For number 1, let
X=1st number
X + 1 = 2nd number
[tex]1.) \: \sqrt{x + (x + 1)} = 7 \\ \sqrt{2x + 1} = 7 \\ { (\sqrt{2x + 1}) }^{2} = {7}^{2} \\ 2x + 1 = 49 \\ 2x = 48 \\ \frac{2x}{2} = \frac{48}{2} \\ x = 24[/tex]
The two integers are 24 and 25.
[tex]2.) \: \: \sqrt{x} + 4 = 5 \\ \sqrt{x } =5 - 4 \\ \sqrt{x} = 1 \\ { \sqrt{x} }^{2} = {1}^{2} \\ x = 1[/tex]
[tex]3.) \: \: \sqrt{x + 6} = x \\ { ( \sqrt{x + 6}) }^{2} = {x}^{2} \\ x + 6 = {x}^{2} \\ {x}^{2} - x - 6 = 0 \\ (x - 3)(x + 2) = 0 \\ \\ x - 3 = 0 \\ x = 3 \\ \\ x + 2 = 0 \\ x = - 2[/tex]
For number 3, x = 3, -2.
#CarryOnLearning
#StudyWell
