Question

sa se rezolve ecuatia
$2 \sqrt{ \frac{2x + 1}{x - 3} } - 3 \sqrt{ \frac{x - 3}{2x + 1} } = 5$
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Answer 1

$\displaystyle{ fie \ \frac{2x+1}{x-3} = t, \ t \geqslant 0 }$

$\displaystyle{ 2t - 3t^{-1} = 5 \rightarrow 2t - \frac{3}{t} = 5 }$

$\displaystyle{ \frac{2t^{2} - 3}{t} = 5 \rightarrow 2t^{2} - 3 = 5t }$

$\displaystyle{2t^{2} - 5t - 3 = 0 \rightarrow a = 2, \ b = -5, \ c = -3 }$

$\displaystyle{ \Delta = b^{2} -4ac = (-5)^{2} + 4 \cdot 2 \cdot 3 }$

$\displaystyle{ \Delta = 25 + 4 \cdot 6 = 25 + 24 = 49 }$

$\displaystyle{ t_{1} = \frac{-b+\sqrt{\Delta}}{2a} = \frac{5 +7}{4} = \frac{12}{4} = 3 }$

$\displaystyle{ t_{2} = \frac{-b-\sqrt{\Delta}}{2a} = \frac{5 - 7}{4} = \frac{-2}{4} = -0,5 }$ nu convine deoarece $\displaystyle{ t \geqslant 0 }$

$\displaystyle{ \frac{2x+1}{x-3} = 3 \rightarrow 2x + 1 = 3x - 9 }$

$\displaystyle{ x - 9 = 1 \rightarrow x = 10 }$