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September 5

Anomalous result

Solve for x:

${\sqrt {x+1}}$ = x - 1.

Here's my approach, step by step:

Square both sides:

x + 1 =

{x^{2}}

- 2x + 1

Cancel 1's:

x =

{x^{2}}

- 2x

Collect x's:

{x^{2}}

- 3x = 0

Factorise:

x (x - 3) = 0

Solution:

x = 0 or 3.

So far, so good. Or so it seems.

Plug 3 back into the original equation:

${\sqrt {3+1}}$ = 3 - 1
${\sqrt {4}}$ = 2 = 3 - 1. Correct

Plug 0 back into the original equation:

${\sqrt {0+1}}$ = 0 - 1
${\sqrt {1}}$ = 1 =/= 0 - 1. Incorrect.

I've gone over this a dozen or more times but cannot see what really basic error I must be making.

Any ideas? -- Jack of Oz ^{[pleasantries]} 22:06, 5 September 2024 (UTC)[reply]

You didn't do anything really wrong, but just discovered that 0 is an "Extraneous solution to the problem. As our article says, they "result from performing operations that are not invertible for some or all values of the variables involved, which prevents the chain of logical implications from being bidirectional."

The problem is that squaring is not a one-to-one function, so its inverse, square-rooting needs to be carefully defined. That is, -5 and 5 squared are both 25. So we must pick one of them as "the" square root if we want to define a function, something that spits out just one value "5" when fed "25". If we defined "square root" as Euler did and said either are square roots, then 0 is perfectly good solution. In modern terms it would amount to solving ±

{\sqrt {x+1}}

= x - 1John Z (talk) 23:19, 5 September 2024 (UTC)[reply]

You proved that

{\sqrt {x+1}}=x-1

implies

(x=0

or

x=3).

Indeed, if

x=3

(the only true solution), it is the case that

x=0

or

x=3.

You appear to assume that the converse implication also holds, but this assumption is unwarranted. The false solution is introduced by the squaring operation; it adds solutions of the equation

~{-}{\sqrt {x+1}}=x-1.

A simpler puzzle based on the same issue is the following:

Solve for $x$ the equation
${\sqrt {x}}=-1.$
Square both sides:
$x=1.$
Plug $1$ for $x$ into the original equation:
${\sqrt {1}}=-1.$
What gives?

What we found is the solution of the equation

~{-}{\sqrt {x}}=-1.

--Lambiam 23:17, 5 September 2024 (UTC)[reply]

Wow, that really is basic. But not obvious. I've been aware forever that the sq rt sign

{\sqrt {X}}

is always taken to be the positive root only of X unless modified by a - or ± in front; whereas, the words "the square root of X" mean both positive and negative roots. What I've never quite focussed on is the dangers of squaring, if I can put it that way. Squaring both sides of an equation is a tool we all learn early in our algebraic studies, but I don't remember this particular hazard ever being brought to my attention. But then, my most recent formal mathematical studies were in 1984 [before my younger son was born; he's now produced three grandchildren for me].

Thanks for a very enlightening set of answers. -- Jack of Oz ^{[pleasantries]} 20:16, 6 September 2024 (UTC)[reply]

Resolved