Question

Care sunt solutiile solutiile ecuatiei?

Answer 1

$\sqrt{2x-1}+\sqrt{10-x} = 4\\ \\ \\ x\geq \dfrac{1}{2}\\ \\ \text{Notam }a = \sqrt{2x-1},~~b = \sqrt{10-x},\quad a\geq 0, b\geq 0\\ \\a+b = 4 \\ a^2+2b^2 = 2x-1+2(10-x) = 19 \\ \\ a+b=4 \\ a^2+2b^2 = 19 \\ \\ b = 4-a \\ a^2+2(4-a)^2=19 \\ \\ \Rightarrow a^2+2(a^2-8a+16)=19 \Rightarrow 3a^2-16a+13 = 0 \\ \\ \Delta =16^2-4\cdot 3\cdot 13 = 16\cdot 16-4\cdot 39 = 4(64-39) = 4\cdot 25 = 10^2$

$a_{1,2} = \dfrac{16\pm 10}{6}\Rightarrow a_1 = 1,\quad a_2 = \dfrac{13}{3} \\ \\ (1)\quad a_1 = 1\Rightarrow b = 4-1 = 3 > 0 \\\\ \Rightarrow \sqrt{2x-1}= 1 \Rightarrow \boxed{x = 1}\\ \\(2)~~a_2 = \dfrac{13}{3} \Rightarrow b = 4-\dfrac{13}{3} < 0 \quad (F) \\ \\ \Rightarrow S = \Big\{1\Big\}$

Answer 2

Explicație pas cu pas:

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