Question

The roots of the quadratic equation $(x - 1)^2 = 16$ are :

• A. $5, 3$
• B. $4, -4$
• C. $5, -3$
• D. $-5, 3$

Hint:

Always remember that taking the square root of a positive real number yields both positive and negative roots.
A common mistake is to write $x - 1 = 4$ only, which leads to losing one of the valid roots of the quadratic equation.
You can also quickly verify your answer by substituting the options back into the original equation:
For $x = 5 \implies (5 - 1)^2 = 4^2 = 16$ (Correct)
For $x = -3 \implies (-3 - 1)^2 = (-4)^2 = 16$ (Correct)

Answer 1

The Correct Option is C

Step 1: Expand the equation into standard quadratic form.
Instead of taking the square root straightaway, let's expand $(x-1)^2=16$ using the identity $(a-b)^2=a^2-2ab+b^2$ and bring everything to one side, so we get an ordinary quadratic we can factor.
\[ (x-1)^2 = 16 \]
\[ x^2 - 2x + 1 = 16 \]

Step 2: Move all terms to one side.
Subtract $16$ from both sides so the equation is in the standard form $ax^2+bx+c=0$.
\[ x^2 - 2x + 1 - 16 = 0 \]
\[ x^2 - 2x - 15 = 0 \]

Step 3: Factor the quadratic by splitting the middle term.
We need two numbers that multiply to give $-15$ (the constant term) and add to give $-2$ (the coefficient of $x$). Testing factor pairs of 15, the pair $3$ and $-5$ works, since $3 \times (-5) = -15$ and $3 + (-5) = -2$.
Rewrite the middle term using these numbers:
\[ x^2 + 3x - 5x - 15 = 0 \]

Step 4: Group and factor.
\[ x(x + 3) - 5(x + 3) = 0 \]
\[ (x - 5)(x + 3) = 0 \]

Step 5: Solve each factor for x.
Setting each bracket to zero gives the two roots:
\[ x - 5 = 0 \implies x = 5 \]
\[ x + 3 = 0 \implies x = -3 \]

Final Answer:
The roots of the equation are $5$ and $-3$, which matches Option (C). \[ \boxed{x = 5, -3} \]