Solve 9(x − 1)² − 4 = 0.

Step-by-step solution

1. 1

   Choose the square-root method, not expansion. Expanding gives $9x^2-18x+5=0$, which then needs factoring or the quadratic formula. Because the variable already appears inside a single squared bracket, isolating that bracket is strictly less work.

2. 2

   Isolate the squared term.

   $9(x-1)^2=4\quad\Longrightarrow\quad (x-1)^2=\frac{4}{9}$

3. 3

   Take square roots of both sides — keeping both signs. $\sqrt{u^2}=|u|$, so

   $x-1=\pm\frac{2}{3}$

   Dropping the $\pm$ here is the classic way to lose half the answer.

4. 4

   Solve the two resulting linear equations.

   $x=1+\frac{2}{3}=\frac{5}{3},\qquad x=1-\frac{2}{3}=\frac{1}{3}$

   $\boxed{x=\tfrac{5}{3}\ \text{or}\ x=\tfrac{1}{3}}$

5. 5

   Check both roots in the original equation. At $x=\tfrac53$: $9\left(\tfrac23\right)^2-4=9\cdot\tfrac49-4=0$. At $x=\tfrac13$: $9\left(-\tfrac23\right)^2-4=9\cdot\tfrac49-4=0$. Both check.

6. 6

   Cross-check with a difference of squares. $9(x-1)^2-4=\big(3(x-1)\big)^2-2^2=(3x-3-2)(3x-3+2)=(3x-5)(3x-1)$, whose zeros are $x=\tfrac53$ and $x=\tfrac13$ — the same pair.