Having difficulty understanding radicals

[MathJakob]

I know how the very basics but then I get given a question like this.

$\sqrt{9-3x}$ and I think I can divide both by 3. $3\sqrt{3-x}$ and so $x=3$

Then I get $\sqrt{4x+12}$ and again I can take 4 from each $4\sqrt{x+3}$ and so $x=-3$

Is this the correct way to be solving these? Because I'm looking down the page and it starts getting real complicated looking, real fast and was just wondering if there is a simple way to go about solving these?

Simplify: ${\sqrt\frac{9a}{8b^2}}$ is the next question I face

 

[mfb]

$\sqrt{9-3x}$ and I think I can divide both by 3. $3\sqrt{3-x}$

$\sqrt{9-3x}$ is not the same as $3\sqrt{3-x}$. Consider x=0, for example, and see what you get in the different expressions.

"and so x=3" does not make sense. There is no equality involved in the expressions.

Is this the correct way to be solving these?

To solve anything, you need an equation. something=somethingelse.

Simplify: ${\sqrt\frac{9a}{8b^2}}$ is the next question I face

There are rules for square roots (more general: rules for exponents) you can use here.

 

[MathJakob]

Also here is another one but I'm not sure it's correct:

${\sqrt\frac{8}{9}}$ = ${\sqrt\frac{8}{9}\cdot{\sqrt\frac{9}{9}}}$ = ${\sqrt\frac{72}{81}}$ =$\frac{\sqrt{72}}{9}$

?

 

[mfb]

That is correct.

 

[symbolipoint]

I know how the very basics but then I get given a question like this.

$\sqrt{9-3x}$ and I think I can divide both by 3. $3\sqrt{3-x}$ and so $x=3$

Then I get $\sqrt{4x+12}$ and again I can take 4 from each $4\sqrt{x+3}$ and so $x=-3$

Is this the correct way to be solving these? Because I'm looking down the page and it starts getting real complicated looking, real fast and was just wondering if there is a simple way to go about solving these?

Simplify: ${\sqrt\frac{9a}{8b^2}}$ is the next question I face

The distinction between the root symbol and grouping symbols is that although both are grouping symbols, the root symbol is a function symbol. Inside the root symbol, you can factor one or more common factors from terms, but you cannot necessarily move them outside of the root symbol; because doing so requires that you have great enough count of the factors as indicated by the root index.

$\sqrt{9-3x}$ you can do $\sqrt{3(3-x)}$ but not further.

([strike]All I did was added the proper parentheses for grouping symbol balancing and the typesetting fails. [/strike] [strike]I NEED the paired parentheses in the second expression! [/strike] The typesetting did work but failed while I first made the post and mathematical notation.)

 

[symbolipoint]

I really want the typesetting to work right but it did not so I do this instead:

(9-3x)^1/2 is the same as (3(3-x))^1/2 but you cannot go any further.

 

[verty]

This is something you will commonly see/use:

$\frac{1}{2} \sqrt{x} = \sqrt{\frac{1}{4}} \; \sqrt{x} = \sqrt{\frac{x}{4}}$
$\sqrt{4x} = \sqrt{4} \; \sqrt{x} = 2 \sqrt{x}$

Also $\sqrt{54} = \sqrt{9 * 6} = 3 \sqrt{6}$.

These are all explained by replacing each radical with the corresponding exponent, like $(\frac{x}{4})^{\frac{1}{2}}$.

 

[HallsofIvy]

Also here is another one but I'm not sure it's correct:

${\sqrt\frac{8}{9}}$ = ${\sqrt\frac{8}{9}\cdot{\sqrt\frac{9}{9}}}$ = ${\sqrt\frac{72}{81}}$ =$\frac{\sqrt{72}}{9}$

?

That's true but $$\sqrt{9}= 3$$ so that $$\frac{\sqrt{8}}{3}$$ would be better. In fact, since $$8= 4(2)$$ and $$\sqrt{8}= 2\sqrt{2}$$, the simplest form is $$\frac{2\sqrt{2}}{3}= \frac{2}{3}\sqrt{2}$$.