Question

Restringeti in patrate a)29+12√5 d)89+36√2 b)91-48√3 f)17-4√15 c)73-40√3
e)9+6√2

Answer 1

$29+12\sqrt5 = (\sqrt a+\sqrt b)^2 = a+b+2\sqrt{ab} \\ \\ a+b = 29 \\ 2\sqrt{ab}=12\sqrt{5} \\ \\ a+b = 29 \\ 4ab = 144\cdot 5 \\ \\ a+b = 29 \\ ab = 180 \\ \\ t^2-St+P = 0 \Rightarrow t^2-29t+180 = 0 \\ \Delta = 29^2-4\cdot 180 = 121 = 11^2 \\ t_{1,2} = \dfrac{29-11}{2} \Rightarrow t_1 = 9,\quad t_2 = 20 \\ \\ \Rightarrow a = 9,\quad b = 20 \Rightarrow 29+12\sqrt5 =(\sqrt{9}+\sqrt{20})^2 \Rightarrow \\ \\ \Rightarrow \boxed{29+12\sqrt5 = (3+2\sqrt5)^2}$

La restul faci la fel.