Solving Square Root Questions: A Math Tutorial for Beginners

[OMGMathPLS]

How does this work? (Also, I am very math illiterate and do not understand even the most basic terms, if you could kindly speak as you would to a child I would GREATLY appreciate it.)

sq rt sign over 13^2 - 12^2 (over both of it together)

Now the answer is 5, because

(13)(13) - (12)(12)

149-144 = 5

But what canceled out the sq rtt? We did the exponents so did that cancel it out?

Thanks for help.

View attachment 3248

 

[mathbalarka]

Several of your steps are invalid.

1) $\sqrt{a^2 - b^2} \neq a - b$, i.e., the square root and the difference of squares do not "cancel out". For example, let $a = 5$ and $b = 4$. $a^2 - b^2 = 5^2 - 4^2 = 25 - 16 = 9$ and $\sqrt{5^2-4^2} = \sqrt{9} = 3$ whereas $4 - 3 = 1$.

2) What you wrote in the attachment was $\sqrt{13^2 - 12^2} = 13^2 - 12^2$. Where did the square root go? This is an invalid step.

3) $13^2 = 13 \times 13$ is actually $169$, not $149$. So your calculations are void.

Can you recalculate the expression now that your have been pointed out?

 

[mathmari]

How does this work? (Also, I am very math illiterate and do not understand even the most basic terms, if you could kindly speak as you would to a child I would GREATLY appreciate it.)

sq rt sign over 13^2 - 12^2 (over both of it together)

Now the answer is 5, because

(13)(13) - (12)(12)

149-144 = 5

But what canceled out the sq rtt? We did the exponents so did that cancel it out?

Thanks for help.

View attachment 3248

You have a mistake at your calculation.
It is $13^2=169$ instead of $149$.

Therefore, $$\sqrt{13^2-12^2}=\sqrt{169-144}=\sqrt{25}=5$$

 

[OMGMathPLS]

Oh. Well I guess that explains it then. God!

Ok, so you leave the sq rt then at 25 but it's ok to put it back to 5?

 

[mathbalarka]

Ok, so you leave the sq rt then at 25 but it's ok to put it back to 5?

Note that $25 = 5^2$. Thus, by commutativity,

$$\sqrt{25} = \sqrt{5^2} = \sqrt{5}^2 = 5$$

 

[mathmari]

Oh. Well I guess that explains it then. God!

Ok, so you leave the sq rt then at 25 but it's ok to put it back to 5?

Since $5\cdot 5=25 \Rightarrow 5^2=25$ we have that $ \sqrt{5^2}=\sqrt{25}\Rightarrow 5=\sqrt{25}$

 

[ineedhelpnow]

Several of your steps are invalid.

1) $\sqrt{a^2 - b^2} \neq a - b$, i.e., the square root and the difference of squares do not "cancel out". For example, let $a = 5$ and $b = 4$. $a^2 - b^2 = 5^2 - 4^2 = 25 - 16 = 9$ and $\sqrt{5^2-4^2} = \sqrt{9} = 3$ whereas $4 - 3 = 1$.

2) What you wrote in the attachment was $\sqrt{13^2 - 12^2} = 13^2 - 12^2$. Where did the square root go? This is an invalid step.

3) $13^2 = 13 \times 13$ is actually $169$, not $149$. So your calculations are void.

Can you recalculate the expression now that your have been pointed out?

Oh. Well I guess that explains it then. God!

Ok, so you leave the sq rt then at 25 but it's ok to put it back to 5?

He was really only trying to help without giving direct answers (Blush)

 

[OMGMathPLS]

He was really only trying to help without giving direct answers (Blush)

Yeah, thanks for the help. It's just embarrassing. I really thought I partially figured it out. So they explained the arithmetic error and it's ok to solve it all the way to a 5 not simplify. Just a dumb mistake I made and I'm praying it won't kill me later. Thanks.

 

[john1231]

You can do this problem in this manner also
Hint: a^2-b^2 =(a+b)(a-b)
13^2-12^2=(13+12)(13-12)
=25*1