Answer to Question #321184 in Algebra for Mhean

Question

√x^2+16=x+4

Answer 1

Consider two possible formulations of the problem:

1. $\sqrt{x^2}+16=x+4$

In order to solve the equation consider two cases:

a). $x>0$. We receive: $16=4$. Thus, the problem has no solutions.

b). $x<0$. We get: $-x+16=x+4$. The latter equation is solved via: $x=6$. But it does not belong to the domain $x<0.$

2. $\sqrt{x^2+16}=x+4$. We point out that the expression under the square root is always positive. From equation we receive: $x^2+16=x^2+8x+16$. The solution is $x=0$.