Question

d. $(x - 3)^2 + 5 = -11$

Explanation:

Step1: Isolate the squared term

Subtract 5 from both sides of the equation:
$$(x - 3)^2 + 5 - 5 = -11 - 5$$
$$(x - 3)^2 = -16$$

Step2: Analyze the square's result

The square of a real number \((x - 3)\) (where \(x\) is real) is always non - negative, i.e., \((x - 3)^2\geq0\) for all real numbers \(x\). But we have \((x - 3)^2=-16\), and \(-16<0\).

Answer:

There is no real solution for the equation \((x - 3)^2+5 = - 11\). If we consider complex numbers, we can solve it as follows:

From \((x - 3)^2=-16\), take the square root of both sides:
\(x - 3=\pm\sqrt{-16}=\pm4i\) (where \(i=\sqrt{-1}\) is the imaginary unit)

Then solve for \(x\):
\(x = 3\pm4i\)

If we are only considering real - valued solutions, the answer is no real solution. If we consider complex - valued solutions, the solutions are \(x = 3 + 4i\) and \(x = 3-4i\).