Solve 2(x - 1.6)2 + (x - 1.5)2 = 0.

Step-by-step solution

1. 1

   Note that every term on the left is nonnegative. For real $x$, $(x-1.6)^2 \ge 0$ and $(x-1.5)^2 \ge 0$, and multiplying the first by the positive constant $2$ preserves that. So the left-hand side is a sum of two nonnegative quantities.

2. 2

   Use the zero-sum principle for nonnegatives. A sum of nonnegative numbers equals $0$ only when every term is $0$. Here that forces $2(x-1.6)^2 = 0 \quad\text{and}\quad (x-1.5)^2 = 0,$ i.e. $x = 1.6$ and $x = 1.5$ simultaneously.

3. 3

   Spot the contradiction. A single number cannot be both $1.6$ and $1.5$, so no real $x$ satisfies both conditions. The equation therefore has no real solution.

4. 4

   Confirm algebraically by expanding. $2(x-1.6)^2 = 2x^2-6.4x+5.12$ and $(x-1.5)^2 = x^2-3x+2.25$, so the equation becomes $3x^2-9.4x+7.37 = 0.$

5. 5

   Check the discriminant of the expanded form. $\Delta = (-9.4)^2-4(3)(7.37) = 88.36-88.44 = -0.08 < 0$, which independently confirms there is no real root. The margin is small, which is exactly why the structural argument in the earlier steps is safer than decimal arithmetic here.

6. 6

   Read the practical meaning. This is the residual of a least-squares fit: the minimum of $3x^2-9.4x+7.37$ occurs at $x = \tfrac{9.4}{6} \approx 1.5667$ with value $\tfrac{0.08}{12} \approx 0.00667 > 0$. The best compromise between $1.6$ and $1.5$ is $x \approx 1.5667$, but the sum never actually reaches zero.