Whats the sum of sqrt-2 and sqrt-18?

Question

possible answers are:
1) 5i sqrt2
2) 4i sqrt2
3) 2i sqrt5
4) 6i

sqrt NEGATIVE 2 and sqrt NEGATIVE 18 ***

Answer 1

Here are some hints well you can rewrite sqrt(-2) and sqrt(-18) as
i sqrt(2) and i sqrt(18)

and 18 has a perfect square as a factor, so you can pull that factor out of the square root sign.

Answer 2

First you must simplify the sqrt-18.

Find a perfect square that is a multiple of 18: in this case it would be 9, because 9 x 2 = 18.

So, you can take a 3 out of the sqrt., because 3^2 is 9.
You then have: 3 sqrt-2.

Now you can add the two sqrts.

1 sqrt.- 2 + 3 sqrt.- 2 = 4 sqrt. 2

So the answer would be 2.

Answer 3

sqrt(-2)

Whenever you have the square root of a negative number, the first step is to pull out an i, or a sqrt(-1)

isqrt(2)

We do the same for the second one:

isqrt(18)

If you break 18 down, it turns out to be 3*3*2, which is 3^2*2:

isqrt(3^2*2)

3isqrt(2)

isqrt(2) + 3isqrt(2)

= 4isqrt(2)

Which is 2.

That's your answer.

Answer 4

No idea what the other answer is about but here's how you do it:
sqrt(-2) + sqrt(-18)
sqrt(-2) + sqrt(-2*9)
sqrt(-2) + 3sqrt(-2)
4 sqrt(-2)
4 sqrt(-1*2)
4i sqrt(2)

Answer 5

Note that i x i = -1
So sqrt(-1) = i

sqrt(-2) = isqrt(2)
sqrt(-18) = i x sqrt(18) = i x sqrt(2 x 9) = i3sqrt(2)

Now isqrt(2) + i3sqrt(2) = 4isqrt(2)

Remember: do not throw away the i as i is sqrt(-1)

Answer 6

sqrt(-2) = sqrt(2)*i
sqrt(18) =3sqrt(2)*i
so the sum = 4*i*sqrt(2)

Answer 7

sqrt(-2) = sqrt( (-1) (2) ) = sqrt(-1) sqrt(2)

By definition, sqrt(-1) = i, so

sqrt(-1) sqrt(2) = i sqrt(2)

sqrt(-18) = sqrt(18)sqrt(-1) = sqrt(9)sqrt(-1)sqrt(2)
= 3i sqrt(2)

Add them together,

i sqrt(2) + 3i sqrt(2) = 4i sqrt(2)

Answer 8

well, sqrt2 is already simplified, but sqrt18 is also sqrt9 by sqrt2 which is 3sqrt2. now go back to original problem and u got 3sqrt2+(1)sqrt2=4sqrt2