Question

solve the equation $x^2 - 8x + 12 = 0$
a) $x = \\{3, 4\\}$ b) $x = \\{-3, -4\\}$ c) $x = \\{-6, -2\\}$ d) $x = \\{6, 2\\}$

1. find the center and radius for the circle with the equation $x^2 + (y + 5)^2 = 9$

a) center $(0, 5)$; radius $81$
b) center $(1, -5)$; radius $3$
c) center $(0, -5)$; radius $3$
d) center $(0, 5)$; radius $3$

1. solve the system of equations $\

$$\begin{cases}2x + y = 5 \\\\ x - 3y = -8\\end{cases}$$

$. write your answer as an ordered pair $(x, y)$
a) $(1, 3)$ b) $(-1, -3)$ c) $(5, -5)$ d) $(-5, 5)$

Explanation:

Question 28 (Solving \(x^2 - 8x + 12 = 0\))

Step 1: Factor the quadratic

We factor \(x^2 - 8x + 12\). We need two numbers that multiply to \(12\) and add to \(-8\). Those numbers are \(-6\) and \(-2\). So, \(x^2 - 8x + 12=(x - 6)(x - 2)\).

Step 2: Solve for \(x\)

Set each factor equal to zero: \(x - 6 = 0\) or \(x - 2 = 0\). Solving these gives \(x = 6\) or \(x = 2\), so \(x=\{2,6\}\) (or in the format of the options, \(x=(2,6)\) or \(x=(6,2)\); the correct option is D).

The standard form of a circle is \((x - h)^2+(y - k)^2 = r^2\), where \((h,k)\) is the center and \(r\) is the radius.
For the equation \(x^2+(y + 5)^2 = 9\), we can rewrite it as \((x - 0)^2+(y - (-5))^2 = 3^2\). So, \(h = 0\), \(k=-5\), and \(r = 3\). The center is \((0,-5)\) and radius \(3\), so the correct option is C.

Step 1: Solve the second equation for \(x\)

From \(x - 3y=-8\), we get \(x = 3y - 8\).

Step 2: Substitute into the first equation

Substitute \(x = 3y - 8\) into \(2x + y = 5\): \(2(3y - 8)+y = 5\).

Step 3: Simplify and solve for \(y\)

Expand: \(6y - 16 + y = 5\). Combine like terms: \(7y - 16 = 5\). Add \(16\) to both sides: \(7y=21\). Divide by \(7\): \(y = 3\).

Step 4: Solve for \(x\)

Substitute \(y = 3\) into \(x = 3y - 8\): \(x = 3(3)-8=9 - 8 = 1\). So the solution is \((1,3)\), which is option A.

Answer:

D) \(x=(6,2)\)

Question 29 (Circle Equation \(x^2+(y + 5)^2 = 9\))