Correct square root factorization?

[Steph Annie]

Hello, could someone tell me if my square root factorization is correct for the following problems?
Thank you.

It's for the square root of √75 +√12 −√3

I said √(3x5²) ⁺√(3x2²) −√3
=5√3 +2√3 −√3
=7√3 −√3 = 6 √3

And also the square root of √200

√100 x 2
√100 √2
10 √2
√200

Should I also write √200 ≈ 14.14215462
Or is that not needed?


And finally the square root for the fraction √32⁄9
Do I multiply the numerator and denominator by 3, because 9 is a perfect square,
thus 32x3=96, 9x3=27, so fraction of 96/27......??

I am not sure how to go about this since the sq. root is in the denominator.
I have only done square root fractions like √9/16 =3/4

 

[serds]

√(32/9) = √32 / √9

 

[Steph Annie]

So you're saying the answer is the same thing as the question?
the square root of 32/9 equals the square root of 32/9? No. That is not it.
What about the denominator, which is the square root of 3 (3²=9)
You can't have a perfect square in the denominator!

I will work it out.

√(32/9) = √32 / √9

 

[DrPhil]

So you're saying the answer is the same thing as the question?
the square root of 32/9 equals the square root of 32/9? No. That is not it.
What about the denominator, which is the square root of 3 (3²=9)
You can't have a perfect square in the denominator!

I will work it out.

The perfect square 9 in the denominator becomes (1/3).
The square root of 32 also has a factor which is a perfect square, 16, leaving 2 as the only number that has to stay under the radical.

\(\displaystyle \displaystyle \sqrt{32/9} = (4/3) \sqrt{2}\)

 

[JeffM]

Hello, could someone tell me if my square root factorization is correct for the following problems?
Thank you.

It's for the square root of √75 +√12 −√3

I said √(3x5²) ⁺√(3x2²) −√3
=5√3 +2√3 −√3
=7√3 −√3 = 6 √3 Correct Well done.

And also the square root of √200

√100 x 2
√100 √2
10 √2 Correct
√200

Should I also write √200 ≈ 14.14215462
Or is that not needed?

It depends on whether the question asks for an approximation or not. Usually if no approximation is asked for, then the exact answer is sufficient.


And finally the square root for the fraction √32⁄9
Do I multiply the numerator and denominator by 3, because 9 is a perfect square,

You are (I think) mixing up simplifying a radical with rationalizing it. Rationalizing a radical is required only if you END UP with a radical in the denominator.

thus 32x3=96, 9x3=27, so fraction of 96/27......??

I am not sure how to go about this since the sq. root is in the denominator.
I have only done square root fractions like √9/16 =3/4

Rule 1 Multiplication

\(\displaystyle \sqrt{ab} = \sqrt{a} * \sqrt{b}.\)

Rule 2 Division

\(\displaystyle \sqrt{\dfrac{a}{b}} = \dfrac{\sqrt{a}}{\sqrt{b}}.\)

Rule 3 Rationalization

\(\displaystyle \dfrac{a}{\sqrt{b}} = \dfrac{a}{\sqrt{b}} * 1 = \dfrac{a}{\sqrt{b}} * \dfrac{\sqrt{b}}{\sqrt{b}} = \dfrac{a\sqrt{b}}{\sqrt{b} * \sqrt{b}} =\dfrac{a\sqrt{b}}{\sqrt{b * b}} = \dfrac{a\sqrt{b}}{\sqrt{b^2}} = \dfrac{a\sqrt{b}}{b}.\)

In this problem apply rule 2 and then rule 1. Rule 3 is not needed for this problem.

\(\displaystyle \sqrt{\dfrac{32}{9}} = \dfrac{\sqrt{32}}{\sqrt{9}} = \dfrac{\sqrt{32}}{3} = \dfrac{16 * 2}{3} = \dfrac{\sqrt{16} * \sqrt{2}}{3} = \dfrac{4\sqrt{2}}{3}.\)

 

[lookagain]

It's for the square root of √75 + √12 − √3

I said √(3x5²)+ √(3x2²) − √3 . . . Good! You included the required grouping symbols for the radicands.
= 5√3 + 2√3 − √3
= 7√3 − √3 = 6√3



And also the square root of √200

√(100 x 2 ) . . . No, you are missing the required grouping symbols, for which I added parentheses in bold.
√100 √2
10 √2
√200


And finally the square root for the fraction √32⁄9 . . . You are missing the required grouping symbols. ---> Make it: √(32/9)


I have only done square root fractions like > > √9/16 < < =3/4 . . . . Same here.

Make it: √(9/16) = 3/4

.