Question

How to find t?

(4t + 3)^(-1/2) = -3

How do you do this if the exponent is a negative fraction?

Answer 1

a^(-1) = 1/a

Answer 2

Yes, I know that...

Answer 3

Or, just put both sides to the power of -2.

Answer 4

If you do that, is the left side left with 4t+3?

Answer 5

Yes, right will be (-3)^(-2)

Answer 6

You can take -3 as -1(3)^1/2

Answer 7

That's so confusing...

Answer 8

If i do what agent0smith said to do, is it correct if i get 4t+3 = -1/9 ?

Answer 9

^no, (-3)^2 = 9 not -9

Answer 10

Sorry
You can take -3 as -1(9)^1/2

Answer 11

Okay. I will solve from 4t+3 = 1/9 then! :)

Answer 12

@esl.0095
the right side is (-3)^(-2) = 1/9

Answer 13

Yes, I got that :) thank you :)

Answer 14

So final answer... t = -13/18. Anyone interested?

Answer 15

yep

Answer 16

Thanks agent0smith and all of you others. I will close the quesiton now! :D

Answer 17

The solution is not \(t = -\dfrac{13}{18} \)

Let's look at the first few steps of dealing with a negative exponent and a fractional exponent.

\(\large (4t + 3)^{-\frac{1}{2}} = -3\)

\(\large \left(\dfrac{1}{4t + 3} \right)^{\frac{1}{2}} = -3\)

\(\large \sqrt{\dfrac{1}{4t + 3}} = -3\)

Do you understand the steps so far?

Answer 18

Oh yeah, you may have to check your solution since we squared both sides

Answer 19

Notice that you have a square root equaling a negative number.
Therefore, there is no real solution.

Answer 20

Yep. Btw @mathstudent55 nothing you wrote shows up.

Answer 21

Correct.
When you raise both sides to the -2 power you are both squaring and getting the reciprocal. Any time you square both sides of an equation, you need to check the solution in the original equation because squaring both sides of an equation may introduce extraneous solutions.

Answer 22

@agent0smith
I see what I wrote in my previous responses, and LaTeX is showing properly on my screen.

Answer 23

What browser are you on? Doesn't work on chrome

Answer 24

Guys, I made a mistake prior to this, so I didn't need to solve for (4t + 3)^(-1/2) = -3 in the first place. Sorry and thanks to all you still for helping me!! We can close this question for good now :)