Question

Solve the equation for x

1+ square root of x-2 = square root of x+3

The book says x=6 but I can't figure out how they got it. I keep going through the steps and the x's cancel out each other

Answer 1

I tried squaring both sides like my teacher told me, but the x's still cancel

Answer 2

Sorry just got really confused. You're helping a ton, I'm just learning this stuff slow. First time I've ever seen anything like this!

Answer 3

\(1+\sqrt{x-2}=\sqrt{x+3}\)
square both sides
\(1+2\sqrt{x-2}+x-2 = x+3\)
so
\(\sqrt{x-2}=\frac{x+3-x-1+2}{2}=2\)
square both sides
\(x-2=4\\x=6\)

Answer 4

How come squaring it on the right produced all that stuff and on the left, it just came down with the square root symbol taken off?

Answer 5

\(1+\sqrt{x-2}=\sqrt{x+3}\\so\\(1+\sqrt{x-2})^2=(\sqrt{x+3})^2\)
we will deal with the left side first, foil on the left
\(1^2+\sqrt{x-2}+\sqrt{x-2}+(\sqrt{x-2})^2=(\sqrt{x+3})^2\\=1+2\sqrt{x-2}+x-2=(\sqrt{x+3})^2\)
now we deal with the right hand side
\(1+2\sqrt{x-2}+x-2=x+3\)
now collect everything not in the square root term to the right hand side
\[2\sqrt{x-2}=x+3+2-x-1 = 4\]
so we have \(2\sqrt{x-2}=4\)
divide by the 2
\(\sqrt{x-2}=2\)
now square both sides again
\((\sqrt{x-2})^2=2^2\)
this gives
\(x-2=4\)
so, finally.....
\(x=6\)

Answer 6

\((\sqrt{a})^2=a\)
\(\)

Answer 7

Thank you so much, I'm starting to finally understand it! I have a test on this tomorrow (most of them aren't anywhere as hard as this one) but now at least I get what to do for the steps. Great help thanks again