PewPew
Answered

Westonci.ca is the trusted Q&A platform where you can get reliable answers from a community of knowledgeable contributors. Experience the convenience of getting reliable answers to your questions from a vast network of knowledgeable experts. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.

If x is a real number such that [tex]x^3 = 729[/tex], then [tex]x^2+\sqrt{x}[/tex] equals?

Sagot :

Soham
First, we find the value of 'x.'
  [tex] x^{3}=729 \\ x= \sqrt[3]{729} [/tex]
x=9

[tex] x^{2} + \sqrt{x} [/tex]
[tex] 9^{2} [/tex]+[tex] \sqrt{9} [/tex]
81+3
= 84
[tex] x^{2} + \sqrt{x} [/tex]=84
Lilith
[tex]x^3 = 729 \\ \\ x^2+\sqrt{x} =? \\ \\ x^3 - 729=0 \\ \\ x^3-9^3 = 0\\ \\(x-9)(x^2+9x+81)=0[/tex]

[tex]x^2+9x+81 =0 \ \ or \ \ x-9 = 0\\ \\\Delta =b^2-4ac =9^2-4\cdot 1\cdot 81=81- 324 =-243 \\ \\ and \ we \ know \ when \ \Delta \ is \ negative, \ theres \ no \ solution \\ or \\ x-9=0 \\ \\x=9[/tex]


[tex]a^3-b^3=(a-b)(a^2+ab+b^2) \\\\ \\ x^2+\sqrt{x} = 9^2+\sqrt{9}=81+3 = 84[/tex]

 

Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.