madelineberry
madelineberry madelineberry
  • 26-04-2017
  • Mathematics
contestada

if x-12√x + 36 = 0 what is the value of x

Respuesta :

gmany
gmany gmany
  • 21-07-2018

We use:


[tex]a^2-2ab+b^2=(a-b)^2[/tex]


We have:


[tex]x-12\sqrt{x}+36=0[/tex]


The domain: [tex]x\geq0[/tex]


[tex]x=(\sqrt{x})^2\\\\12\sqrt{x}=2\cdot\sqrt{x}\cdot6\\\\36=6^2[/tex]


Therfore:


[tex]x-12\sqrt{x}+36=0\\\\\underbrace{(\sqrt{x})^2-2\cdot\sqrt{x}\cdot6+6^2}_{a^2-2ab+b^2=(a-b)^2}=0\\\\(\sqrt{x}-6)^2=0\iff\sqrt{x}-6=0\ \ \ \ |+6\\\\\sqrt{x}=6\ \ \ \ |^2\\\\(\sqrt{x})^2=6^2\to\boxed{x=36}[/tex]

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