Question

$\sqrt{2x-3}+6= 5\sqrt[4]{2x-3}$

Answer 1

Notam $\sqrt[4]{2x-3}=y. cu. y. pozitiv ,deci \sqrt{2x-3}= y^{2},$ conditia de existenta a radicalului : 2x-3≥0 adica x ≥ 3/2. Se obtine ecuatia:

$y^{2}-5y+6=0$, Δ=$5^{2}-4*6=1 . deci. radacinile : y_{1}= \frac{5-1}{2}=2 .si. y_{2} = 3,$, ambele pozitive. Vom avea: $\sqrt[4]{2x-3}=2$, ridicam ambii memrii la puterea a 4-a, rezulta 2x-3=16 deci $x_{1}= \frac{19}{2},$, analog , $\sqrt[4]{2x-3}=3, rezulta:2x-3=81, 2x=84, x_{2}= 42$. Ambele . corespund la x ≥ $\frac{3}{2}$