Question

SUBIECTUL 2 ex 3!
✨dau de toate ✨

Answer 1

Răspuns:

3) 4$x^{2}$-4x = 8+$4\sqrt{2}$

4$x^{2}$-4x-8-$4\sqrt{2}$ = 0 /:4

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

$ax^{2}$+bx+c=0

Formula: x= (-b ± $\sqrt{b^{2}-4ac }$)/2a

x= [-(-1) ± $\sqrt{(-1)^{2}-4*1(-2-\sqrt{2)} }$]/2

x= (1 ± $\sqrt{1+8+4\sqrt{2} }$)/2

x=(1 ± $\sqrt{(1+\sqrt{8})^{2} }$)/2

x= (1 ± 1+$\sqrt{8}$)/2

x= (1 ± 1+$2\sqrt{2}$)/2

x= (1+1+$2\sqrt{2}$)/2

x=  [1-(1+$2\sqrt{2}$)]/2

x1 = 1+$\sqrt{2}$

x2 = $-\sqrt{2}$

Raspuns: B. {$-\sqrt{2}$;$\sqrt{2}+1$}