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Square roots

  • Thread starter TSN79
  • Start date
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0
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

arildno
Science Advisor
Homework Helper
Gold Member
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Why should it??
What you're basically saying in line 2, is that (a+b)^2=a^2+b^2
 
480
0
Yes, take a look at line two, you have to square the entire left side of the equation.
 
732
0
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:
VietDao29
Homework Helper
1,417
1
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,
 
732
0
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:
Curious3141
Homework Helper
2,830
86
--------------------
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.
 
732
0
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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