Radical equation

Question

Hi, im having troubling solving : √8-x² + x=0 

 

** 8-x² is all under one radical. The + x is not part of the radical**

Answer 1

hey Alex

sqrt(8-x^2)+x=0

sqrt(8-x^2)=-x

square both sides

8-x^2=x^2

2x^2-8=0

x^2-4=0

x^2=4

x=±2

check

sqrt(8-2^2)+2=sqrt(4)+2=0 not working so might be extraneous solution

try -2

sqrt(8-(2)^2)-2=sqrt(4)-2=0 works!!

Answer 2

√(8-x²) + x = 0

 

√(8-x²) = -x

 

Square both sides to get 8-x² = x²

 

                                      2x² = 8

 

                                        x² = 4

 

                                        x = ±2

 

Checking in the original equation, we find that -2 satisfies the equation

but 2 does not.

 

So, the only solution is -2.