x=x2+11−−−−−−√3−2,x=x2+113−2,x = \root 3 \of {{x^2} + 11} - 2{\text{,}} then the value of x3 + 5x2 + 12x is?" /> x=x2+11−−−−−−√3−2,x=x2+113−2,x = \root 3 \of {{x^2} + 11} - 2{\text{,}} then the value of x3 + 5x2 + 12x is?" /> x=x2+11−−−−−−√3−2,x=x2+113−2,x = \root 3 \of {{x^2} + 11} - 2{\text{,}} then the value of x3 + 5x2 + 12x is?" />

India’s No.1 Educational Platform For UPSC,PSC And All Competitive Exam

Download Our App

If $x = \sqrt[3]{x^{2} + 11} - 2,$ $x = \root 3 \of {{x^2} + 11} - 2{\text{,}}$ then the value of x³ + 5x² + 12x is?

10
23
37
411

Answer:- 1

Explanation:-

Solution:

\begin{aligned} x = \sqrt[3]{x^{2} + 11} - 2 \\ \Rightarrow x + 2 = \sqrt[3]{x^{2} + 11} \\ \Rightarrow Taking cube on both side \\ \Rightarrow {(x + 2)}^{3} = x^{2} + 11 \\ \Rightarrow x^{3} + 8 + 6 x (x + 2) = x^{2} + 11 \\ \Rightarrow x^{3} + 8 + 6 x^{2} + 12 x = x^{2} + 11 \\ \Rightarrow x^{3} + 5 x^{2} + 12 x = 3 \end{aligned}

$\eqalign{ & x = \root 3 \of {{x^2} + 11} - 2 \cr & \Rightarrow x + 2 = \root 3 \of {{x^2} + 11} \cr & \Rightarrow {\text{Taking cube on both side}} \cr & \Rightarrow {\left( {x + 2} \right)^3} = {x^2} + 11 \cr & \Rightarrow {x^3} + 8 + 6x\left( {x + 2} \right) = {x^2} + 11 \cr & \Rightarrow {x^3} + 8 + 6{x^2} + 12x = {x^2} + 11 \cr & \Rightarrow {x^3} + 5{x^2} + 12x = 3 \cr}$

Workspace Report

Post your Comments

then the value of x³ + 5x² + 12x is?", "text": "If

x = \sqrt[3]{x^{2} + 11} - 2,

$x = \root 3 \of {{x^2} + 11} - 2{\text{,}}$ then the value of x³ + 5x² + 12x is?", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" }, "answerCount": "4", "acceptedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} x = \sqrt[3]{x^{2} + 11} - 2 \\ \Rightarrow x + 2 = \sqrt[3]{x^{2} + 11} \\ \Rightarrow Taking cube on both side \\ \Rightarrow {(x + 2)}^{3} = x^{2} + 11 \\ \Rightarrow x^{3} + 8 + 6 x (x + 2) = x^{2} + 11 \\ \Rightarrow x^{3} + 8 + 6 x^{2} + 12 x = x^{2} + 11 \\ \Rightarrow x^{3} + 5 x^{2} + 12 x = 3 \end{aligned}

$\eqalign{ & x = \root 3 \of {{x^2} + 11} - 2 \cr & \Rightarrow x + 2 = \root 3 \of {{x^2} + 11} \cr & \Rightarrow {\text{Taking cube on both side}} \cr & \Rightarrow {\left( {x + 2} \right)^3} = {x^2} + 11 \cr & \Rightarrow {x^3} + 8 + 6x\left( {x + 2} \right) = {x^2} + 11 \cr & \Rightarrow {x^3} + 8 + 6{x^2} + 12x = {x^2} + 11 \cr & \Rightarrow {x^3} + 5{x^2} + 12x = 3 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} x = \sqrt[3]{x^{2} + 11} - 2 \\ \Rightarrow x + 2 = \sqrt[3]{x^{2} + 11} \\ \Rightarrow Taking cube on both side \\ \Rightarrow {(x + 2)}^{3} = x^{2} + 11 \\ \Rightarrow x^{3} + 8 + 6 x (x + 2) = x^{2} + 11 \\ \Rightarrow x^{3} + 8 + 6 x^{2} + 12 x = x^{2} + 11 \\ \Rightarrow x^{3} + 5 x^{2} + 12 x = 3 \end{aligned}

$\eqalign{ & x = \root 3 \of {{x^2} + 11} - 2 \cr & \Rightarrow x + 2 = \root 3 \of {{x^2} + 11} \cr & \Rightarrow {\text{Taking cube on both side}} \cr & \Rightarrow {\left( {x + 2} \right)^3} = {x^2} + 11 \cr & \Rightarrow {x^3} + 8 + 6x\left( {x + 2} \right) = {x^2} + 11 \cr & \Rightarrow {x^3} + 8 + 6{x^2} + 12x = {x^2} + 11 \cr & \Rightarrow {x^3} + 5{x^2} + 12x = 3 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

Test

Classes

E-Book