Solve (5x − 4)² − 49x² = 0.

Step-by-step solution

1. 1

   Rewrite the second term as a square. Since $49x^2 = (7x)^2$, the left side is a difference of two squares:

   $(5x-4)^2 - (7x)^2 = 0$

   Spotting this avoids expanding to $25x^2 - 40x + 16 - 49x^2 = -24x^2 - 40x + 16$ and then dividing through — factoring is shorter and less error-prone.

2. 2

   Apply the identity A² − B² = (A − B)(A + B). With $A = 5x - 4$ and $B = 7x$:

   $\bigl((5x-4) - 7x\bigr)\bigl((5x-4) + 7x\bigr) = 0$

3. 3

   Simplify each bracket.

   $(5x - 4) - 7x = -2x - 4, \qquad (5x - 4) + 7x = 12x - 4$

   so the equation is

   $(-2x - 4)(12x - 4) = 0$

4. 4

   Use the zero-product property on each factor.

   $-2x - 4 = 0 \ \Rightarrow \ x = -2$

   $12x - 4 = 0 \ \Rightarrow \ x = \frac{4}{12} = \frac{1}{3}$

5. 5

   Check both roots in the original equation. At $x = -2$: $(-10-4)^2 - 49(4) = 196 - 196 = 0 \ \checkmark$. At $x = \tfrac13$: $\left(\tfrac53 - 4\right)^2 - 49\cdot\tfrac19 = \tfrac{49}{9} - \tfrac{49}{9} = 0 \ \checkmark$. Both satisfy it exactly, and a quadratic can have at most two roots, so the solution set is complete.