Tutor profile: Dylan H.

$$a^2–26a+69=0$$. If $$a–5>0$$, what is the value of a? A. 3 B. 5 C. 18 D. 23

There are two methods for solving this problem. First there is the "plug-and-chug" method, which involves testing the answer choices. From $$a–5>0$$, we have that $$a>5$$ so answers A and B are not correct. So now we plug $$18$$ and $$23$$ into $$a^2–26a+69=0$$. By calculation, $$a=23$$ so D is the answer. The second method involves solving the polynomial $$a^2–26a+69=0$$ for $$a$$. There are several methods to factor this including using the quadratic equation or completing the square. We will use completing the square because it involves little computation: $(a^2-26a=-69$) We add $$(26/2)^2=169$$ to both sides then factor the left-hand side of the equation: $(a^2-26a+169=-69+169$) $((a-13)^2=100$) Taking the square root, we have $$a-13=\pm10$$. Thus, $$a=3$$ and $$a=23$$. The second equation tells us that $$a>5$$ so $$a=23$$ (option D) is the solution.

Show that $$1+2+3+...+n=\frac{n(n+1)}{2}$$.

We prove this by induction. First, the base case of $$n=1$$ has that $$1=\frac{1(1+1)}{2}$$. This is true, so we assume $$1+2+3+...+n=\frac{n(n+1)}{2}$$ for the purpose of induction. We need to show that $$1+2+3+...+n+(n+1)=\frac{(n+1)(n+2)}{2}$$. By our inductive assumption, we have that $$1+2+3+...+n+(n+1)=\frac{n(n+1)}{2}+(n+1)$$. Using algebra, we simplify: $(=\frac{n^2+n}{2}+\frac{2n+2}{2}=\frac{n^2+3n+2}{2}=\frac{(n+1)(n+2)}{2}$) This closes the induction and we have shown what was needed.

Derive the derivative for $$y = x^3$$.

The derivative is the instantaneous change in a function at a point. The change in a function is the estimated by the linear change between two nearby points on that line: $(\frac{(x+x_0)^3-x^3}{(x+x_0)-x}$) where $$x_0$$ is the difference between our initial value of $$x$$ and our other value for $$x$$ on the function. It can be visually shown that as $$x_0$$ becomes smaller, the slope is closer to the slope at $$x$$ itself. (I can't show this here because the website doesn't integrate graphing.) We now simplify the equation from above using algebra: $(=\frac{3x^2x_0+3xx_0^2+x_0^3}{x_0}=3x^2+3xx_0+x_0^2$) This allows us to take the limit as $$x_0$$ approaches $$0$$. (Notice that this is different from dividing by $$0$$ because $$x_0$$ only approaches $$0$$ but never IS $$00$$.) So, we have: $(=3x^2+3x(0)+0^2=3x^2$) Thus, the derivative of $$y=x^3$$ is $$\frac{dy}{dx}=3x^2$$.