paala paala
  • 20-09-2018
  • Mathematics
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What is the solution of (4x+3)^2=18

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gmany
gmany gmany
  • 20-09-2018

[tex](4x+3)^2=18\iff4x+3=\pm\sqrt{18}\\\\4x+3=-\sqrt{9\cdot2}\ \vee\ 4x+3=\sqrt{9\cdot2}\\\\4x+3=-\sqrt9\cdot\sqrt2\ \vee\ 4x+3=\sqrt9\cdot\sqrt2\\\\4x+3=-3\sqrt2\ \vee\ 4x+3=3\sqrt2\ \ \ \ \ \ \ \ |-3\\\\4x=-3-3\sqrt2\ \vee\ 4x=-3+3\sqrt2\ \ \ \ \ \ \ |:4\\\\\boxed{x=\dfrac{-3-3\sqrt2}{4}\ \vee\ x=\dfrac{-3+3\sqrt2}{4}}[/tex]

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