Question

Deteeminati nr rationale a si b, stiind ca
$( \sqrt{2}+ \sqrt{6})^2 =a-b \sqrt{3}$

Answer 1

[tex]\it (\sqrt2+\sqrt6)^2 = (\sqrt2)^2+2\sqrt2\sqrt6+(\sqrt6)^2 =2+2\sqrt{12}+6= \\\;\\ =8+2\sqrt{4\cdot3} =8+2\cdot2\sqrt3 =8+4\sqrt3 [/tex]

Ecuația din enunț devine:


[tex]\it 8+4\sqrt3 = a-b\sqrt3 \Rightarrow \begin{cases}\it a=8 \\\;\\ \it -b=4|_{\cdot(-1)} \Rightarrow b = -4\end{cases}[/tex]

Answer 2

(√2+√6)^2=a-b√3
(√2+√6)^2=(√2)^2+2×√2×√6+(√6)^2=
√4+2√12+√36=2+2×2√3+6=
2+6+4√3=8+4√3
=>8+4√3=a-b√3
=>a=8
=>-b√3=4√3 |:√3
-b=4 |×(-1)
b= -4