Find the domain and solve.
$\sqrt{-x^2+3 x-2}=4$
To find the domain and solve the equation \(\sqrt{-x^2+3x-2}=4\), we first identify the domain of the function. The expression inside the square root must be non-negative for the value to be defined in the real number system. Thus, we require \(-x^2+3x-2 \geq 0\). We can rewrite this as \(x^2-3x+2 \leq 0\). Factoring the quadratic, we have \((x-1)(x-2) \leq 0\). This inequality holds when \(x\) is between the roots of the quadratic equation \(x^2-3x+2=0\), which are \(x=1\) and \(x=2\). Therefore, the domain of the expression is the interval \([1, 2]\).
Now we solve the equation \(\sqrt{-x^2+3x-2}=4\). By squaring both sides of the equation, we eliminate the square root:
\[-x^2+3x-2 = 4^2\]
\[-x^2+3x-2 = 16\]
Rearranging the terms to form a standard quadratic equation, we get:
\[-x^2+3x-18 = 0\]
or equivalently,
\[x^2-3x+18 = 0\]
To find the roots of \(x^2-3x+18=0\), we examine the discriminant \(D = b^2 - 4ac\), where \(a=1\), \(b=-3\), and \(c=18\):
\[D = (-3)^2 - 4(1)(18) = 9 - 72 = -63\]
Since the discriminant is negative (\(D < 0\)), there are no real numbers that satisfy this equation. Furthermore, note that the maximum value of the expression \(-x^2+3x-2\) occurs at the vertex \(x = -b/(2a) = -3/(2 \times -1) = 1.5\). Evaluating the expression at \(x=1.5\):
\[-(1.5)^2 + 3(1.5) - 2 = -2.25 + 4.5 - 2 = 0.25\]
Since the maximum value of the expression under the radical is \(0.25\), it is impossible for the square root to equal 4. Thus, there are no real solutions.
This problem is a great example of why it is essential to check the domain and feasibility of a solution before concluding. In engineering or physics, such as calculating the potential energy of a system restricted to certain spatial boundaries, solving for an impossible state (like finding an energy level outside of the possible physical constraints) is a common hurdle. By verifying the domain, we ensure that our mathematical solutions represent real-world physical possibilities.