1) Must repair notation. \(\displaystyle (\sqrt{}x+5)^{2}\) is NOT the same as \(\displaystyle [\sqrt{}(x+5)]^2\). Make sure it's clear what is under the radical and what is not. Of course, the tiniest bit of LaTeX would help. There is nothing unclear about \(\displaystyle \sqrt{x+5}^{2}\)
2) You MUST worry about Domain issues and Existence before you go around squaring things willy-nilly. Solve this \(\displaystyle \sqrt{x+2} = -4\). You WILL make a mistake if you just square it and don't worry about the Domain or other issues. In this case, we haev a positive square root and a negative number. No solution. Done.
3) You should have studied quadratic equations before you got to this section. Did you? Solve your quadratic equations using any method available to you. Factoring. Completing the Square. Quadratic Formula. Other?
Here's how I'd do the first one.
\(\displaystyle \sqrt{x+23}^{2} = (x+11)^{2}\)
Domain Considerations: x + 23 >= 0 ==> x >= -23. If we get anything less than -23, we'll just discard it.
\(\displaystyle x+23 = x^{2}+22x+121\)
I'll make an error on purpose!
\(\displaystyle 0 = x^{2}+21x+99\)
This has no positive solutions, but it might have two negative solutions. I'd be happy to tell you how I know that, but it's a lesson for another day.
Quadratic Formula
\(\displaystyle x = \frac{-21+\sqrt{21^{2}-4\cdot 99}}{2}\) or \(\displaystyle x = \frac{-21-\sqrt{21^{2}-4\cdot 99}}{2}\)
\(\displaystyle x = \frac{1}{2}(3\cdot\sqrt{5}-21)\) or \(\displaystyle x = -\frac{1}{2}(3\cdot\sqrt{5}+21)\)
That's around x = -7.146 or x = -13.854 so everything is greater than -23. I expect them to work in the ORIGINAL equation.
\(\displaystyle \sqrt{\frac{1}{2}(3\cdot\sqrt{5}-21)+23}^{2} = \left(\frac{1}{2}(3\cdot\sqrt{5}-21)+11\right)^{2}\)
\(\displaystyle \sqrt{\frac{1}{2}(3\cdot\sqrt{5}+25)}^{2} = \left(\frac{1}{2}(3\cdot\sqrt{5}+1)\right)^{2}\)
\(\displaystyle \frac{1}{2}(3\cdot\sqrt{5}+25) = \frac{1}{2}(3\cdot\sqrt{5}+23)\) --What?!?!?! How could this be wrong? Remember that error? It should have been 98, not 99. See how important CHECKING is? Did you catch the error when I made it and even TOLD you about it?
Start back here.
\(\displaystyle x+23 = x^{2}+22x+121\)
\(\displaystyle 0 = x^{2}+21x+98\)
Look at that. It makes so much more sense.
7*14 = 98
7+14=21
0=(x+7)(x+14)
And we get x = -7 or x = -14, both of which are greater than -23. Go check these in the ORIGINAL equation and see if they work. Who knows, I may have put another error in there.
