Ok so I'm suppose to do this
Square root of 6
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Square root of 2
I know when you divide a radical over radical and they have the same what ever it's called (the same number outside of the radical that determines if it's cube root squared root etc.) you can just divide the number and put it in the radical right?
So I can just do square root 3 because 6/2 is three
the only problem is that my text book says to put it into exponential form first so i get
6^(1/2)
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2^(1/2)
now i divide and i get three and when exponenet over exponent you subtrac them so 1/2 - 1/2 = 0?!?!?!?!?!?!?!?!
3^0 = 1
so i don't get it please assist me.
sqrt6/sqrt2 = 6^(1/2)/2*(1/2) = 3^1/2
= sqrt3
I don't understand why they wanted you to put in into exponent form first. You could have written sqrt6/sqrt2 as
sqrt3*sqrt2/sqrt2 and cancelled out the sqrt2
I'd be happy to assist you with your question!
To simplify the expression (square root of 6) divided by (square root of 2), let's break it down step by step using exponential form.
1. Start with the expression (square root of 6) divided by (square root of 2).
2. Rewrite the expression in exponential form: 6^(1/2) divided by 2^(1/2).
3. To divide exponents, subtract the exponent in the denominator from the exponent in the numerator. In this case, (1/2) - (1/2) = 0.
4. Simplify the expression to 6^0.
5. Any number raised to the power of 0 is equal to 1. Therefore, 6^0 = 1.
So the final answer is 1.
You correctly performed all the steps, but the confusion arises from the concept of dividing exponents with the same base. When you divide exponents, the result is not always 0. In this case, the exponents in the numerator and denominator are both 1/2, but their difference is 0, which means the result is not 0.
It's important to remember that when dividing radicals with the same index, you divide the numbers outside the radicals and also simplify the numbers inside the radicals if possible. In this case, the square roots of 6 and 2 can't be further simplified, so the answer is simply 1.
I hope this explanation helps you understand the concept better! Let me know if you have any further questions.