may i please have help solving with the following 2 equations:

$\displaystyle

3b^3 \sqrt{20 a^3 b^2 c^5}

$

and

$\displaystyle \sqrt{3}(4+\sqrt{3}- \sqrt{15})$

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- Jun 3rd 2009, 02:32 PMgriffincesurds
may i please have help solving with the following 2 equations:

$\displaystyle

3b^3 \sqrt{20 a^3 b^2 c^5}

$

and

$\displaystyle \sqrt{3}(4+\sqrt{3}- \sqrt{15})$ - Jun 3rd 2009, 02:49 PMaidan
- Jun 3rd 2009, 03:16 PMgriffince
yes and to solve them

- Jun 3rd 2009, 05:44 PMProve It
What you have are NOT equations. You only have an equation when you have an equals sign.

I think you're asking to simplify these expressions.

For the first one:

$\displaystyle 3b^3\sqrt{20 a^3 b^2 c^5} = 3b^3\times\sqrt{20}\times\sqrt{a^3}\times\sqrt{b^2 }\times\sqrt{c^5}$

$\displaystyle = 3b^3\times\sqrt{4\times 5}\times\sqrt{a^2\times a}\times b \times \sqrt{c^4\times c}$

$\displaystyle = 3b^4 \times \sqrt{4}\times\sqrt{5}\times\sqrt{a^2}\times\sqrt{ a}\times \sqrt{c^4} \times \sqrt{c}$

$\displaystyle = 3b^4 \times 2 \times \sqrt{5} \times a \times \sqrt{a} \times c^2 \times \sqrt{c}$

$\displaystyle = 6ab^4c^2 \times \sqrt{5} \times \sqrt{a}\times \sqrt{c}$

$\displaystyle = 6ab^4c^2\sqrt{5ac}$

The second one looks already simplified. You can't break up a surd over addition and subtraction (only multiplication and division) and there aren't any other common factors besides the 3.

I think the only other ways it could be written are

$\displaystyle \sqrt{3}\sqrt{4 + \sqrt{3} - \sqrt{15}}$

or

$\displaystyle \sqrt{12 + 3\sqrt{3} - 3\sqrt{15}}$. - Jun 3rd 2009, 08:43 PMyeongil
The OP's 2nd question is different than what has been discussed:

$\displaystyle \sqrt{3}(4+\sqrt{3}- \sqrt{15})$

Assuming that this is the actual question, to solve this you distribute the square root of 3:

$\displaystyle 4\sqrt{3} + \sqrt{3}\sqrt{3} - \sqrt{3}\sqrt{15}$

$\displaystyle 4\sqrt{3} + 3 - \sqrt{45}$

$\displaystyle 4\sqrt{3} + 3 - 3\sqrt{5}$

$\displaystyle 3 + 4\sqrt{3} - 3\sqrt{5}$

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