Question

Given the equation −4Square root of x minus 3 = 12, solve for x and identify if it is an extraneous solution.

Answer 1

\[\sqrt[-4]{x-3}\]

Answer 2

ok, let's rewrite the question into

(x-3)^(-1/4) = 12

Since the exponent is negative flip the inside

(1/(x-3))^(1/4) = 12

Let's take out the (1/4) by putting the power of 4 on both sides

((1/(x-3))^(1/4))^4 = 12^4

The exponents cancel out on the left side

So, 1/(x-3) = 12^4

Send the x-3 to the other side and 12^4 to left side

1/12^4 = x-3

thus, x = (1/12^4)+3

The solution is not extraneous solution because there is only one value for x.

Answer 3

wait is x = to 0 or 12 @Darkmirror

Answer 4

neither. x = (1/12^4)+ 3

Answer 5

c x = 0, solution is not extraneous

x = 0, solution is extraneous

x = 12, solution is not extraneous

x = 12, solution is extraneous

Answer 6

@Darkmirror

Answer 7

Hmm... can you please recheck your equation? @lovelycharm

Answer 8

u right i forget to put =12 at the end sorry

Answer 9

@Darkmirror

Answer 10

\[\sqrt[-4]{x-3=12}\]

Answer 11

You are miswriting it.

it should be -4 * sqrt(x-3) = 12

so, it's

x = 12, solution is not extraneous

Answer 12

Srry, x=12 solution is extraneous because it doesn't equal to 12 when you plug back in

Answer 13

it fine thank u