1. **Stating the problem:** We are given the equation $$3(x-4)^2=27$$ and need to identify mistakes made by Andre and Clare in solving it (Part A), and then find the correct solution(s) to the equation (Part B). 2. **Part A: Analyze Andre's work** - Andre starts with $$3(x-4)^2=27$$. - He divides both sides by 3 to get $$(x-4)^2=9$$. This is correct. - Then he writes $$x^2 - 42 = 9$$, which is incorrect because he incorrectly distributed the square over the subtraction inside the parentheses. - The correct expansion of $$(x-4)^2$$ is $$x^2 - 8x + 16$$, not $$x^2 - 42$$. - Andre then solves $$x^2 = 25$$ and finds $$x=5$$ or $$x=-5$$, which are not the correct solutions to the original equation. 3. **Part A: Analyze Clare's work** - Clare also starts with $$3(x-4)^2=27$$. - She divides both sides by 3 to get $$(x-4)^2=9$$. - Then she takes the square root of both sides and writes $$x-4=3$$, ignoring the negative root. - The correct step should be $$x-4=3$$ or $$x-4=-3$$. - Clare finds $$x=7$$ but misses $$x=1$$. 4. **Summary of mistakes:** - Andre incorrectly distributed the exponent over subtraction (which is not allowed). - Clare ignored the negative root when taking the square root. 5. **Part B: Find the correct solution(s)** - Starting from $$(x-4)^2=9$$. - Taking square roots: $$x-4=3$$ or $$x-4=-3$$. - Solving each: - $$x=7$$ - $$x=1$$ 6. **Final answer:** The solutions to the equation $$3(x-4)^2=27$$ are $$x=1$$ and $$x=7$$. **Therefore:** - Mistakes: B) Andre distributed the exponent incorrectly. - Mistakes: E) Clare left out the negative root. - Solutions: D) $$x=1$$ or $$x=7$$.