = 0\] ?? what do i do next?

Answer 25

i can't find any factors that add up to 8 and have product of -3

Answer 26

No, now you solve \[16u^2 + 8u - 3 = 0\]What are the solutions?

Answer 27

Use the quadratic formula

Answer 28

so this equation with the u's does not factor?

Answer 29

Oh, you can factor it.

To factor that, you multiply 16 and -3 = -48. Now find a pair of factors of -48 that add to 8. -4 and 12 would work.
\[(4u -1 )(4u+3 ) = 16u^2+12u-4u-3 = 0\]
Pair of real roots:\[4x^2 -1 = 0\]
Pair of complex roots:\[4x^2+3=0\]

Answer 30

I just didn't want to explain how to factor it :-)

Answer 31

haha! okay, well thank you so much for your help :)

Answer 32

Can you find the roots from the remaining two equations?

Answer 33

i think so.

Answer 34

I'll check your work if you want when you've got them.

Answer 35

i got 4x^2 +1, not 4x^2 - 1. also, how do i find x? i know that the correct answer is that x=1/2 (from the back of my math book) but i'm not sure why.

Answer 36

based on \[4x ^{2} +1 = 0\] ....how do i find x? i can't find the square root of \[-\frac{ 1 }{ 4 }\]

Answer 37

OH nevermind. you were right, it's negative 1. okay, i got 1/2 now. THANK YOU!!!

Answer 38

Well, it's both 1/2 and -1/2, because (-1/2)^2 = 1/4...

Answer 39

Now how about the (4x^2 +3) = 0? You've got a pair of roots there, too.

Answer 40

Hint: remember that i^2 = -1, so that can be written as (4x^2 - (-1)(3)) = 0 or 4x^2 - 3i^2 = 0 or 4x^2 = 3i^2

Answer 41

@shawnomnom whenever you square or cube or whatever both sides of the equation, you need to check your answers to make sure that they actually solve the original equation. often, you will get some solutions that do not work.