Решите уравнение: корень из (x^2+3x-10)+корень из (x^2-10x+16)=0.

\[\sqrt{x^{2} + 3x - 10} \geq 0;\ \ \]

\[\sqrt{x^{2} - 10x + 16} \geq 0 \Longrightarrow сумма\ \]

\[равна\ 0,\ если\ оба\ корня\ \]

\[равны\ 0.\]

\[x^{2} + 3x - 10 = 0\]

\[D = 3^{2} - 4 \cdot 1 \cdot ( - 10) =\]

\[= 9 + 40 = 49\]

\[x_{1} = \frac{- 3 + \sqrt{49}}{2} = \frac{- 3 + 7}{2} =\]

\[= \frac{4}{2} = 2\]

\[x_{2} = \frac{- 3 - \sqrt{49}}{2} = \frac{- 3 - 7}{2} =\]

\[= - \frac{10}{2} = - 5\]

\[x^{2} - 10x + 16 = 0\]

\[D = ( - 10)^{2} - 4 \cdot 1 \cdot 16 =\]

\[= 100 - 64 = 36\]

\[x_{1} = \frac{10 + \sqrt{36}}{2} = \frac{10 + 6}{2} =\]

\[= \frac{16}{2} = 8\]

\[x_{2} = \frac{10 - \sqrt{36}}{2} = \frac{10 - 6}{2} =\]

\[= \frac{4}{2} = 2\]

\[Ответ:x = 2.\]