# Thread: Algebraic manipulations and solutions

1. ## Algebraic manipulations and solutions

Hi,

Can someone help clarify why the solution to this expression does not result in both solutions equating to 2 when substituted in for x? I'm referring to when x = 4, since (4) + (4)^(1/2) does not equal 2. What is going on here?

- otownsend

2. ## Re: Algebraic manipulations and solutions

What expression are you talking about?

3. ## Re: Algebraic manipulations and solutions

Sorry, I forgot to include it. The expression is x + (x)^(1/2) = 4

4. ## Re: Algebraic manipulations and solutions

Originally Posted by otownsend
The expression is x + (x)^(1/2) = 4
That's an equation, not an expression...
It simplifies to x^2 - 9x + 16 = 0

5. ## Re: Algebraic manipulations and solutions

Originally Posted by DenisB
That's an equation, not an expression...
It simplifies to x^2 - 9x + 16 = 0

@otownsend, just so you can see how to obtain that simplified equation, here are the steps:

$x+x^{1/2} = 4$

Subtract x from both sides:

$x^{1/2} = 4-x$

Square both sides:

$x = (4-x)^2 = x^2-8x+16$

Subtract x from both sides:

$x^2-9x+16=0$

6. ## Re: Algebraic manipulations and solutions

my question is why is it that the solutions for the equation, which are x = 1 and x = 4, do not both work when substituting them into the original equation before algebraic manipulation? x = 1 works, but x = 4 does not work. I forget why this is the case and would appreciate someone to explain this.

Also I am referring to the equation x + (x)^(1/2) = 2. I mentioned x + (x)^1/2 = 4 in my previous post, but that was a mistake. Please take this into account.

7. ## Re: Algebraic manipulations and solutions

Originally Posted by otownsend
my question is why is it that the solutions for the equation, which are x = 1 and x = 4, do not both work when substituting them into the original equation before algebraic manipulation? x = 1 works, but x = 4 does not work. I forget why this is the case and would appreciate someone to explain this.

Also I am referring to the equation x + (x)^(1/2) = 2. I mentioned x + (x)^1/2 = 4 in my previous post, but that was a mistake. Please take this into account.
Then, $x^{1/2} = 2-x$

When we square both sides, we introduce another solution. When we square both sides, we obtain $x = (2-x)^2$. If we were to go backwards and take the square root of both sides, we could get $\pm \sqrt{x} = 2-x$. But, the original problem was only $\sqrt{x} = 2-x$. So, by squaring, we actually added another possible solution. You find the correct solution(s) by plugging the possible solutions into the original equation to determine which one(s) is/are correct.

8. ## Re: Algebraic manipulations and solutions

Gotcha! Thanks a lot