Solve the square root of (4x squared - 8x + 4) equals 5.

Step-by-step solution

1. 1

   Factor the radicand before squaring. $4x^2-8x+4 = 4\left(x^2-2x+1\right) = 4(x-1)^2.$ Recognising the perfect square turns the problem into a one-line absolute-value equation.

2. 2

   Simplify the radical using |a| not a. $\sqrt{4(x-1)^2} = 2\left|x-1\right|$ - the absolute value is essential, because $\sqrt{t^2}=|t|$, not $t$. So the equation is $2|x-1| = 5.$

3. 3

   Isolate the absolute value. $|x-1| = \frac52.$

4. 4

   Split into the two cases. $x-1 = \frac52 \quad\text{or}\quad x-1 = -\frac52,$ giving $x = \frac72 \quad\text{or}\quad x = -\frac32.$

5. 5

   Verify both. At $x=\tfrac72$: $4(12.25)-8(3.5)+4 = 49-28+4 = 25$ and $\sqrt{25}=5$. At $x=-\tfrac32$: $4(2.25)+12+4 = 9+12+4 = 25$ and $\sqrt{25}=5$. Both are genuine.

6. 6

   Note what dropping the absolute value would cost. Writing $\sqrt{4(x-1)^2} = 2(x-1)$ would give only $x=\tfrac72$ and silently lose the negative solution.