How do you solve and check for extraneous solutions in $sqrt(x^2+5)=3-x$ ?

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Answer 1 · 2015-08-01T16:37:56

$color(red)(x = 2/3)$
There are $color(red)("no")$ extraneous solutions.

$sqrt(x^2+5) = 3-x$

Square each side.

$x^2+5=(3-x)^2$

Remove parentheses.

$x^2+5=9-6x+x^2$

Move all terms in $x$ the left hand side.

$x^2+5+6x-x^2 =9$

$5+6x =9$

$6x = 9-5 = 4$

$x = 4/6$

$x = 2/3$

There are no extraneous solutions.

Check:

$sqrt(x^2+5) = 3-x$

$sqrt((2/3)^2+5) = 3-2/3$

$sqrt(4/9+5) = (9-2)/3$

$sqrt((45+4)/9) = 7/3$

$sqrt(49/9) = 7/3$

$7/3 = 7/3$

How do you solve and check for extraneous solutions in sqrt(x^2+5)=3-xsqrt(x^2+5)=3-x?