Pentru a rezolva asemenea ecuatii, trebuie respectati urmatorii pasi:
- Se pun conditiile de existenta (C.E.) pentru numerele de sub radicali
- Se ridica la putere, pentru a scapa de radicali
- Se rezolva ecuatia si se afla x
- Se verifica daca x indeplineste conditiile de existenta stabilite
- Se noteaza solutia ecuatiei
Punctul A
[tex]\displaystyle{ \sqrt{2x -1} = 3 }[/tex]
CE: [tex]\displaystyle{ 2x - 1 \geqslant 0 \rightarrow 2x \geqslant 1 \rightarrow x \geqslant \frac{1}{2} \rightarrow x \in [0,5, \infty) }[/tex]
[tex]\displaystyle{ (\sqrt{2x-1})^{2} = 3^{2} }[/tex]
[tex]\displaystyle{ 2x - 1 = 9}[/tex]
[tex]\displaystyle{ 2x = 10 }[/tex]
[tex]\displaystyle{ x = 5 }[/tex]
[tex]\displaystyle{ 5 \in [0,5, \infty) \rightarrow 5 \in S }[/tex]
S = {5}
Punctul B
[tex]\displaystyle{ \sqrt[3]{2x - 1} = 3 }[/tex]
CE: [tex]\displaystyle{ 2x -1 \in R \rightarrow x \in R }[/tex]
[tex]\displaystyle{ (\sqrt[3]{2x-1})^{3} = 3^{3} }[/tex]
[tex]\displaystyle{ 2x - 1 = 27 }[/tex]
[tex]\displaystyle{ 2x = 28 }[/tex]
[tex]\displaystyle{ x = 14 }[/tex]
[tex]\displaystyle{ 14 \in R \rightarrow 14 \in S }[/tex]
S = {14}
Punctul C
[tex]\displaystyle{ \sqrt{2x + 3} = 4x }[/tex]
CE: [tex]\displaystyle{ 2x + 3 \geqslant 0 \rightarrow 2x \geqslant -3 \rightarrow x \geqslant -\frac{3}{2} \rightarrow x \in [-1,5, \infty) }[/tex]
[tex]\displaystyle{ (\sqrt{2x + 3})^{2} = (4x)^{2} }[/tex]
[tex]\displaystyle{ 2x + 3 = 16x^{2} }[/tex]
[tex]\displaystyle{ 16x^{2} - 2x - 3 = 0 }[/tex]
Observam ca avem o ecuatie de gradul al doilea.
[tex]\displaystyle{ a = 16, b = -2, c = -3 }[/tex]
[tex]\displaystyle{ \Delta = b^{2} - 4ac = 4 + 4 \cdot 16 \cdot 3 }[/tex]
[tex]\displaystyle{ \Delta = 4 + 4 \cdot 48 = 4 + 192 = 196}[/tex]
Δ > 0, deci avem doua radacini reale.
[tex]\displaystyle{ x_{1} = \frac{-b + \sqrt{\Delta}}{2a} = \frac{2 + 14}{32} = \frac{1}{2} }[/tex]
[tex]\displaystyle{ x_{2} = \frac{-b - \sqrt{\Delta}}{2a} = \frac{2-14}{32} = -\frac{3}{8} }[/tex]
[tex]\displaystyle{ \frac{1}{2} \in [-1,5, \infty) \rightarrow \frac{1}{2} \in S }[/tex]
[tex]\displaystyle{ -\frac{3}{8} \notin [-1,5, \infty) \rightarrow -\frac{3}{8} \notin S }[/tex]
S = {[tex]\dfrac{1}{2}[/tex]}
Punctul D
[tex]\displaystyle{ \sqrt[3]{x + 2\sqrt{2} } = \sqrt{2} }[/tex]
CE: [tex]\displaystyle{ x + 2\sqrt{2} \in R \rightarrow x \in R }[/tex]
[tex]\displaystyle{ ( \sqrt[3]{ x + 2\sqrt{2} } )^{3} = (\sqrt{2})^{3} }[/tex]
[tex]\displaystyle{ x + 2\sqrt{2} = 2\sqrt{2} }[/tex]
[tex]\displaystyle{ x = 0 }[/tex]
[tex]\displaystyle{ 0 \in R \rightarrow 0 \in S }[/tex]
S = {0}
Punctul E
[tex]\displaystyle{ \sqrt{3-x} = -2x }[/tex]
CE: [tex]\displaystyle{ 3 - x \geqslant 0 \rightarrow x \leqslant 3 \rightarrow x \in (-\infty, 3] }[/tex]
[tex]\displaystyle{ (\sqrt{3-x})^{2} = (-2x)^{2} }[/tex]
[tex]\displaystyle{ 3 - x = 4x^{2} }[/tex]
[tex]\displaystyle{ -4x^{2} -x + 3 = 0 }[/tex]
[tex]\displaystyle{ a = -4, b = -1, c = 3 }[/tex]
[tex]\displaystyle{ \Delta = b^{2} - 4ac = 1 + 4 \cdot 4 \cdot 3 }[/tex]
[tex]\displaystyle{ \Delta= 1 + 16 \cdot 3 = 1 + 48 = 49 }[/tex]
Δ > 0 deci avem doua radacini reale.
[tex]\displaystyle{ x_{1} = \frac{-b + \sqrt{\Delta}}{2a} = \frac{1 + 7}{-8}= \frac{8}{-8} = -1 }[/tex]
[tex]\displaystyle{ x_{2} = \frac{-b -\sqrt{\Delta}}{2a} = \frac{1 - 7}{-8} = \frac{-6}{-8} = \frac{3}{4} }[/tex]
Iar acum, este foarte important sa facem verificarea, deoarece la inceput am ridicat la patrat un numar negativ (-2x), ceea ce inseamna ca s-ar putea sa avem solutii false.
Dupa ce am facut verificarea, va rezulta ca singura solutie este x = -1.
S = {-1}
