Find the root of the equation (5x - 8)^2 - 25x^2 = 0.

Step-by-step solution

1. 1

   Expand the square.

   $(5x-8)^2 = 25x^2 - 80x + 64$

   so the equation becomes $25x^2 - 80x + 64 - 25x^2 = 0$.

2. 2

   Watch the quadratic terms cancel. $25x^2 - 25x^2 = 0$, leaving

   $-80x + 64 = 0$

   The equation only looked quadratic; because both squared terms have the same coefficient $25$, it is really linear and has exactly one root, not two.

3. 3

   Solve the linear equation.

   $80x = 64 \quad\Longrightarrow\quad x = \frac{64}{80} = \frac{4}{5} = 0.8$

4. 4

   Confirm with the difference-of-squares route. Factoring instead of expanding:

   $(5x-8)^2 - (5x)^2 = \left[(5x-8) - 5x\right]\left[(5x-8) + 5x\right] = (-8)(10x - 8)$

   Setting this to zero requires $10x - 8 = 0$, i.e. $x = 0.8$ — the constant factor $-8$ can never vanish, which is another way to see there is only one root.

5. 5

   Verify. At $x = 0.8$: $(4 - 8)^2 - 25(0.64) = 16 - 16 = 0$. Correct.