Solving a Radical Equation

[e^x]

sqrtx + 1 = sqrt(x + 5)

I'm confused because if I square everything then I get x + 1 = x +5. Subtracting both xs yields 1 = 5... what am I missing?

 

[Dr.Peterson]

sqrtx + 1 = sqrt(x + 5)

I'm confused because if I square everything then I get x + 1 = x +5. Subtracting both xs yields 1 = 5... what am I missing?

If you square [imath]\sqrt{x}+1[/imath], you don't get [imath]\sqrt{x}^2+1^2=x+1[/imath]; you get [imath](\sqrt{x}+1)^2=\sqrt{x}^2+2\sqrt{x}+1^2[/imath].

Keep in mind that you are squaring a binomial, not just each term. (The square of [imath]1+1[/imath] is not [imath]1^2+1^2[/imath]!)

 

[Steven G]

If in fact you correctly (or even incorrectly) obtained x+ 1 = x+5, then there is no solution. When you get something like 1=5, it doesn't mean that 1 actually equals 5 and that you must have made a mistake, but rather it means that there is no solution.

After all, there is no number that when you add 1 or 5 to it that you get the same result - UNLESS, 1=5. And that is absurd! So there is no solution.

 

[lookagain]

sqrtx + 1 = sqrt(x + 5)

I'm confused because if I square everything then I get x + 1 = x +5. Subtracting both xs yields 1 = 5... what am I missing?

You should use grouping symbols on the left as well, especially as you are writing a function.

So, if you are indicating the square root of x, then added to one, as seen on your left-hand side, write it as
sqrt(x) + 1 in this non-Latex style. Or, write it in the Latex style as seen in post # 2.