Recheck please

**courteous** · September 18th 2008, 01:12 AM

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?

**mr fantastic** · September 18th 2008, 03:37 AM

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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