User:Kingpomba/math

Surd Troubles
I've tried and i thought i did it correctly but as it seems i did not =[. Some help would be appreciated. $$=4\sqrt\times(-2\sqrt)=(-8\sqrt)$$ $$=\sqrt{7}\times\sqrt{7}=\sqrt{49}$$ $$=-8\sqrt{49}$$ $$=8\sqrt{9}-\sqrt{18}$$ $$=8\times 3\times 3 - 9 \times 3$$ $$=24\sqrt{3}- 3\sqrt{3}$$ $$=21\sqrt{3}$$ textbook says $$=24 -  6\sqrt{2}$$ Thanks for the help =] !. —Preceding unsigned comment added by 121.219.227.217 (talk) 07:52, 14 February 2008 (UTC)
 * Problem 1 $$\sqrt{63}-\sqrt{28}=\sqrt{x}$$ $$=\sqrt{9}\times\sqrt{7}-\sqrt{4}\times\sqrt{7}$$ $$=3\sqrt{7}-2\sqrt{7}$$ $$=1\sqrt{7}$$ $$=\sqrt{7}$$ but the textbook says $$=5\sqrt{7}$$
 * Problem 2 - Simplify $$4\sqrt{7}\times(-2\sqrt{7})$$
 * Problem 3 - Simplify $$2\sqrt{3}(4\sqrt{3}-\sqrt{6})$$


 * In problem 1, it's probably supposed to be a plus sign. In problem 2, since 49 is a square you can simplify it further. In problem 3, I think you should just do it more carefully. Black Carrot (talk) 08:02, 14 February 2008 (UTC)
 * Ah i figured out problem 2 and possibly 1( before i noticed your comment that'll teach me to refresh eh) [Note:In problem 1 a minus sign is written in the text book]

$$=\sqrt{7}\times\sqrt{7}=\sqrt{7}^2=7$$ $$=4\sqrt\times(-2\sqrt)=(-8\sqrt)$$ $$=-8\times 7$$ $$=(-56)$$ (this is the answer in the text book) $$=\sqrt{9}\times\sqrt{7}-\sqrt{4}\times\sqrt{7}$$ $$=9\sqrt{7}-4\sqrt{7}$$ ( i know it isnt simplifying it but its all i could think of) $$=5\sqrt{7}$$
 * Problem 2 - Simplify $$4\sqrt{7}\times(-2\sqrt{7})$$
 * Problem 1 $$\sqrt{63}-\sqrt{28}=\sqrt{x}$$
 * And i got futher on problem three but still hit a snag..

$$=8\sqrt{9}-\sqrt{18}$$ $$=\sqrt8\times \sqrt3\times \sqrt3 - \sqrt9 \times \sqrt2$$ $$=24 - 3\sqrt{2}$$ (Why isnt this 6!) textbook says $$=24 -  6\sqrt{2}$$  —Preceding unsigned comment added by 121.219.227.217 (talk) 08:25, 14 February 2008 (UTC)
 * Problem 3 - Simplify $$2\sqrt{3}(4\sqrt{3}-\sqrt{6})$$
 * For problem 1 - just because the textbook made a mistake, doesn't mean you have to, too. $$\sqrt{63}-\sqrt{28}$$ is equal to $$\sqrt{7}$$. If the textbook says something else then it is wrong, and you shouldn't invent new math rules to cover up for it.
 * For problem 3 - take a closer look at your very first step. -- Meni Rosenfeld (talk) 08:33, 14 February 2008 (UTC)
 * Ah i see the problem i didn't know you had to do that, well i did learn something new now.

$$=8\sqrt{9}-2\sqrt{18}$$ $$=\sqrt8\times \sqrt3\times \sqrt3 -\sqrt2 \times \sqrt9 \times \sqrt2$$ $$=24 - 6\sqrt{2}$$

Thanks Guys =] you've been a big help! —Preceding unsigned comment added by 121.219.227.217 (talk) 08:49, 14 February 2008 (UTC)


 * Of the three expressions above, the first and last are equal, but the middle one is different. There are too many root signs there. --Lambiam 18:37, 14 February 2008 (UTC)


 * Something else seems wrong, problem 1 says: $$\sqrt{63}-\sqrt{28}=\sqrt{x}$$, do they want you to simplify the left side or solve for x on the right (they're very different problems). If the book answer is to be believed then I suspect they asked you to simplify $$\sqrt{63}-\sqrt{28}$$. A math-wiki (talk) 02:17, 15 February 2008 (UTC)
 * I suspect this was indeed the question, but $$\sqrt{63}-\sqrt{28}$$ is not equal to $$5\sqrt{7}$$. -- Meni Rosenfeld (talk) 09:21, 15 February 2008 (UTC)


 * You seem to be having trouble formatting your equations. Do you need any help? Black Carrot (talk) 07:57, 15 February 2008 (UTC) 07:48, 15 February 2008 (UTC)