Question

$\ \textless \ br /\ \textgreater \ x + \sqrt{x} + 1 = 5$
Resolva a equação irracional nos conjuntos

dos reais.​

Answer 1

Resposta:

$\\\boxed{\boxed{\underbrace{x2=\dfrac{9-\sqrt{17} }{2} }}}$

Explicação passo-a-passo:

$x+\sqrt{x} +1=5\\x+\sqrt{x} =5-1\\x+\sqrt{x} =4\\(\sqrt{x}) ^{2}=(4-x)^2\\x=16-8x+x^2\\x^2-9x+16=0 \\ \\\Delta= b^2-4ac\\\Delta=(-9)^2-4(1)(16)\\\Delta=81- 64\\\Delta=17\\\\x=\dfrac{-b+-\sqrt{\Delta} }{2a} \\x=\dfrac{-(-9)+-\sqrt{17} }{2} \\\boxed{\boxed{\underbrace{x1=\dfrac{9+\sqrt{17} }{2}}}} \\\boxed{\boxed{\underbrace{x2=\dfrac{9-\sqrt{17} }{2} }}}$

Verificando =>>

$\dfrac{9+\sqrt{17} }{2} +\sqrt{\dfrac{9+\sqrt{17} }{2} } +1=5\\4,5+\dfrac{\sqrt{17}}{2} + \dfrac{3+\sqrt[4]{17} }{\sqrt{2} } =4\\4,5+\dfrac{\sqrt{17}}{2} + \dfrac{3+\sqrt[4]{17} }{\sqrt{2} } =4\\Falso$

Veja que 4,5 já é maior que 4, refutando portanto a igualdade.

=>>

$\dfrac{9-\sqrt{17} }{2} +\sqrt{\dfrac{9-\sqrt{17} }{2} } +1=5\\\dfrac{9-\sqrt{17} }{2} +\dfrac{\sqrt{9-\sqrt{17} }\times\sqrt{2} }{\sqrt{2} \times\sqrt{2} } +1=5\\\dfrac{9-\sqrt{17} }{2} +\dfrac{\sqrt{(9-\sqrt{17}) \times2} }{2 } +1=5\\\dfrac{9-\sqrt{17} }{2} +\dfrac{\sqrt{(9-\sqrt{17}) \times2} }{2 } +1=5\\\dfrac{9-\sqrt{17} }{2} +\dfrac{\sqrt{(18-2\sqrt{17}) } }{2 } +1=5 \\\dfrac{9-\sqrt{17} }{2} +\dfrac{\sqrt{(1-\sqrt{17})^2 } }{2 } +1=5$

$\dfrac{9-\sqrt{17} }{2} +\dfrac{\sqrt{(1-\sqrt{17})^2 } }{2 } +1=5 \\\dfrac{9-\sqrt{17} }{2} +\dfrac{(\sqrt{17}-1) }{2 } +1=5 \\4,5-\dfrac{\sqrt{17} }{2} +\dfrac{\sqrt{17} } {2 } -\dfrac{1}{2} +1=5\\4,5-\dfrac{1}{2} +1=5\\\boxed{\boxed{\Rightarrow 5=5}}$