The real roots of the equation \[{x^{2/3}} + {x^{1/3}} - 2 = 0\] are
A. 1, 8
B. \[ - 1\], \[ - 8\]
C. \[ - 1\], 8
D. 1, \[ - 8\]
Answer
251.4k+ views
Hint: Here, we will first take \[{x^{1/3}} = y\] in the given equation and then find the real roots of the obtained equation by factorization. Then we will put the value of \[y\] back to find the real roots of the given equation.
Complete step by step answer:
It is given that the equation is \[{x^{2/3}} + {x^{1/3}} - 2 = 0\].
Taking \[{x^{1/3}} = y\] in the above equation, we get
\[{y^2} + y - 2 = 0\]
We will now factor the above equation to find the root of the equation.
Factoring the above equation to find the value of \[y\], we get
\[
\Rightarrow {y^2} - y + 2y - 2 = 0 \\
\Rightarrow y\left( {y - 1} \right) + 2\left( {y - 1} \right) = 0 \\
\Rightarrow \left( {y + 2} \right)\left( {y - 1} \right) = 0 \\
\]
Taking \[y + 2 = 0\] and \[y - 1 = 0\] in the above equation, we get
\[ \Rightarrow y + 2 = 0\] or \[y - 1 = 0\]
\[ \Rightarrow y = - 2\] or \[y = 1\]
Replacing \[{x^{1/3}}\] for \[y\] in these above equations, we get
\[ \Rightarrow {x^{1/3}} = 1\] or \[{x^{1/3}} = - 2\]
Taking the square in the above equations, we get
\[
\Rightarrow x = {\left( 1 \right)^3} \\
\Rightarrow x = 1 \\
\] or \[
x = {\left( { - 2} \right)^3} \\
x = - 8 \\
\]
Thus, we have found that the real roots of the given equation are 1 and \[ - 8\].
Hence, the option D is correct.
Note: In this question, the equation can also be solved using the quadratic formula. Students should also know the concept of real roots before solving this question. Also, we are supposed to write the values properly to avoid any miscalculation.
Complete step by step answer:
It is given that the equation is \[{x^{2/3}} + {x^{1/3}} - 2 = 0\].
Taking \[{x^{1/3}} = y\] in the above equation, we get
\[{y^2} + y - 2 = 0\]
We will now factor the above equation to find the root of the equation.
Factoring the above equation to find the value of \[y\], we get
\[
\Rightarrow {y^2} - y + 2y - 2 = 0 \\
\Rightarrow y\left( {y - 1} \right) + 2\left( {y - 1} \right) = 0 \\
\Rightarrow \left( {y + 2} \right)\left( {y - 1} \right) = 0 \\
\]
Taking \[y + 2 = 0\] and \[y - 1 = 0\] in the above equation, we get
\[ \Rightarrow y + 2 = 0\] or \[y - 1 = 0\]
\[ \Rightarrow y = - 2\] or \[y = 1\]
Replacing \[{x^{1/3}}\] for \[y\] in these above equations, we get
\[ \Rightarrow {x^{1/3}} = 1\] or \[{x^{1/3}} = - 2\]
Taking the square in the above equations, we get
\[
\Rightarrow x = {\left( 1 \right)^3} \\
\Rightarrow x = 1 \\
\] or \[
x = {\left( { - 2} \right)^3} \\
x = - 8 \\
\]
Thus, we have found that the real roots of the given equation are 1 and \[ - 8\].
Hence, the option D is correct.
Note: In this question, the equation can also be solved using the quadratic formula. Students should also know the concept of real roots before solving this question. Also, we are supposed to write the values properly to avoid any miscalculation.
Recently Updated Pages
JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Isoelectronic Definition in Chemistry: Meaning, Examples & Trends

Ionisation Energy and Ionisation Potential Explained

Iodoform Reactions - Important Concepts and Tips for JEE

Introduction to Dimensions: Understanding the Basics

Instantaneous Velocity Explained: Formula, Examples & Graphs

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Hybridisation in Chemistry – Concept, Types & Applications

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Electric Field of a Uniformly Charged Ring

Derivation of Equation of Trajectory Explained for Students

Other Pages
JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Electron Gain Enthalpy and Electron Affinity Explained

Understanding the Angle of Deviation in a Prism

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

