Question

solve the equation
x+1=sqrt/19-x

Answer 1

So I think you are trying to solve the equation:
\[x+1=\sqrt{19-x}\]
Right?

If not, let me know...
To solve this equation you need to square both sides...
the square cancels out the square root but you have to be careful!
Squaring both sides of an equation can give solutions that aren't of the original equation. These are called extraneous solutions...
\[(x+1)^2=(\sqrt{19-x})^2\]

Answer 2

i dont kno how to write it step by step i dont get it

Answer 3

O.k.,
\[(x+1)^2= (\sqrt{19-x})^2\]
\[(x+1)(x+1)=19-x\]
\[x^2+x+x+1=19-x\]
\[x^2+2x+1=19-x\]
-19 +x -19 +x
\[x^2+3x-18=0\]
(x-3)(x+6)=0
so either x = 3 or x = -6

Now you have to check your answer in the original equation

Answer 4

if x = 3,
\[x+1 = \sqrt{19-x}\] becomes
\[(3)+1 = \sqrt{19-(3)}\] which reduces to
\[4 = \sqrt{16}\] which is true so x = 3 is a solution.

if x = -6,
\[x+1 = \sqrt{19-x}\] becomes
\[(-6)+1 = \sqrt{19-(-6)}\] which reduces to
\[x+1 = \sqrt{19-x}\] which becomes
\[-5 = \sqrt{25}\] which is
-5 = 5 which isn't true so x is not equal to -6
Therefore your only solution is
x = 3

Answer 5

Is that enough steps?
:)