Решите уравнение: (корень из x-4)(12x^2+17x-5)=0.

Ответ:

\[\left( \sqrt{x} - 4 \right)\left( 12x^{2} + 17x - 5 \right) = 0\]

\[\sqrt{x} = 4\]

\[x = 16.\]

\[D = 17^{2} - 4 \cdot 12 \cdot ( - 5) =\]

\[= 289 + 240 = 529\]

\[x_{1} = \frac{- 17 + \sqrt{529}}{2 \cdot 12} = \frac{- 17 + 23}{24} =\]

\[= \frac{6}{24} = \frac{1}{4}\]

\[x_{2} = \frac{- 17 - \sqrt{529}}{2 \cdot 12} = \frac{- 17 - 23}{24} =\]

\[= - \frac{40}{24} = - \frac{5}{3}\]

\[Ответ:x = 16;x = \frac{1}{4};x = - 1\frac{2}{3}.\]