Question

Am I correct?

Answer 1

c is a much nicer option... i'd pick c

Answer 2

Am I correct this time?

Answer 3

i'm not 100% on how to add sqr rts though, so i just punched them in a calculator and found out which answer was closest (darn rounding)

Answer 4

Can anyone help me on this one?

Answer 5

are you guessing?

Answer 6

No. I feel that it's 4x due to the 4 in front of the square root.

Answer 7

IN maths there is no feeling but a method lol so lets try it again

Answer 8

actually none of the answer math. But b) is the closet

Answer 9

To do this, we need to factor things out of the square roots. The first one is simplified as much as it can be, but the other two are not. We can factor things out of square roots like this:

\[\sqrt{a^2b} = a \sqrt{b}\]

Using one of your two as an example:
\[\sqrt{40} = \sqrt{2^2 \cdot 10}=2 \sqrt{10}\]

We can then add the first two terms:
\[5 \sqrt{10} + 2 \sqrt{10} = 7 \sqrt{10}\]

Now we just need to simplify and add in the last term. Can you do that?

Answer 10

@ankit042 ... r u sure...?
i think it does math c is the closest in this, yeah?

Answer 11

smh @Jack1 how are you solving this besides the use of calculator?

Answer 12

\(\sqrt (32x^3) - \sqrt (16x^3) + 4 \sqrt (x^3) - \sqrt (2x^3)\)
\(\sqrt (32x^3) - \sqrt (4^2x^3) + 4 \sqrt (x^3) - \sqrt (2x^3)\)
\(\sqrt (32x^3) - 4\sqrt (x^3) + 4 \sqrt (x^3) - \sqrt (2x^3)\)
\(\sqrt (32x^3) - \cancel{\color{red}{4\sqrt (x^3)}} + \cancel {\color{red}{4 \sqrt (x^3)}} - \sqrt (2x^3)\)
hang on @nincompoop im on a roll here, @Vandreigan reminded me how to do these

Answer 13

\(\sqrt (32x^3) - \sqrt (2x^3)\)
\(\sqrt (16 \times 2\times x^3) - \sqrt (2x^3)\)
\(\sqrt (4^2 \times 2\times x^3) - \sqrt (2x^3)\)
\(4\sqrt (2\times x^3) - \sqrt (2x^3)\)
\(\large 4 \times 2^{1/2}\times x^{3/2} - 2^{1/2}x^{3/2}\)
so assume \(\Large2^{1/2}\times x^{3/2} = y\)
\(\large 4 \times 2^{1/2}\times x^{3/2} - 2^{1/2}x^{3/2}\)
\(\large 4 y - y\)
\(\large =3 y \)
transform back
\(\large =3 y \)
\(\large =3 \times 2^{1/2}\times x^{3/2} \)
\(\large =3 \times \sqrt2\times \sqrt {x^3} \)

Answer 14

so closest answer would be... (c) yeah?

Answer 15

You can factor the root of x^3 as well:
\[\sqrt{x^3} = \sqrt{x^2 \cdot x} = x \sqrt{x}\]