# When is negative square valid when checking a graph?

The problem was to solve for x in

$$x = (1/2)\sqrt{1 -x} + 1.$$

I came up with possible solutions: $x = 3/4$; and $x = 1$.

Checking my answers, $x = 1$ checks out; it is a valid solution. The answer key agrees.

Checking my answers, $x = 3/4$ does not check out, or does it?

$${3\over 4} = {1\over 2} \sqrt{1/4} + 1.$$

So, if we take the positive square root, $1/2$, then clearly the answer is wrong.

Then I went down the path of over-thinking the problem; what about the negative square root?!

So, if we take the negative square root, $-1/2$, then the solution checks out. But it's not. The graph clearly shows that $3/4$ is not a solution.

When is it proper to take into account the negative square root?

• Marty, I reformatted your post with latex. You should learn how to use this; there is a nice explanation you can find in the meta area. – ncmathsadist Dec 29 '13 at 19:44
• @ncmathsadist, Marty: Is the original equation $x = \frac{\sqrt{1-x}}{2} + 1$ It's how I read the question before it was edited but I could be wrong. Please confirm. – Warren Hill Dec 29 '13 at 20:32
• Yes, I fixed the typo. – Marty B. Dec 29 '13 at 23:12

If someone simply uses the square root symbol, they mean the positive one. Your second answer is a solution to the quadratic equation that you get by rearranging and squaring, but not an answer to the original equation.

• Also, if you square both sides of an equation, you are destroying information about signs. This can cause extraneous roots. – ncmathsadist Dec 29 '13 at 19:46

When you square an equation, it's possible to create new solutions. Therefore, when squaring, you always have to check your answers.
Example
Suppose we want to solve $x=\sqrt{1}$. Squaring both sides gives $x^2=1$, so $x=\pm 1$. But we know that (by definition) $\sqrt{1}$ is nonnegative, so only $x=1$ is a solution to the original equation. (Of course, in this example, you can just calculate $\sqrt{1}$, but in your equation, you can't.)

One important note: There is no such thing as "negative" square root.Square root is always taken as positive.If we want to indicate a negative sign also ,then we put $\pm$ before it.

While solving equations, we tend to square them which leads to extra roots which may not satisfy the original equation. Thus we must always check the roots.

It is nothing absurd that you are getting.The only thing is that you have tot check them.

Thank you for the responses. Yes, it's true, when playing with equations, there are cases where you might introduce extraneous roots. Hence, it's important to check the solutions.

But that's not the issue here. I think the respondents gave me to much credit; the actual problem here is far more mundane and comical.

The concern of this question is the square root in the equation, while checking the possible result, 3/4.

$$3/4 = (1/2)\sqrt{1 - 3/4} + 1$$

which is

$$3/4 = (1/2)\sqrt{1/4} + 1$$

The best head-in-hand provoking answer is the one that tells me that I was making up square root signs that didn't exist.

When you have

$$x$$

and in the course of deriving a solution you decide to take a square root, then you must account for possible negative square and thus prepend $\pm$ like

$$\pm \sqrt{x}$$

However, when the equation already comes equipped with its own square root straight from the factory floor, as in

$$3/4 = (1/2)\sqrt{1/4} + 1$$

You don't get to decorate the equation with your own arbitrary $\pm$ symbols, and making up solutions like I did.

This is wrong:

$$3/4 = (1/2)\pm\sqrt{1/4} + 1$$

Hence, the only correct next step was

$$3/4 = (1/2)^2 + 1$$

which proves that $3/4$ is not a valid answer.