Question

How to solve: square root of r+4 + the square root of 2 is equal to 2?

Answer 1

\[\sqrt{r+4} + \sqrt 2 = 2\]
yes?

Answer 2

put the sqrt 2 on the other side and square both sides

Answer 3

square both sides.
\[(\sqrt{r+4}+\sqrt{2})^{2}=2^{2}\]
\[r+4+2+2(\sqrt{r+4})(\sqrt{2})=4\]...........by applying identity.
\[r+6+2({r+4})({2})=4\]
\[r+6+{4r+16}=4\]
\[5r+20=4\]
\[5r+20-20=4-20\]
\[5r=-16\]
\[r=\frac {-16}{5}\]

Answer 4

I am working on other types of equetions, equations with two radicals. I was trying to facro 5r+16=0 to get two answers. In this case I have only one solution?? Can a solution of radical be negative?

Answer 5

never

Answer 6

sol. of a radical is always positive.

Answer 7

so this equation has no solution?

Answer 8

or if this checks into the equation i can say this satisfy the equation?