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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]

 

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