# Thread: Recheck please

1. ## Recheck please

I've done math problems on some paper, and there are no solutions given.
1. Simplify $\sqrt{3-\sqrt5}-\sqrt{3+\sqrt5}$
(My) solution: $expression^2$
$(3-\sqrt5)+(3+\sqrt5)-2\sqrt{(3-\sqrt5)(3+\sqrt5)}$
$6-2\sqrt{4}$
$6-4=2$
Two solutions?
$-\sqrt2$
$\sqrt2$
Now, what if you've "gained" some (and which?) solution by squaring the original?

2. Simplify
a) $\frac{1}{2}*\sqrt2*\sqrt[3]{4}:\sqrt[4]{8}\Rightarrow \frac{2^{\frac{1}{2}}*2^{\frac{2}{3}}}{2*2^{\frac{ 3}{4}}}\Rightarrow 2^{-\frac{7}{12}}\Rightarrow {\frac{1}{\sqrt[12]{2^7}}}$
b) (For what $a\in\mathbb{R}$ does the following expression exist?) $a^{-1}*(1+a^{-2})^{-\frac{1}{2}}*(1+a^2)^{\frac{1}{2}} \Rightarrow \frac{(1+a^2)^{\frac{1}{2}}}{a*(1+a^{-2})^{\frac{1}{2}}} \Rightarrow \frac{(1+a^2)^{\frac{1}{2}}}{(a^2+1)^{\frac{1}{2}} } = 1$(If the solution is correct) what is now the answer (on "expression exists for what a?"), every a?

3. Solve
$x+2\sqrt{1-x}=1 \Rightarrow expression^2 \Rightarrow 4*(1-x)=1+x^2-2x \Rightarrow x_1=-3, x_2=1$
Again, what if you've "gained" some solution by squaring the original? What are considered best practises in solving such?

2. Originally Posted by courteous
I've done math problems on some paper, and there are no solutions given.
1. Simplify $\sqrt{3-\sqrt5}-\sqrt{3+\sqrt5}$
(My) solution: $expression^2$
$(3-\sqrt5)+(3+\sqrt5)-2\sqrt{(3-\sqrt5)(3+\sqrt5)}$
$6-2\sqrt{4}$
$6-4=2$
Two solutions?
$-\sqrt2$
$\sqrt2$
Now, what if you've "gained" some (and which?) solution by squaring the original?

Mr F says: You have to go back to the original expression and test each potential solution. Clearly ${\color{red}\sqrt{3-\sqrt5}-\sqrt{3+\sqrt5} < 0}$ therefore discard the positive solution.

[snip]

3. Solve
$x+2\sqrt{1-x}=1 \Rightarrow expression^2 \Rightarrow 4*(1-x)=1+x^2-2x \Rightarrow x_1=-3, x_2=1$
Again, what if you've "gained" some solution by squaring the original? What are considered best practises in solving such?

Mr F says: You have to test each potential solution in the original equation. Both solutions satisfy the original equation so you keep both.
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