Algebra2 extraneous solution

Question

2X-1= square root of 8-X

Thank you.

Answer 1

2x-1 = sqr(8-x)

square both sides to get

4x^2 -4x +1 = 8-x

4x^2 -3x -7 = 0

factor

(4x -7)(x+1) = 0

set each factor = 0 and solve for x

x = 7/4 or -1

since we multiplied by a term involving a variable, there is a possible extraneous solution which is not a solution to the original problem, so check each answer, plug them into the original problem and see if they work.

2x-1 = sqr(8-x)

2(7/4)-1 = 7/2-1= 5/2 = sqr(8-7/4) = sqr((32-7)/4) = sqr(25/4) = 5/2 This is a real solution to the original problem

2x-1 = sqr(8-x)

2(-1)-1 = sqr(8- -1)

-2-1 = sqr(9)

-3 = 3 this is the extraneous solution

sqr9= +3, not -3, as square root means the principal square root = the positive square root, unless expressly indicated otherwise

another minor detail: the original problem might have been intended to be

2x-1 = (sqr8) -x, as 2x-1 = sqr8-x is ambiguous, if so, then

3x -1 = sqr8

9x^2 -6x +1 = 8

9x^2 -6x -7 = 0

x = 6/18 + or - (1/18)sqr(36+36(7))

= 1/3 +/-(2/3)sqr2

again that's 2 answers, one is likely extraneous

again the extraneous answer is the negative one 1/3 -(2/3)sqr2 <0

Answer 2

Hello, thank you for posting your question!

An extraneous solution would mean one where it solves the underlying quadratic equation but doesn’t actually fit into the original equation.

So I would start on this one by squaring both sides to set up the quadratic! Then once you get the values for that you can go back and plug them back in.

It should end up being x = 7/4 and x = -1 for the two values that solve the quadratic ... then if you go back to the original function, x = -1 ends up being extraneous because it doesn't mathematically solve the first expression.

I hope that helps get you moving in the right direction! Feel free to reach out if you have any more questions beyond that :)

Answer 3

2x - 1 = √ (8 - x)

Square both sides to get 4x² - 4x + 1 = 8 - x

So, 4x² - 3x - 7 = 0

(4x - 7)(x + 1) = 0

So, x = 7/4 or -1

Check: if x = -1, then we have -3 = 3. So, -1 is extraneous.

if x = 7/4, then we get 5/2 = 5/2

The only solution is 7/4.

Answer 4

2x-1=√(8-x)

square both sides to remove radical

(2x-1)²=(√(8-x))²

4x²-4x+1=8-x

4x²-3x-7=0

Use the quadratic equation to get two answers for x. Only one is a valid solution.