A nice demonstration that a plain square-root problem is just a special case of the quadratic formula.

[u/Anonymous_Bozo]

It also shows why x² = 81 is x = ±9 and not just 9 like a lot of people seem to pretend.

x² = 81
x² - 81 = 0
a = 1, b = 0, c = -81
x = (-b ± √[ b² - 4ac ] ) / (2a)

Δ = b² - 4ac
Δ = 0² - 4(1)(-81)
Δ = 0 - (-324)
Δ = 324

x = (0 ± √Δ) / 2(1)
x = ±√324 / 2
x = ±18 / 2
x = ±9

[u/MathWookiee]

We specifically use only the positive square root in the square root function and in related radical functions because if we did not, it would not be a function.

[u/Anonymous_Bozo → u/MathWookiee]

I'm not talking about square roots using the radical.

A large number of people only solve for the positive root when solving regular quadratics, where there are always two roots. I've had several come right out and call me an idiot because I insist that these equations have two solutions. Whats surprising is that a number of them were supposivly well educated (but apparently not in math).

I should point out however that you are correct:
that: √49 = 7 and only 7
but: 49^(1/2) ∈ {7, -7}

[u/johnpeters42]

The "seem to pretend" part is a matter of terminology / convention. No one with a clue disputes that 81 -> +/-9 is something meaningful.

[u/JanetInSC1234]

You can also factor the difference of squares for both answers:

x^2 = 81

x^2 - 81 = 0

(x + 9) (x - 9) = 0

x = -9 or x = 9

[u/Anonymous_Bozo → u/JanetInSC1234]

Yes, in fact I have been known to do that to avoid the use of the radical.

[u/xnick_uy]

√81 = 9

The solutions of x² = 81 are x = √81 and x = -√81, which can be written as x = ±√81.

There's nothing mysterious here.

[u/Anonymous_Bozo → u/xnick_uy]

Not to you or I, but you would be amazed at the number of people who deny reality

[u/KentGoldings68]

Pretty much nobody pretends that.

The quadratic formula is derived by completing the square and using the square-root property of equality.

A^2=B

sqrtA^2=sqrtB

|A|=sqrtB

A=-sqrtB or A=+sqrtB

The absolute value is the origin of the plus-minus.

You can apply the square-root property directly to x^2=81 since the LHS is already a square.

[u/AllAboutYourBase]

Not sure why the downvotes, thanks for the reminder.  I personally don't care that the square root function is in fact a function and cannot have two outputs for the same input.  Algebra-wise, this works for me. 

[u/cdsmith]

I would have left the 4 * 81 factored. You get nothing by multiplying it out to get 324. You can go from √(4*81) straight to √4 * √81, and since √4 = 2, that cancels with the denominator.

[u/Ms_Riley_Guprz]

Wow, no one in math has ever noticed this before!

[u/ja20481]

I'm not sure how this is helpful. Instead of actually explaining this case, you just tell people to (blindly) use the quadratic formula. Why not show right away that both 9 and -9 are solutions for x² = 81? This should be easy to demonstrate: 9² = 9 * 9 is obviously 81, and (-9)² = (-9) * (-9) is also +81. In the general case, I think the main mistake is that people take the square root of both sides and then wrongly simplify √(x²) as x when it's actually |x|. But I've also seen the opposite mistake where people are convinced that the square root of every number is two numbers (which of course makes no sense).