Question

Resolva a equação irracional:
$\sqrt{2x + 1 } - \sqrt{x - 3} =2$
me ajudem pfvr com cálculo ​

Answer 1

Resposta:

$\textsf{Leia abaixo}$

Explicação passo a passo:

$\mathsf{\sqrt{2x + 1} - \sqrt{x - 3} = 2}$

$\mathsf{\sqrt{2x + 1} = 2 + \sqrt{x - 3}}$

$\mathsf{(\sqrt{2x + 1})^2 = (2 + \sqrt{x - 3})^2}$

$\mathsf{2x + 1 = 4 + 4\sqrt{x - 3} + (x - 3)}$

$\mathsf{x = 4\sqrt{x - 3}}$

$\mathsf{(x)^2 = (4\sqrt{x - 3})^2}$

$\mathsf{x^2 = 16(x - 3)}$

$\mathsf{x^2 = 16x - 48}$

$\mathsf{x^2 - 16x + 48 = 0}$

$\mathsf{x^2 - 16x + 48 + 16 = 0 + 16}$

$\mathsf{x^2 - 16x + 64 = 16}$

$\mathsf{(x - 8)^2 = 16}$

$\mathsf{(x - 8) = \pm\:\sqrt{16}}$

$\mathsf{(x - 8) = \pm\:4}$

$\mathsf{x' = 4 + 8 = 12}$

$\mathsf{x'' = -4 + 8 = 4}$

$\boxed{\boxed{\mathsf{S = \{12;4\}}}}$

Answer 2

$\boxed{\begin{array}{l}\sf \sqrt{2x+1}-\sqrt{x-3}=2\\\sf \sqrt{2x+1}=2+\sqrt{x-3}\\\sf (\sqrt{2x+1})^2=(2+\sqrt{x-3})^2\\\sf 2x+1=4+4\sqrt{x-3}+x-3\\\sf 2x-x+1-4+3=4\sqrt{x-3}\\\sf x=4\sqrt{x-3}\\\sf x^2=(4\sqrt{x-3})^2\\\sf x^2=16\cdot(x-3)\\\sf x^2=16x-48\\\sf x^2-16x+48=0\end{array}}$

$\boxed{\begin{array}{l}\sf\Delta=b^2-4ac\\\sf\Delta=(-16)^2-4\cdot1\cdot48\\\sf\Delta=256-192\\\sf\Delta=64\\\sf x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\\\sf x=\dfrac{-(-16)\pm\sqrt{64}}{2\cdot1}\\\\\sf x=\dfrac{16\pm8}{2}\begin{cases}\sf x_1=\dfrac{16+8}{2}=\dfrac{24}{2}=12\\\\\sf x_2=\dfrac{16-8}{2}=\dfrac{8}{2}=4\end{cases}\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\rm verificac_{\!\!,}\tilde ao\!:}\\\sf para~x=12\\\sf \sqrt{2\cdot12+1}-\sqrt{12-3}\\\sf \sqrt{24+1}-\sqrt{9}\\\sf\sqrt{25}-\sqrt{9}=5-3=2\\\sf x=12~\acute e~soluc_{\!\!,}\tilde ao.\\\sf para~x=4\\\sf \sqrt{2\cdot4+1}-\sqrt{4-3}\\\sf\sqrt{9}-\sqrt{1}\\\sf3-1=2\\\sf x=4~\acute e~soluc_{\!\!,}\tilde ao\\\sf S=\{12,4\}\end{array}}$