+817 Find the domain of the function $$f(x)=\sqrt{-10x^2-11x+6+9x^2-5x}.$$
+121 To find the domain of the function \(f(x) = \sqrt{-10x^2-11x+6+9x^2-5x}\), we need to determine the values of \(x\) for which the expression under the square root is defined.
Simplify the expression under the square root:
\[-10x^2 - 11x + 6 + 9x^2 - 5x = -x^2 - 16x + 6.\]
The expression under the square root must not be negative, as the square root of a negative number is not a real number. Therefore, we need to find the values of \(x\) for which \(-x^2 - 16x + 6 \geq 0\).
To solve this inequality, we can factor the quadratic expression:
\[-x^2 - 16x + 6 = -(x^2 + 16x - 6).\]
Now, we want to find the values of \(x\) that make the quadratic expression \(x^2 + 16x - 6\) non-negative. We can do this by finding the roots of the quadratic equation \(x^2 + 16x - 6 = 0\) and determining the intervals where the expression is positive or zero.
Factoring the quadratic equation \(x^2 + 16x - 6 = 0\) is a bit tricky, so we can use the quadratic formula:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.\]
In this case, \(a = 1\), \(b = 16\), and \(c = -6\), so the solutions are:
\[x = \frac{-16 \pm \sqrt{16^2 - 4 \cdot 1 \cdot (-6)}}{2 \cdot 1}.\]
Simplifying this gives:
\[x = -8 \pm \sqrt{130}.\]
Since the quadratic expression \(x^2 + 16x - 6\) opens upwards (the coefficient of \(x^2\) is positive), it is non-negative in the interval between its roots. Therefore, the values of \(x\) that satisfy \(-x^2 - 16x + 6 \geq 0\) are given by:
\[-8 - \sqrt{130} \leq x \leq -8 + \sqrt{130}.\]
In interval notation, the domain of the function \(f(x)\) is:
\(\boxed{[-8 - \sqrt{130}, -8 + \sqrt{130}]}.\)
