square root Multiplication: sqrt(13) X sqrt(5)

[dkarolasz]

1) sqrt(13) X sqrt(5)

neither of them have a sqrt so I just times the number and got 65

another one I'm stuck on

2) sqrt7(2sqrt3 + 3sqrt7)

I got

sqrt7 (1sqrt3 + 1(3sqrt7)

Not sure after that

 

[jwpaine]

sqrt(13) * sqrt(5) = sqrt(13 * 5) = sqrt(65)

Both numbers are prime, so you can't reduce the radicals from there....might as well just combine them into one radical.

sqrt(7)(2sqrt(3) + 3sqrt(7))

use the distributive property:

sqrt(7)*2sqrt(3) = 2sqrt(7*3) = 2sqrt(21)

sqrt(7)3sqrt(7) = sqrt((7^2)*(3^2)) = (7*3) = 21

add both together and you get: 2sqrt(21) + 21

 

[Mrspi]

Re: square root Multiplication

sqrt(13) X sqrt(5)

neither of them have a sqrt so I just times the number and got 65




another one I'm stuck on

sqrt7(2sqrt3 + 3sqrt7)

I got

sqrt7 (1sqrt3 + 1(3sqrt7)

Not sure after that

On the first one, they both are square roots.

remember this property of radicals: sqrt(a) * sqrt(b) = sqrt(a*b)

So.....

sqrt(13) * sqrt(5) = sqrt(13*5)
sqrt(3) * sqrt(5) =sqrt(65)

For the second one, you have

sqrt(7) * [2 sqrt(3) + 3 sqrt(7) ]

Multiply each term inside the square brackets by the "thing" outside:

sqrt(7)*2 sqrt(3) + sqrt(7) * 3 sqrt(7)

2*sqrt(7)*sqrt(3) + 3*sqrt(7)*sqrt(7)

Now...when you multiply two square roots together, you get the square root of the product of the radicals. The rule says: sqrt(a)*sqrt(b) = sqrt(ab).3

2 sqrt(7)*sqrt(3) becomes 2 * sqrt(7*3), or 2*sqrt(21)
3*sqrt(7)*sqrt(7) becomes 3* sqrt(7*7), or 3 * sqrt(49)....but sqrt(49) is just 7, so you have 3*7 or 21

2 sqrt(7)*sqrt(3) + 3 sqrt(7)*sqrt(7), then, is 2 sqrt(21) + 21

So.....
your "final answer" is

2 sqrt(21) + 21

 

[Denis]

Re: square root Multiplication

> sqrt(13) X sqrt(5)
> neither of them have a sqrt so I just times the number and got 65

Yes: but it's sqrt(65)
NOTE: do NOT use x or X for multiplication; use *

> another one I'm stuck on
> sqrt7(2sqrt3 + 3sqrt7)
> I got sqrt7 (1sqrt3 + 1(3sqrt7)

sqrt(7) * 2sqrt(3) = 2sqrt(21)
sqrt(7) * 3sqrt(7) = 21
Remember that sqrt(x) * sqrt(x) = x

 

[dkarolasz]

sqrt???

square root

sqrt-121

I know that the sqrt121 is 11 but not sure what to do with the neg sign

would it be -sqrt11

___________________________________________________________

this I think I got but then again I'm not sure

18+sqrt567/9

18+9sqrt7/9

27sqrt7/9 reduced

3sqrt7 <-- my final answer????

 

[jwpaine]

Re: sqrt???

square root

sqrt-121

sqrt(-121) is a non-real number: you could write it as 11i

I know that the sqrt121 is 11 but not sure what to do with the neg sign

would it be -sqrt11

___________________________________________________________

this I think I got but then again I'm not sure

18+sqrt567/9

18+9sqrt7/9

27sqrt7/9 reduced

3sqrt7 <-- my final answer????

18+ sqrt(567/9)
=18+ sqrt(567) / sqrt(9)
=18 +sqrt((3^4)*7) / 3
=18 +sqrt(81 * 7) / 3
=18 + [9sqrt(7) / 3]
=18 + (9/3)sqrt(7)
=18 + 3sqrt(7)

 

[Denis]

Re: sqrt???

sqrt-121
I know that the sqrt121 is 11 but not sure what to do with the neg sign
would it be -sqrt11

Further to what John told you on this:
sqrt(-121)
= sqrt( 121 * -1)
= 11sqrt(-1)
= 11i

i (for imaginary) is used to represent sqrt(-1); less typing!

Soooo...sqrt(-25) = 5i ; kapish?