How do you solve and check for extraneous solutions in #2(x + 8) ^ (4/5) - 12 = 150#?

Answer 1

#x = 235 " "# or #" "x = -251#

Start by rewriting your equation using the radical form for the term that has the fractional exponent

#2root(5)( (x+8)^4) - 12 = 150#
Next, isolate the radical term on the left-hand side of the equation by adding #12# to both sides and dividing all the terms by #2#
#2root(5)( (x+8)^4) - color(red)(cancel(color(black)(12))) + color(red)(cancel(color(black)(12))) = 150 + 12#
#(color(red)(cancel(color(black)(2)))root(5)( (x+8)^4))/color(red)(cancel(color(black)(2))) = 162/2#
#root(5)( (x+8)^4) = 81#

This can be rewritten as

#(root(5)(x+8))^4 = 81#
Take the fourth root from both sides of the equation - do not forget that you have positive and negative roots for the fourth root of #81#!
#root(4)((root(5)(x+8))^4) = root(4)(81)#
#root(5)(x+8) = +-3#

Raise both sides of the equation to the fifth power

#(root(5)(x+8))^5 = (+-3)^5#

This equation will now produce two solutions

#x+8 = 3^5#
#x = 243 - 8 = color(green)(235)#

and

#x + 8 = (-3)^5#
#x = -243 - 8 = color(green)(-251)#

Your original equation will thus have two valid solutions and no extraneous solutions.

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Answer 2

To solve and check for extraneous solutions in the equation 2(x + 8)^(4/5) - 12 = 150, follow these steps:

Step 1: Add 12 to both sides of the equation: 2(x + 8)^(4/5) = 162

Step 2: Divide both sides of the equation by 2: (x + 8)^(4/5) = 81

Step 3: Raise both sides of the equation to the power of 5/4: [(x + 8)^(4/5)]^(5/4) = 81^(5/4)

Step 4: Simplify both sides of the equation: x + 8 = 3^5

Step 5: Subtract 8 from both sides of the equation: x = 243 - 8

Step 6: Simplify: x = 235

Step 7: Check the solution by substituting x = 235 back into the original equation: 2(235 + 8)^(4/5) - 12 = 150

The left side of the equation equals 150, which matches the right side. Therefore, the solution x = 235 is valid and not an extraneous solution.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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