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a. Kevin said that if the index of a radical is even and the radicand is positive, then the radical has two real roots. Do you agree with Kevin? Explain why or why not.b. Kevin said that if the index of a radical is odd, then the radical has one real root and that the root is positive if the radicand is positive and negative if the radicand is negative. Do you agree with Kevin? Explain why or why not.

$\sqrt[n]{k}=x^{n} o r-x^{n}$$\sqrt[n]{k}=x^{n} : \sqrt[n]{-k}=-x^{n}$

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Chapter 3

REAL NUMBERS AND RADICALS

Section 2

Roots and Radicals

Whole which of Numbers

Fractions and Mixed Numbers

Decimals

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Kevin says that if the index of a radicals, even in the Radicals, positive that we get to solutions, And for that we say Kevin is absolutely correct. Okay, here's why Kevin is correct on this with RN Through of K, we get to solutions. Here's a real example. If N is even like to in case positive, like nine, we have the square root of nine, and we know that the square root of nine is either. Three. You're negative three for part B. Uh, coming says that if the Radic, if the index is odd in the radical, has a Onley one real root. And for this we should say that Kevin is correct again. Here's an example. Why again, with the end through of K in this instance, either we have one route either a positive or a negative root. Let's take an example of the cube root of eight. So N is odd, and K is positive. The cube root of eight is positive, too, whereas if K is negative, like the Cuba of negative eight that is equal to negative two. So we get only one solution. It depends on if K is positive or K is negative

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