Question

x^2+2x+1=3+2 radical din 2

Answer 1

(x+1)²=(√2+1)²

x+1=√2+1...x=√2

x+1= -√2-1....x=-2-√2

S= {-2-√2; √2}, care verifica ambele

verificare

(-2-√2+1)²=(-1-√2)²=(1+√2)²=1+2√2+2=3+2√2

(√2+1)²=2+2√2+1=3+2√2

Answer 2

${x}^{2} + 2x + 1 = 3 + 2 \sqrt{2}$

${x}^{2} + 2x + 1 - 3 - 2 \sqrt{2} = 0$

${x}^{2} + 2x - 2 - 2 \sqrt{2} = 0$

$a = 1$

$b = 2$

$c = - 2 - 2 \sqrt{2}$

$\Delta = {b}^{2} - 4ac$

$\Delta = {2}^{2} - 4 \times 1 \times ( - 2 - 2 \sqrt{2} )$

$\Delta = 4 - 4( - 2 - 2 \sqrt{2} )$

$\Delta = 4 + 8 + 8 \sqrt{2}$

$\Delta = 12 + 8 \sqrt{2}$

$x_{1,2}=\frac{-b\pm\sqrt{\Delta}}{2a}$

$x_{1,2}=\frac{-2\pm\sqrt{12 + 8 \sqrt{2} }}{2 \times 1}$

$x_{1,2}=\frac{-2\pm\sqrt{12 + 8 \sqrt{2} }}{2}$

$x_{1,2}= - 1\pm\sqrt{12 + 8 \sqrt{2} }$

$x_{1}= - 1 + \sqrt{12 + 8 \sqrt{2} }$

$x_{2}= - 1 - \sqrt{12 + 8 \sqrt{2} }$