Simplifying radical with Fractions question

[Speckington]

I have a question about simplifying the radical with fractions in which both the numerator and denominator are a perfect square


Take this as an example:


√9/900 this can be broken down into three perfect squares √3^2/3^2*10^2


Should I simplify by squaring the entire fraction or should I simply factor the digits?


Simplify by squaring the entire fraction: (√9/900) ^2

Simplify by factoring out the digits √3^2/3^2*10^2 = 3/ 3 * 10 = 3/30


If I do these two things out, they will be different.
(√9/900) ^2 = 9/900 = 1/100
√9/900 = √ 3 ^2/ 3^2 * 10^2 = 3/3*10 = 3/30 = 1/10


I would like to know which of the two methods is correct and why the other one is wrong. As a side note would your answer be any different if there was a variable squared in the denominator/numerator?



Thank you so much for your help!

 

[Harry_the_cat]

Where you have sqrt 9/900, do you really mean sqrt (9/900)? Brackets are very important here. It is very hard to interpret what you are trying to say if you don't use brackets correctly.

 

[Steven G]

sqrt(9/900) = sqrt(1/100) = sqrt(1)/sqrt(100) = 1/10

OR

sqrt(9/900) = sqrt(9)/sqrt(900) = sqrt(9)/[(sqrt(9)*sqrt(100)] = 1/ sqrt(100) = 1/10

OR

sqrt(9)/sqrt(900) = 3/30 = 1/10

There is not just one way of doing these type problems. The first way I think is the fastest way of getting to the answer. The last way is also a good method.

 

[JeffM]

I have a question about simplifying the radical with fractions in which both the numerator and denominator are a perfect square


Take this as an example:


√9/900 this can be broken down into three perfect squares √3^2/3^2*10^2


Should I simplify by squaring the entire fraction or should I simply factor the digits?


Simplify by squaring the entire fraction: (√9/900) ^2

Simplify by factoring out the digits √3^2/3^2*10^2 = 3/ 3 * 10 = 3/30


If I do these two things out, they will be different.
(√9/900) ^2 = 9/900 = 1/100
√9/900 = √ 3 ^2/ 3^2 * 10^2 = 3/3*10 = 3/30 = 1/10


I would like to know which of the two methods is correct and why the other one is wrong. As a side note would your answer be any different if there was a variable squared in the denominator/numerator?



Thank you so much for your help!

As has been said by others, your notation is poor.

$$\sqrt{x} = x \iff x = 1 \text { or } x = 0$$ ... Corrected

The square root of a number is NEVER equal to the number except in the special cases of zero and one. Thus

$$\sqrt{\dfrac{9}{900}} \ne \dfrac{1}{100}.$$
That method is WRONG. When you simplify an expression, you must not change its value. But squaring a number changes the value (except in the special cases of zero and one).

What I think is mixing you up is something else, which is valid.

$$a \ge 0 \text { and } b > 0 \implies \sqrt{\dfrac{a}{b}} = \dfrac{\sqrt{a}}{\sqrt{b}}.$$
Consequently

$$0 \le u = \sqrt{a} \text { and } 0 < v = \sqrt{b} \implies \sqrt{\dfrac{a}{b}} = \dfrac{\sqrt{a}}{\sqrt{b}} = \dfrac{u}{v}.$$
To take a concrete example.

$$\sqrt{\dfrac{9}{900}} = \sqrt{\dfrac{1}{100}} = \sqrt{\left ( \dfrac{1}{10} \right )^2} = \dfrac{1}{10}.$$
$$\sqrt{\dfrac{9}{900}} = \dfrac{\sqrt{9}}{\sqrt{900}} = \dfrac{\sqrt{3^2}}{\sqrt{30^2}} = \dfrac{3}{30} = \dfrac{1}{10}.$$
Both methods are valid; both lead to the same result.

 

[Harry_the_cat]

JeffM, I think you have a typo in your second line. Should be x=1 or x=0.

 

[JeffM]

JeffM, I think you have a typo in your second line. Should be x=1 or x=0.

Thank you. Here is a soup bowl of cream.

 

[Harry_the_cat]

Purrrrfect! Thanks!

 

[Speckington]

As has been said by others, your notation is poor.

$$\sqrt{x} = x \iff x = 1 \text { or } x = 0$$ ... Corrected

The square root of a number is NEVER equal to the number except in the special cases of zero and one. Thus

$$\sqrt{\dfrac{9}{900}} \ne \dfrac{1}{100}.$$
That method is WRONG. When you simplify an expression, you must not change its value. But squaring a number changes the value (except in the special cases of zero and one).

What I think is mixing you up is something else, which is valid.

$$a \ge 0 \text { and } b > 0 \implies \sqrt{\dfrac{a}{b}} = \dfrac{\sqrt{a}}{\sqrt{b}}.$$
Consequently

$$0 \le u = \sqrt{a} \text { and } 0 < v = \sqrt{b} \implies \sqrt{\dfrac{a}{b}} = \dfrac{\sqrt{a}}{\sqrt{b}} = \dfrac{u}{v}.$$
To take a concrete example.

$$\sqrt{\dfrac{9}{900}} = \sqrt{\dfrac{1}{100}} = \sqrt{\left ( \dfrac{1}{10} \right )^2} = \dfrac{1}{10}.$$
$$\sqrt{\dfrac{9}{900}} = \dfrac{\sqrt{9}}{\sqrt{900}} = \dfrac{\sqrt{3^2}}{\sqrt{30^2}} = \dfrac{3}{30} = \dfrac{1}{10}.$$
Both methods are valid; both lead to the same result.

As has been said by others, your notation is poor.

$$\sqrt{x} = x \iff x = 1 \text { or } x = 0$$ ... Corrected

The square root of a number is NEVER equal to the number except in the special cases of zero and one. Thus

$$\sqrt{\dfrac{9}{900}} \ne \dfrac{1}{100}.$$
That method is WRONG. When you simplify an expression, you must not change its value. But squaring a number changes the value (except in the special cases of zero and one).

What I think is mixing you up is something else, which is valid.

$$a \ge 0 \text { and } b > 0 \implies \sqrt{\dfrac{a}{b}} = \dfrac{\sqrt{a}}{\sqrt{b}}.$$
Consequently

$$0 \le u = \sqrt{a} \text { and } 0 < v = \sqrt{b} \implies \sqrt{\dfrac{a}{b}} = \dfrac{\sqrt{a}}{\sqrt{b}} = \dfrac{u}{v}.$$
To take a concrete example.

$$\sqrt{\dfrac{9}{900}} = \sqrt{\dfrac{1}{100}} = \sqrt{\left ( \dfrac{1}{10} \right )^2} = \dfrac{1}{10}.$$
$$\sqrt{\dfrac{9}{900}} = \dfrac{\sqrt{9}}{\sqrt{900}} = \dfrac{\sqrt{3^2}}{\sqrt{30^2}} = \dfrac{3}{30} = \dfrac{1}{10}.$$
Both methods are valid; both lead to the same result.

Thank you for your wonderful explanation! I will try and use the correct notation in the future

 

[Steven G]

sqrt(x)=x
Squaring both sides we get x = x^2. Then x^2 - x = 0
x(x-1) = 0
So x= 0 or x=1.
That is why the only numbers where sqrt(x)=x is 1 or 0.