I have my Algebra 1 SOL tomorrow. Stressed and worried are understatements. I never really understood how to find the square root of numbers and variables with exponents.. Can anyone give me some examples of how to solve this type of* problem?
Preparing for End of Course test..
- Thread starter Dannielle
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chalaman1403
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Dannielle,
You are not being very specific with your question, but let me give you some hints on the topic.
First, radicals (i.e square roots, third roots, ect) are really exponents, fraction exponents. And in many situations it helps to see them as such.
The general identity is \(\displaystyle ^n\sqrt{a^m} = a^{m/n}\). And all the same properties that apply for exponents, aslo apply for radicals.
For example that nested exponents get multiplied, this is \(\displaystyle (a^n)^m = a^{n*m}\) and exponents get distributed in multiplications and divisions, this is
\(\displaystyle (a*b)^n = a^n*b^n\) and \(\displaystyle (a/b)^n= a^n/b^n\). Using these properties, and this other ones \(\displaystyle a^{-n} = 1/a^n\) and \(\displaystyle a^n*a^m= a^{n+m}\) you can work out almost any expression with radicals, say you have for example:
\(\displaystyle \sqrt{4a^6b^8}\)
if you write the sqrt as a 1/2 exponent (because square root's order is 2, we never write it though) you have
\(\displaystyle (4a^6b^8)^{1/2}\) and you can disrtibute the exponent because the terms inside are multiplied, you have then
\(\displaystyle 4^{1/2}*(a^6)^{1/2}*(b^8)^{1/2}\). You can always go back to writing the square root if you feel more comfortable.
\(\displaystyle 4^{1/2}= \sqrt4 = 2\)
\(\displaystyle (a^6)^{1/2}\) for this we use that nested exponents get multiplied, so we have 6*1/2 that is 3.
And similar for the \(\displaystyle (b^8)^{1/2}\) that is \(\displaystyle b^4\)
so the simplified form would be
\(\displaystyle 2a^3b^4\)
I hope that helped, feel free to ask again if you have any question about that.
Regards,
Damián Vallejo
You are not being very specific with your question, but let me give you some hints on the topic.
First, radicals (i.e square roots, third roots, ect) are really exponents, fraction exponents. And in many situations it helps to see them as such.
The general identity is \(\displaystyle ^n\sqrt{a^m} = a^{m/n}\). And all the same properties that apply for exponents, aslo apply for radicals.
For example that nested exponents get multiplied, this is \(\displaystyle (a^n)^m = a^{n*m}\) and exponents get distributed in multiplications and divisions, this is
\(\displaystyle (a*b)^n = a^n*b^n\) and \(\displaystyle (a/b)^n= a^n/b^n\). Using these properties, and this other ones \(\displaystyle a^{-n} = 1/a^n\) and \(\displaystyle a^n*a^m= a^{n+m}\) you can work out almost any expression with radicals, say you have for example:
\(\displaystyle \sqrt{4a^6b^8}\)
if you write the sqrt as a 1/2 exponent (because square root's order is 2, we never write it though) you have
\(\displaystyle (4a^6b^8)^{1/2}\) and you can disrtibute the exponent because the terms inside are multiplied, you have then
\(\displaystyle 4^{1/2}*(a^6)^{1/2}*(b^8)^{1/2}\). You can always go back to writing the square root if you feel more comfortable.
\(\displaystyle 4^{1/2}= \sqrt4 = 2\)
\(\displaystyle (a^6)^{1/2}\) for this we use that nested exponents get multiplied, so we have 6*1/2 that is 3.
And similar for the \(\displaystyle (b^8)^{1/2}\) that is \(\displaystyle b^4\)
so the simplified form would be
\(\displaystyle 2a^3b^4\)
I hope that helped, feel free to ask again if you have any question about that.
Regards,
Damián Vallejo
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