1. ## Square root help...

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2. Originally Posted by alwaysalillost
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Hello,

are you supposed to expand the brackets?

If so:

to #1: The result is a diffrence of squares:

(√(2) - 3)*(√(2) + 3) = (√(2))² - 3² = 7

to #2: This is a complete square. Use the binomal formula:

(√(3) - √(5))*(√(3) - √(5)) = (√(3))² - 2*√3 * √5 + (√(5))² = 8 - 2*√(15)

3. Originally Posted by earboth
to #1: The result is a diffrence of squares:

(√(2) - 3)*(√(2) + 3) = (√(2))² - 3² = 7

If you can please explain to me how you got this answer I'd be very grateful. Thanks in advance!

4. Originally Posted by alwaysalillost
If you can please explain to me how you got this answer I'd be very grateful. Thanks in advance!
do you know what "the difference of two squares" refers to?

as it's name suggests, we have the difference of two squares when we subtract one squared number from another. now it turns out, that this operation has a pretty neat factorization. in general:

x^2 - y^2 = (x + y)(x - y)

meaning, if you expand the brackets on the right, you get the thing on the left. (do you know how to expand brackets? if that's your problem, you should say so). so all earboth did, was to work this equation backwards. that is, he went from the factorized version to the expanded version:

(x + y)(x - y) = x^2 - y^2

in this particular question, your x was √(2) and your y was 3, so all he had to do was to find what (√(2))² was and what 3² was and subtract the latter from the first.

know, when we square a number that is square-rooted, we simply remove the square root, why is that? well, i believe you asked a question about going from radicals to exponents a while back (see http://www.mathhelpforum.com/math-help/pre-algebra-algebra/14933-exponential-form.html#post51855 ) so you should recall, that we can think of a squareroot of a number as that number to the 1/2 power. now if we raise a number to a power, we multiply the power, so we multiply the 1/2 by 2 and get 1, so we end up with the number to the 1 power, which is just the number. that is:

(√(2))² = [(2)^(1/2)]^2 = 2^[(1/2) * 2] = 2^1 = 2

and of course you know that 3^2 = 9

so we get:

(..x..-.y).*.(..x..+..y.).=........-..
(√(2) - 3)*(√(2) + 3) = (√(2))² - = 2 - 9 = -7

Challenge:

what do you think (cuberoot(5))^3 = ?

what do you think (sqrt(3))^3 = ? as a radical? as a number raised to an exponent?

5. Thanks a lot Jhevon!