Question

(4+radical din 15)^x+(4-radicaldin 15)^x=8

Answer 1

Răspuns

1 , -1

Explicație pas cu pas:

Mai intai hai sa observam ca :

$(4+\sqrt{15})^x = \dfrac{1}{(4-\sqrt{15})^x}$

Prin urmare:

$(4+\sqrt{15})^x+ \dfrac{1}{(4+\sqrt{15})^x}=8$

Facem notatia : $(4+\sqrt{15})^x=t , t>0$

Ecuatia devine:

$t+\dfrac{1}{t}=8 |\cdot t\\t^2+1=8t\\t^2-8t+1=0\\\Delta= 64-4= 60\Rightarrow \sqrt{\Delta}=2\sqrt{15}\\t_1= \dfrac{8+2\sqrt{15}}{2}=4+\sqrt{15}\\t_2=\dfrac{8-2\sqrt{15}}{2}=4-\sqrt{15}$

Se observa ca amandoua solutii sunt convenabile.

$(4+\sqrt{15})^x=4+\sqrt{15} \Rightarrow x=1\\(4+\sqrt{15})^x=4-\sqrt{15}\Rightarrow x=-1\\S:x\in \{\pm 1\}$