Square roots

  • Thread starter TSN79
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  • #1
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I'm just wondering about something, look at this:

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

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?!
 

Answers and Replies

  • #2
arildno
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Why should it??
What you're basically saying in line 2, is that (a+b)^2=a^2+b^2
 
  • #3
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Yes, take a look at line two, you have to square the entire left side of the equation.
 
  • #4
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TSN79, remember that [tex] ( a + b ) ^ 2 = a^2 + 2ab + b^2 [/tex] :redface:

Therefore, fixing the second line, you might have:

[tex] \sqrt {3 + x} + \sqrt {2 - x} = 3 \Rightarrow 2\sqrt {\left( {3 + x} \right)\left( {2 - x} \right)} = 4 \Rightarrow [/tex]

[tex] \sqrt {\left( {3 + x} \right)\left( {2 - x} \right)} = 2 \Rightarrow \left( {3 + x} \right)\left( {2 - x} \right) = 4 \Rightarrow [/tex]

[tex] x^2 + x - 2 = 0 \Rightarrow \left( {x + 2} \right)\left( {x - 1} \right) = 0 \Rightarrow \boxed{x = - 2,1} [/tex]
 
Last edited:
  • #5
VietDao29
Homework Helper
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That's wrong, bomba923. This line is wrong:
bomba923 said:
[tex] 4 - x^2 = 0 \Rightarrow x = \pm 2 [/tex]
It should look like this:
[tex]x ^ 2 + x - 2 = 0 \Rightarrow \left[ \begin{array}{l} x = 1 \\ x = - 2 \end{array} \right.[/tex] :wink:
Viet Dao,
 
  • #6
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VietDao29 said:
That's wrong, bomba923. This line is wrong:

It should look like this:
[tex]x ^ 2 + x - 2 = 0 \Rightarrow \left[ \begin{array}{l} x = 1 \\ x = - 2 \end{array} \right.[/tex] :wink:
Viet Dao,
Indeed! :redface:, stupid mistakes are the bane of my math education!!!

Anyway,
Thanks for pointing that out-->
Mistake correct, see updated post! :smile:
-------
 
Last edited:
  • #7
Curious3141
Homework Helper
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--------------------
However, :blushing:
[tex] \because \forall x > 2,\;\sqrt {2 - x} \notin \mathbb{R} \, , {so} [/tex]
[tex] \therefore \boxed{x = - 2} [/tex]​
Both of your solutions are admissible.
 
  • #8
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Curious3141 said:
Both of your solutions are admissible.
Oh yes, that's right! [tex] 1 < 2 [/tex], :cool:
See updated post!
 

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