Find all points of the form $(x, −1)$, which are $4$ units from the point
$(3, 2)$.
I understand the distance formula, I think. Also, I don't know how to format for math on here yet so I apologize for that.
\begin{align}
&\text{dist} = 4 = \sqrt{(x-3)^2 + (2-(-1))^2)} \\
&\implies 16 = (x-3)^2 + (2 + 1)^2 \tag{square both sides} \\
&\implies 16 = x^2 -6x + 18 \tag{expand} \\
&\implies 0 = x^2 -6x +2 \tag{zero on the left}
\end{align}
Next should be finding the factors which would give me the answers but I have no idea what would work for this.
$$(x + \text{something})(x – \text{something}).$$
I must be doing something wrong or missing something. Please correct me.
Best Answer
Your computations are correct up to the factoring $$ (x+\text{something})(x-\text{somethign}).$$ In general, there is no reason to expect that your polynomial will factor like a difference of squares. In fact, the presence of a linear term (the $-6x$) indicates that this won't work. Instead, you need to either factor the polynomial somehow, or solve the quadratic using some other technique. In this case, completing the square seems reasonable.
Every quadratic polynomial can be written in the form $$ A(x-h)^2 + k, $$ where $A$, $h$, and $k$ are real numbers. Hence we have $$x^2 - 6x + 2 =A(x-h)^2 + k = Ax^2 - 2Ahx + Ah^2 + k. \tag{1}$$ Since two polynomials are equal to each other if and only if they have the same coefficients, we can equate the coefficients on the left with the coefficients on the right in order to obtain $$ \begin{cases} 1 = A, \\ -6 = -2Ah, \\ 2 = Ah^2 + k. \end{cases} $$ Solving this system, we immediately have that $A = 1$. Substituting this into the second equation, we get $-6 = -2h$, which implies that $h = 3$. Finally, plugging these into the last equation, we get $2 = 3^2 + k$, which implies that $k = -7$. Putting these back into (1), we get $$ x^2 - 6x + 2 = (x-3)^2 - 7.$$ Since we are looking for the roots of this polynomial, we set it equal to zero, then solve by extracting roots. This gives us \begin{align} 0 = (x-3)^2 - 7 &\implies (x-3)^2 = 7 \\ &\implies x-3 = \pm \sqrt{7} \tag{extract the roots} \\ &\implies x = 3\pm \sqrt{7}. \end{align} Hence the two points on the line $x=-1$ that are a distance of 4 from the point $(3,2)$ are the points $$\left( 3+\sqrt{7}, -1 \right) \qquad\text{and}\qquad \left( 3-\sqrt{7}, -1 \right).$$