[tex] \sqrt{x + 2} + \sqrt{3 - x} = 3 \\ x + 2 + 2 \sqrt{(x + 2)(3 - x)} + 3 - x = 9 \\ 5 + 2 \sqrt{( x + 2)(3 - x)} = 9 \\ 2 \sqrt{(x + 2)(3 - x)} = 4 \\ \sqrt{(x + 2)(3 - x)} = 2 \\ (x + 2)(3 - x) = 4 \\ 3x - {x}^{2} + 6 - 2x - 4 = 0 \\ - {x}^{2} + x + 2 = 0 \\ {x}^{2} - x - 2 = 0 \\ D = ( - 1) {}^{2} - 4 \times ( - 2) = 1 + 8 = 9 \\ x_{1} = \frac{1 - 3}{2} = - \frac{2}{2} = - 1 \\x _{2} = \frac{1 + 3}{2} = \frac{4}{2} = 2[/tex]
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[tex] \sqrt{x + 2} + \sqrt{3 - x} = 3 \\ x + 2 + 2 \sqrt{(x + 2)(3 - x)} + 3 - x = 9 \\ 5 + 2 \sqrt{( x + 2)(3 - x)} = 9 \\ 2 \sqrt{(x + 2)(3 - x)} = 4 \\ \sqrt{(x + 2)(3 - x)} = 2 \\ (x + 2)(3 - x) = 4 \\ 3x - {x}^{2} + 6 - 2x - 4 = 0 \\ - {x}^{2} + x + 2 = 0 \\ {x}^{2} - x - 2 = 0 \\ D = ( - 1) {}^{2} - 4 \times ( - 2) = 1 + 8 = 9 \\ x_{1} = \frac{1 - 3}{2} = - \frac{2}{2} = - 1 \\x _{2} = \frac{1 + 3}{2} = \frac{4}{2} = 2[/tex]