Radicals Help?

Question

what is the answer to this (√2+√3)*(√2-√3)
and explain please because i don't get it

Answer 1

Multiply it out:

√2*√2
√2*(-√3)
√2*(+√3)
√3*(-√3)

the first line becomes 2; the 2 middle lines added together is 0; and the last line becomes -3.

Answer 2

Distribute.

2 - 3 = -1

Answer 3

(a+b)(a-b) = a^2 - b^2

here a = √2 and b = √3

(√2 +√3)(√2 - √3) = (√2)^2 - (√3)^2 =>2 - 3 => -1

Answer 4

foil
First √2 * √2 =2
Outside √2 * -√3= - √6
Inside √3 * √2 =√6
Last √3 * -√3 = -3

2 -√6 + √6 -3
-1

Answer 5

lets (a + b)*(a - b) = a^2 -b^2

(√2+√3)*(√2-√3) = (√2)^2 - (√3)^2 = 2 - 3 = -1

Answer 6

i would use the FOIL method to multiply them (First Outer Inner Last) so it would be rad(2)*rad(2) + rad(2)*-rad(3) + rad(3)*rad(2) +rad(3)-rad(3). Since you multiply the number in the radicals together you reduce to rad(4) - rad(6) +rad(6) -rad(9), so the two rad(6) cancel out and you're left with rad(4)-rad(9) which reduces to 2-3 = -1

Answer 7

this is a difference of two squares problems. (a+b)(a-b)= a^2 - b^2

(sq.rt 2)^2 - (sq rt 3)^2 = 2-3= -1

Answer 8

use foil
(√2)*(√2)=2
(√3)*(√3)=3
the inner ones cancel out b/c they are like terms so that is all you are left with
2-3=-1
*** I am pretty sure***

Answer 9

Im not gunna do the math out because I left my calc in the car. I'm not sure, but did you try factoring it?

Answer 10

(sqrt(2) + sqrt(3) * (sqrt(2) - sqrt(3))

= sqrt(2) * sqrt(2) + sqrt(6) -sqrt(6) -sqrt(3) * sqrt(3)

= sqrt(4) + 0 -sqrt(9)

= 2 - 3

= -1