1. **State the problem:** We are given the equation $$(3x - d)^2 = ax^2 - 3kx + 49$$ and need to find the values of $a$, $k$, and $d$. 2. **Expand the left side:** Use the formula for the square of a binomial $$(A - B)^2 = A^2 - 2AB + B^2$$. $$ (3x - d)^2 = (3x)^2 - 2 \cdot 3x \cdot d + d^2 = 9x^2 - 6dx + d^2 $$ 3. **Match coefficients:** The expanded form must equal the right side: $$ 9x^2 - 6dx + d^2 = ax^2 - 3kx + 49 $$ Equate coefficients of like terms: - Coefficient of $x^2$: $9 = a$ - Coefficient of $x$: $-6d = -3k$ which simplifies to $6d = 3k$ or $k = 2d$ - Constant term: $d^2 = 49$ 4. **Solve for $d$:** $$ d^2 = 49 \implies d = \pm 7 $$ 5. **Find $k$ using $k = 2d$:** - If $d = 7$, then $k = 2 \times 7 = 14$ - If $d = -7$, then $k = 2 \times (-7) = -14$ 6. **Final values:** $$ a = 9 $$ $$ d = \pm 7 $$ $$ k = \pm 14 $$ The values of $a$, $k$, and $d$ satisfy the given equation with $d$ and $k$ having two possible sign pairs.