Checking to see if the solution is extraneous or not!

Question

anonymous · Answer

$$\sqrt{16+9} -4 = 1$$

So the "solution" is 16 in this case 

@phi

jdoe0001 · Answer

I don't see a variable for one... so... not sure what we're solving :/

jdoe0001 · Answer

though the equation does EQUATEs :)

anonymous · Answer

the original expression was: $$\sqrt{x+9} - 4 =1$$

anonymous · Answer

i had solved it and got x = 16, im just not good at checking to see if it is extraneous or not

anonymous · Answer

@phi ?

mathstudent55 · Answer

You are correct that it is not extraneous.

mathstudent55 · Answer

x = 16 worked in the original equation, so it is a true solution, not an extraneous one.

jdoe0001 · Answer

ditto

anonymous · Answer

but i don't know how to check it exactly. i'm not sure as to where i should start first

phi · Answer

the order you do this is
$$ \sqrt{16+9} -4 = 1 $$
first do the add inside the square root
$$ \sqrt{25} - 4 = 1$$
next do the sqrt(25) (which gives 5)
$$ 5 -4 = 1$$
finally do the subtraction

phi · Answer

you get 
1=1
which is true, so that means that 16 "works"  i.e. is a solution

anonymous · Answer

thank you again @phi could you help me with the other question i had? i can make a new post!

anonymous · Answer

thank you all actually @mathstudent55 @jdoe0001 :)

mathstudent55 · Answer

$\sqrt{x+9} - 4 =1$

Add 4 to both sides:

$\sqrt{x+9}  = 5$

Square both sides:

$(\sqrt{x+9})^2  = 5^2$

$x+9  = 25$

Subtract 9 from both sides:

$x = 16$

Check:
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