How do you solve the equation and identify any extraneous solutions for # sqrt(x^2+5)=3-x#?

1 Answer
Jun 11, 2015

Answer:

We can square both sides of the equation. But we have to remember that the outcome of a radical is allways #>=0#

Explanation:

This means that #3-x>=0->x<=3#

Now for the squaring (which annulls the root at the left):
#x^2+5=(3-x)^2->#
#x^2+5=x^2-6x+9->#
#5=-6x+9->6x=4->x=2/3#

And this fits with the condition that #x<=3#
(it also checks if you put the number back into the equation)