Unraveling the Mystery of Solving Square Roots

[TSN79]

I'm just wondering about something, look at this:

$$\[<br /> \begin{array}{l}<br /> \sqrt {3 + x} + \sqrt {2 - x} = 3 \\ <br /> \left( {\sqrt {3 + x} } \right)^2 + \left( {\sqrt {2 - x} } \right)^2 = 3^2 \\ <br /> 3 + x + 2 - x = 9 \\ <br /> 5 = 9 \\ <br /> \end{array}<br /> \]$$

Obviously something isn't right here, and I know that if I isolate one of the square roots on the other side, then it all works out. But why doesn't this way work?!

 

[arildno]

Why should it??
What you're basically saying in line 2, is that (a+b)^2=a^2+b^2

 

[moose]

Yes, take a look at line two, you have to square the entire left side of the equation.

 

[bomba923]

TSN79, remember that $$( a + b ) ^ 2 = a^2 + 2ab + b^2$$

Therefore, fixing the second line, you might have:

$$\sqrt {3 + x} + \sqrt {2 - x} = 3 \Rightarrow 2\sqrt {\left( {3 + x} \right)\left( {2 - x} \right)} = 4 \Rightarrow$$

$$\sqrt {\left( {3 + x} \right)\left( {2 - x} \right)} = 2 \Rightarrow \left( {3 + x} \right)\left( {2 - x} \right) = 4 \Rightarrow$$

$$x^2 + x - 2 = 0 \Rightarrow \left( {x + 2} \right)\left( {x - 1} \right) = 0 \Rightarrow \boxed{x = - 2,1}$$

 

[VietDao29]

That's wrong, bomba923. This line is wrong:

$$4 - x^2 = 0 \Rightarrow x = \pm 2$$

It should look like this:
$$x ^ 2 + x - 2 = 0 \Rightarrow \left[ \begin{array}{l} x = 1 \\ x = - 2 \end{array} \right.$$
Viet Dao,

 

[bomba923]

That's wrong, bomba923. This line is wrong:

It should look like this:
$$x ^ 2 + x - 2 = 0 \Rightarrow \left[ \begin{array}{l} x = 1 \\ x = - 2 \end{array} \right.$$
Viet Dao,

Indeed! , stupid mistakes are the bane of my math education!!

Anyway,
Thanks for pointing that out-->
Mistake correct, see updated post!
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[Curious3141]

--------------------
However,
$$\because \forall x > 2,\;\sqrt {2 - x} \notin \mathbb{R} \, , {so}$$

$$\therefore \boxed{x = - 2}$$​

Both of your solutions are admissible.

 

[bomba923]

Both of your solutions are admissible.

Oh yes, that's right! $$1 < 2$$,
See updated post!