Question

the first two steps in determining the solution set of the system of equations, $y = x^2 - 6x + 12$ and $y = 2x - 4$, algebraically are shown in the table. \

step	equation	\
---	---	\
step 1	$x^2 - 6x + 12 = 2x - 4$	\
step 2	$x^2 - 8x + 16 = 0$	\

which represents the solution(s) of this system of equations?\
\\(\circ\\) $(4, 4)$\
\\(\circ\\) $(-4, -12)$\
\\(\circ\\) $(4, 4)$ and $(-4, 12)$\
\\(\circ\\) $(-4, 4)$ and $(4, 12)$

Explanation:

Step1: Solve the quadratic equation

The quadratic equation from Step 2 is \(x^{2}-8x + 16=0\). We can factor this quadratic. Notice that \(x^{2}-8x + 16=(x - 4)^{2}\) (since \((a - b)^{2}=a^{2}-2ab + b^{2}\), here \(a=x\), \(b = 4\), and \(-2\times x\times4=-8x\), \(4^{2}=16\)). So, \((x - 4)^{2}=0\).

Step2: Find the value of x

Setting \((x - 4)^{2}=0\), we take the square root of both sides. \(\sqrt{(x - 4)^{2}}=\sqrt{0}\), which gives \(x-4 = 0\), so \(x = 4\).

Step3: Find the value of y

Now that we have \(x = 4\), we can substitute \(x = 4\) into one of the original equations. Let's use \(y=2x - 4\). Substituting \(x = 4\) into \(y=2x - 4\), we get \(y=2\times4-4=8 - 4=4\). So the solution is \((4,4)\).

Answer:

\((4, 4)\)