Question

name
standards practice week 6
selected response

1. solve $2x^2 + 16x + 10 = 0$ by completing the square.

a $x = -8 + 3\sqrt{16}$ and $x = -8 - 3\sqrt{6}$
b $x = -4 + \sqrt{21}$ and $x = -4 - \sqrt{21}$
c $x = -4 + \sqrt{11}$ and $x = -4 - \sqrt{11}$
d $x = 4 + \sqrt{11}$ and $x = 4 - \sqrt{11}$

1. factor the expression $36x^2 + 64$.

a $(6x + 8)(6x - 8)$
b $(6x + 8i)(6x - 8i)$
c $(6x + 8)(6x + 8)$
d $(100x + 8i)(x + 8i)$
constructed response

1. solve $6x^2 - 8x = -18$ using the quadratic formula. show your work.

extended response

1. classify each linear - quadratic system as having either 0 solutions, 1 solution, or 2 solutions.

i. $\

$$\begin{cases}y = -2x^2 - 4x - 6\\\\-x - 2y = -16\\end{cases}$$

$
ii. $\

$$\begin{cases}y = 2x^2\\\\y = -100\\end{cases}$$

$
iii. $\

$$\begin{cases}y = -\\frac{1}{2}x^2 + 1\\\\y = -2x + 3\\end{cases}$$

$
iv. $\

$$\begin{cases}y = \\frac{1}{6}x^2 - 1\\\\y = -1\\end{cases}$$

$
v. $\

$$\begin{cases}y = -4x^2 + 2\\\\y = -2x - 14\\end{cases}$$

$
vi. $\

$$\begin{cases}y = x^2\\\\y = -2x + 10\\end{cases}$$

$
\

$$\begin{tabular}{|c|c|c|} \\hline 0 solutions & 1 solution & 2 solutions\\\\ \\hline & & \\\\ \\hline\\end{tabular}$$

Explanation:

Question 1

Step1: Divide by 2

$x^2 + 8x + 5 = 0$

Step2: Move constant term

$x^2 + 8x = -5$

Step3: Complete the square

$x^2 +8x +16 = -5 +16$ → $(x+4)^2=11$? Correction: $(x+4)^2=11$ wrong, correct: $x^2+8x+16=11$ → $x=-4±√11$, so option C.

Question 2

Step1: Recognize difference of squares

$36x^2 - (-64) = (6x)^2 - (8i)^2$

Step2: Factor using formula

$(6x +8i)(6x -8i)$

Question 3

Step1: Rewrite in standard form

$6x^2 -8x +18=0$ → $3x^2 -4x +9=0$

Step2: Apply quadratic formula

$x=\frac{4±√(16-108)}{6}=\frac{4±√(-92)}{6}=\frac{4±2i√23}{6}$ Correction: $√(-92)=2i√23$, so $x=\frac{2±i√23}{3}$

Question 4

Answer:

C. $x=-4+\sqrt{17}$ and $x=-4-\sqrt{17}$