why is there no X intercept for y = sqtr(x-2) + 6???

[jtingato]

I have graphed the equation

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f(x) = sqrt(x - 2) + 6


and have come up with the domain x=> 2. I also have the pairs (2,6)(3,7)(6,8)(11,9(18,10). Graphing this makes it obvious that there is not an x intercept.

But if I take the equation and set the numerator to zero and solve...

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 0 = sqrt(x - 2) + 6


, I should get the x intercept, which I do get x = 34. I know this is impossible, but why do I get this answer.

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0 = sqrt(x - 2) + 6
-6 = sqrt(x - 2)
-6^2] = (  sqrt(x - 2)  ) ^2
36 = x - 2
34 = x


Anyone???

[stapel_eliz]

...But if I take the equation and set the numerator to zero and solve...

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 0 = sqrt(x - 2) + 6


, I should get the x intercept....

How? Since when can a negative six be equal to a positive square root of anything?

(This is why they always tell you to check your solutions when you've squared both sides to get that solution!)

[jtingato]

Yeah, I guess I am having a hard time with that concept.

I guess I thought that the square root of any positive number was positive and negative , as in ... x = +/- ( sqrt(y) ).
Kinda like in the quadratic equation. So sqrt(9) = ( 3, -3).

Like I said, I am sure I am confusing things.

[stapel_eliz]

I guess I thought that the square root of any positive number was positive and negative , as in ... x = +/- ( sqrt(y) ).

Solving an equation and simplifying a square root are two very different things.

When solving, you are finding all possible values which could work. When simplifying, you need to simplify to one value; you can't start with one expression (for instance, the square root of nine) and end up with two values (for instance, positive and negative three).

If you could start with one value and then end up (after merely simplifying) with two, then major portions of mathematics would stop working.