math sq rooots?

Question

please help on simplyfing expressions.
1. (sq.root)12 - 2(sq.root)3
2. (sq.root)20 + (sq.root)80
show whether the expression is a solution of the equation.
3.x^2 - 6x +2 =0; 3 + (sq.root)7

Answer 1

[1]
sqrt(3 . 4) - 2sqrt(3) =
sqrt(3) . sqrt(4) - 2sqrt(3) =
sqrt(3) (sqrt(4) - 2) =
sqrt(3) . 0 = 0

[2]
sqrt(5. 4) + sqrt(5 . 4 . 4) =
2sqrt(5) + 4sqrt(5) =
6sqrt(5)

[3]
(x - 3)(x - 3) - 7 = 0
(x - 3)^2 = 7
(x - 3) = sqrt(7)
x = sqrt(7) + 3 ; it's a solution

Answer 2

1. (sq.root)12 - 2(sq.root)3
\/12 - 2\/3 = 0

2. (sq.root)20 + (sq.root)80
\/20 + \/80 = 2\/5 - 4\/5 = -2\/5

show whether the expression is a solution of the equation.
3.x² - 6x +2 =0; 3 + (sq.root)7 => Incorrect.
(-6)² - 4.3.2 = 36 -24 = 12
x = (6 +/-\/12) : 2.3
x' = (6 + 2\/3): 6
x' = 1 + \/3/3
x" =1 - \/3/3
Solution:{x' = 1 + \/3/3; x" = 1 - \/3/3}
.::.

Answer 3

1. sqrt 12 - 2 sqrt 3
= sqrt (4 * 3) - 2 sqrt 3
= sqrt {(2^2) * 3} - 2 sqrt 3
= 2 sqrt 3 - 2 sqrt 3
= 0

2. sqrt 20 + sqrt 80
= sqrt (4 * 5) + sqrt (16 * 5)
= sqrt {(2^2) * 5} + sqrt {(4^2) * 5}
= 2 sqrt 5 + 4 sqrt 5
= 6 sqrt 5

3. x^2 - 6x + 2 = 0; 3+ sqrt 7
using the quadratic formula x = {-b +/- sqrt (b^2 - 4ac)} / 2a where
a = 1; b = -6 and c = 2

x = [ -(-6) +/- sqrt {(-6)^2 - (4*1*2)}] / 2 (1)
x = {6 +/- sqrt (36 - 8)} / 2
x = (6 +/- sqrt 28) / 2
x = {6 +/- sqrt (4 * 7)} / 2
x = {(6 +/- sqrt (2^2 * 7)} / 2
x = (6 +/- 2 sqrt 7 ) / 2
x = 3 +/- sqrt 7

therefore 3 + sqrt 7 is a solution or root of the equation.

Answer 4

1) 2*sqrt(3) - 2*sqrt(3) = 0
2) sqrt(20) + 2*sqrt(20) = 3*sqrt(20) = 6*sqrt(5)
3) -b +- sqrt(b^2 - 4ac)/2a = 6 +- sqrt(36 - 4*3*2)/6 = 1 +- sqrt(12)/6 = 1+- sqrt(3)/3
oops, I though the 3 was part of the equation
6 +- sqrt(36 - 4*1*2) / 2 = 3 +- sqrt(28)/2 = 3 +- sqrt(7)

Answer 5

I am not allowed to do some one else's homework but check out homework.com

Answer 6

huh? im not against this but i think that you will learn more if you stop your computer and start doing your assignment on your own.