Addition of radicals are treated similar to polynomials, but instead of multiples of x's and x2's, we have multiples of things that look very much like root x and root x - 2
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Addition of radicals are treated similar to polynomials, but instead of multiples of x's and x2's, we have multiples of things that look very much like root x and root x - 2. When adding such expressions together, no arithmetic can be done underneath the radical. However, like radicals can be combined
together.
Question 1)
-7 root of (2r-1) + 3 root of (2r-1)
=(-7+3) *root of (2r-1)
=-4 root of (2r-1)
Question 2)
8 root of (3y-2) + 2 root of (3y-2)
=(8+2) *root of (3y-2)
=10 root of (3y-2)
I believe you are on the right track but you didn't quite say what it means for use to have like radicals.
What parts of the radical must be the same in order for us to have like radicals? Do the coefficients need to be the same?
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Solution Summary
The solution solves the radical problems.
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In Question 1, we have two terms added together. One is -7 root of (2r-1), and the other is 3 root of (2r-1). We can think of the radical "root of ...
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- BSc , Wuhan Univ. China
- MA, Shandong Univ.
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- "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
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