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The equation x - 16x³ = 0 O (A) has two (and only two) distinct real roots O (B) has four (and only four) distinct real roots O (C) has nine distinct real roots O (D) has three (and only three) distinct real roots O(E) cannot be solved

The equation x - 16x³ = 0 O (A) has two (and only two) distinct real roots O (B) has four (and only four) distinct real roots O (C) has nine distinct real roots O (D) has three (and only three) distinct real roots O(E) cannot be solved

Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Power of 25/3 and 1/3
8 answered ed out of 00 Flag question The equation x - 16x = 0 O (A) has two (and only two) distinct real roots O(B) has four (and only four) distinct real roots O (C) has nine distinct real roots O (D) has three (and only three) distinct real roots O (E) cannot be solved Previous Next
Transcribed Image Text:8 answered ed out of 00 Flag question The equation x - 16x = 0 O (A) has two (and only two) distinct real roots O(B) has four (and only four) distinct real roots O (C) has nine distinct real roots O (D) has three (and only three) distinct real roots O (E) cannot be solved Previous Next
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yes but the negative radical 2 is not a real number and in the options they are asking for a real distinct solution

3:38 Solutions Solving Steps The expressionIS UNTUTTined in the set oricar numbers Simplify Undefined 0=0 0=0 x#-√3 0=0 0=0 Since the expression is undefined in the set of real numbers, x= -√3 is not a solution of the equation x= -√3 x=0 x = √3 The equation has 2 solutions Solution x₁ = 0₁x₂ = √3 0, Alt ive Form .4G Next Step 5 X₁−u, X₂~1,73200 X
Transcribed Image Text:3:38 Solutions Solving Steps The expressionIS UNTUTTined in the set oricar numbers Simplify Undefined 0=0 0=0 x#-√3 0=0 0=0 Since the expression is undefined in the set of real numbers, x= -√3 is not a solution of the equation x= -√3 x=0 x = √3 The equation has 2 solutions Solution x₁ = 0₁x₂ = √3 0, Alt ive Form .4G Next Step 5 X₁−u, X₂~1,73200 X
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yes i understand how to solve it and i know we get 3 values but dont you automatically reject the negative because the exponents are fractions?

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but isnt it 2 distinct roots?

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