Targeted Review Solutions

Determine any missing roots, then write a possible polynomial equation.
$\pm \sqrt{6}, - 2$

$\begin{matrix} x = - \sqrt{6}, x = \sqrt{6}, x = - 2 \end{matrix} (x - (- \sqrt{6})) (x - \sqrt{6}) (x - (- 2)) = 0 (x + \sqrt{6}) (x - \sqrt{6}) (x + 2) = 0 (x^{2} - x \sqrt{6} + x \sqrt{6} - \sqrt{36}) (x + 2) = 0 (x^{2} - 6) (x + 2) = 0$

$x^{3} + 2 x^{2} - 6 x - 12 = 0$

Determine the missing value that will make the expression a perfect trinomial square.
$x^{2} + b x + \frac{49}{25}$

$b = 2 \sqrt{a c} b = 2 \sqrt{(1) (\frac{49}{25})} = 2 \sqrt{\frac{49}{25}} = 2 (\frac{7}{5})$

$b = \frac{14}{5}$

Graph. (piecewise function)

$y = \{\begin{cases} - \frac{1}{3} x + 2 & when - 6 < x < 0 \\ \sqrt{x} & when 0 \leq x < 4 \\ - {(x - 5)}^{2} + 4 & when 4 \leq x \leq 7 \end{cases}$

Graph: $y = 2 {(x - 1)}^{2} - 3$

$a = 2, h = 1, k = - 3 vertex : (1, - 3)$

Solve $2 x^{2} - 4 x = 1$ by completing the square.

$\begin{array}{rcl} \frac{2}{2} x^{2} - \frac{4}{2} x & = & \frac{1}{2} \\ x^{2} - 2 x + {(- 1)}^{2} & = & \frac{1}{2} + {(- 1)}^{2} \\ {(x - 1)}^{2} & = & \frac{3}{2} \\ \sqrt{{(x - 1)}^{2}} & = & \pm \sqrt{\frac{3}{2}} \\ x - 1 & = & \frac{\pm \sqrt{3}}{\sqrt{2}} (\frac{\sqrt{2}}{\sqrt{2}}) \\ x - 1 & = & \frac{\pm \sqrt{6}}{2} \end{array}$

$x = 1 \pm \frac{\sqrt{6}}{2}$

Match both the equation in vertex form and the domain and range to the function name below.

$y = a {(x - h)}^{3} + k$

$y = a {(x - h)}^{2} + k$

$y = \frac{a}{x - h} + k; x \neq h$

$domain : {x | x \in ℝ} range : {y | y \in ℝ, y > 0}$

$domain : {x | x \in ℝ, x \neq 0} range : {y | y \in ℝ, y \neq 0}$

$domain : {x | x \in ℝ} range : {y | y \in ℝ}$

B, D Quadratic
$y = x^{2}$

A, F Cubic
$y = x^{3}$

C, E Rational
$y = \frac{1}{x}$

Multiple Choice

Determine the perimeter of an isosceles right triangle with a hypotenuse that measures $4 \sqrt{2} meters .$

$\pm 4$
4 meters
$8 + 4 \sqrt{2} meters$
cannot be determined

Note

A side of a triangle cannot be negative. This option is also not the perimeter.
This option is the length of each leg. This value is needed to calculate the perimeter, but is not the perimeter.
Because the triangle is an isosceles right triangle, it is possible to solve for the measure of both missing legs.

$a^{2} + b^{2} = c^{2} a = b = x, c = 4 \sqrt{2} \begin{matrix} x^{2} + x^{2} = {(4 \sqrt{2})}^{2} \\ 2 x^{2} = 32 \\ x^{2} = 16 \\ \sqrt{x^{2}} = \sqrt{16} \\ x = 4 \end{matrix} P = a + b + c P = 4 + 4 + 4 \sqrt{2} P = 8 + 4 \sqrt{2}$

Solve:
$x^{4} - 2 x^{2} - 9 x^{2} + 18 = 0$

$x = \pm \sqrt{2}$
$x = \pm 5$
$x = \pm 2, \pm 3$
$x = \pm \sqrt{2}, \pm 3$

Note

This answer is incomplete.
This answer tries to combine the values.
This answer drops the square root symbol with the number 2.

$(x^{4} - 2 x^{2}) + (- 9 x^{2} + 18) = 0 x^{2} (x^{2} - 2) - 9 (x^{2} - 2) = 0 (x^{2} - 2) (x^{2} - 9) = 0 (x - \sqrt{2}) (x + \sqrt{2}) (x - 3) (x + 3) = 0 \begin{matrix} x = \pm \sqrt{2}, \pm 3 \end{matrix}$

The work provided for solving a quadratic equation contains an error.
Select the correct roots.

$x = 1 \pm 3 i \sqrt{2}$
$x = - 3, 5 \begin{matrix} \end{matrix}$
$x = \pm 4 i$
$x = 1 \pm 4 i$

$\begin{matrix} Step 1 & x^{2} - 2 x + 17 = 0 \\ \begin{array}{c} \begin{array}{c} Step 2 \end{array} \end{array} & x^{2} - 2 x = - 17 \\ \begin{array}{c} \begin{array}{c} Step 3 \end{array} \end{array} & x^{2} - 2 x + 1 = - 16 \\ \begin{array}{c} \begin{array}{c} Step 4 \end{array} \end{array} & {(x - 1)}^{2} = - 16 \\ \begin{array}{c} Step 5 \end{array} & \sqrt{{(x - 1)}^{2}} = \pm \sqrt{- 16} \\ \begin{array}{c} \begin{array}{c} Step 6 \end{array} \end{array} & x - 1 = \pm 4 \end{matrix}$

$\begin{matrix} Step 1 & x^{2} - 2 x + 17 = 0 \\ Step 2 & x^{2} - 2 x = - 17 \\ Step 3 & x^{2} - 2 x + 1 = - 16 \\ Step 4 & {(x - 1)}^{2} = - 16 \\ Step 5 & \sqrt{{(x - 1)}^{2}} = \pm \sqrt{- 16} \\ Step 6 & x - 1 = \pm 4 i \\ Step 7 & x = 1 \pm 4 i \end{matrix}$

Note

This answer occurs if the right side of the equation is equal to –18, in step 3.
This answer does not fix the error and solves the problem with the error in step 6.
This answer occurs when –1 is dropped from the left side of the equation in step 6.

Classify the expression: $- 4 \pm \sqrt{6}$

rational
irrational
real
imaginary
complex

Note

The square root of six is irrational; therefore, the expression cannot be rational. The imaginary number is not in the expression; therefore, it cannot be classified as an imaginary number.

Problem	1	2	3	4	5	6	7	8	9	10	11	12
Origin	L23	L24	L21	L18	L24	L17	L17	L17	–	L23	L24	L15

L = Lesson in this level, A1 = Algebra 1: Principles of Secondary Mathematics