algebra2 extraneous solution

Question

2X-1=Square root of 8-X

Thank you.

Answer 1

When solving an equation with a square root in it, we know we will have to square the entire equation. Ideally, we do this once we get the radical on one side of the equal sign by itself. This is already how our problem is formatted, so we can begin by squaring both sides of the equation.

2x -1 = √(8 - x)

(2x - 1)^2 = ( √(8 - x ) )^2

When we square the left hand side, we can write that as two binomials multiplied together and then we can distribute to multiply. On the right side, the square root and the square will cancel since they are opposite operations of each other.

(2x - 1)(2x - 1) = 8 - x

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

Now that we've eliminated the radical, we can move all terms to the same side of the equation so that we can factor and solve.

4x^2 - 3x - 7 = 0

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

4x - 7 = 0 x + 1 = 0

4x = 7 x = -1

x = 7/4

We found two solutions: x = 7/4 and x = -1. To check for extraneous solutions, we need to substitute each of these solutions back into the original equation and see if they work.

Checking x = 7/4:

2(7/4) - 1 = √(8 - 7/4)

7/2 -1 = √(25/4)

5/2 = 5/2

Since both sides of the equation agree, x = 7/4 is a viable solution.

Checking x = -1:

2(-1) - 1 = √(8 - (-1))

-2 - 1 = √9

-3 = 3

Since the sides of our equation do NOT agree, x = -1 is an extraneous solution.

Our final answer is

x = 7/4

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 and plug those values back in, x = -1 ends up being extraneous because it doesn't logically hold for the original equation.

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 = sqr(8-x)

square both sides, potentially adding in an extraneous solution

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

4x^2 -3x -7=0

factor

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

set each factor =0, solve for x

x =-1 x=7/4

check them, plug into original equation to see if they work

2x-1 = sqr(8-x)

2(-1)-1= -3 = sqr(8+1) = sqr9 = 3 x=-1 is the extraneous solution

2(7/4)-1 = 5/2 = sqr(8-7/4) = sqr25/4= 5/2 It works. 1 and only 1 solution is x=7/4

Answer 4

Isolate the term containing the radical (this is already the case).

Since this is a square root square both sides to eliminate the radical.

4x² - 4x + 1 = 8 - x (this is the step that possibly introduces an extraneous solution).

Looks like a quadratic equation with a linear term so set equal to zero.

4x² - 3x - 7 = 0 (solve by factoring -- split the middle works)

4x² - 7x + 4x - 7 = 0

Factor by grouping:

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

(x+1)(4x - 7) = 0 (common binomial factor)

x + 1 = 0 OR 4x - 7 = 0 (zero product property)

x = -1 OR x = 7/4

Now check each possible solution in the original equation.

Check -1:

2(-1) - 1 =? √(8 - (-1))

-3 =? 3 (No)-this extraneous solution was introduced when we squared both sides -- (-3)² = (√9)²

Check 7/4:

2(7/4) - 1 =? √[(32/4) - (7/4)]

7/2 - 2/2 =? √25/4

5/2 =? 5/2 (Yes)

So the solution to the original equation is x = 7/4.

Answer 5

2X-1 = √(8-X)

(2X-1)² = 8-X

4X² - 4X +1 = 8-X

4X² - 3X - 7 = 0

Use quadratic formula

X = 1.75 or -1

You find the extraneous solution by plugging the values back into the original expression.

X = 1.75: 2(1.75)-1 = √(8-1.75) = 2.5

Since both sides equal each other, X = 1.75 is a viable solution.

X = -1: 2(-1)-1 = -3

√(8-(-1)) = 3

Since both sides do not equal one another, X = -1 is the extraneous solution.