I have a question that has two parts:
The first part is relatively straightforward, this is my approach to part a:
However, for part b, I am unsure as to how to approach the problem. I can see that the resulting equation for the area of the sum of the two areas is a quadratic and perhaps finding the minimum value vertex of the function would arrive at the value for x and area. However, I am unsure as to how to go about that. Any help appreciated.
This question could be done fairly easily with calculus but let's take a purely alebraic approach. First write your expression as $ax^2+bx+c.$ Note that $a>0.$ Then we take out the factor $a$ and complete the square. In detail, $$ax^2+bx+c=a(x^2+\frac{b}{a}x+\frac{c}{a})$$ $$=a((x+\frac{b}{2a})^2-\frac{b^2}{4a^2}+\frac{c}{a})$$ $$=a((x+\frac{b}{2a})^2+(\frac{-b^2+4ac}{4a^2}))$$ $$=a(x+\frac{b}{2a})^2+(\frac{-b^2+4ac}{4a})$$ The term $a(x+\frac{b}{2a})^2\ge0$ and is 0 iff $x=-\frac{b}{2a}$. Thus your wloe expression has minimum value $$\frac{-b^2+4ac}{4a}$$ and this minimum value occurs when $x=-\frac{b}{2a}$.