x²+(√2+1)x+√2=0 Помогите, пожалуйста!. Created by Sweets555. algebra-ru

x²+(√2+1)x+√2=0 Помогите, пожалуйста!...

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$x^2+( \sqrt{2}+1)x+ \sqrt{2}=0 \\ D=( \sqrt{2}+1)^2-4 \sqrt{2}=2+2 \sqrt{2}+1-4 \sqrt{2}=2-2 \sqrt{2}+1=( \sqrt{2}-1)^2 \\ x_1= \dfrac{- \sqrt{2}-1+ \sqrt{2}-1 }{2}=- \dfrac{2}{2}=-1 \\ x_2= \dfrac{- \sqrt{2}-1- \sqrt{2}+1 }{2}= \dfrac{-2 \sqrt{2}}{2}=- \sqrt{2}$

Ответ: $- \sqrt{2}; -1$

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Sweets555 Большое спасибо!

TARTILLO

Verified answer

X²+(√2+1)x+√2=0

x1=-1
x2=-√2 по теореме Виетта

x1=[-(√2+1)-√((√2+1)²-4·√2)]/2 =[-(√2+1)-(√2-1)]/2= -√2
x2=[-(√2+1)+√((√2+1)²-4·√2)]/2 =[-(√2+1)+(√2-1)]/2= -1

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Sweets555 Спасибо)

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