Question

name: ____________________________ date: _____________ using tables on your calculator common core algebra ii homework fluency 1. use the store feature on your calculator to evaluate each of the following. no work needs to be shown. (a) $7x + 18$ for $x = -8$ (b) $3x^2 - 2x + 5$ for $x = 3$ (c) $x^3 + 5x^2 - 4x - 20$ for $x = -5$ (d) $left|x^2 - 2x - 8
ight|$ for $x = 1$ (e) $\frac{5x - 3}{4x^2 + 5}$ for $x = 2$ (f) $sqrt{\frac{4 - x}{x + 9}}$ for $x = -5$ 2. the store features is particularly helpful in checking to see if a value is a solution to an equation. let’s see how this works in this problem. consider the relatively easy linear equation: $6x - 3 = 4x + 9$ (a) solve this equation for $x$. (b) using store, determine the value of both the left hand expression, $6x - 3$, and the right hand expression, $4x + 9$, at the value of $x$ you found in (a). (c) why does what you found in part (b) verify that your solution is correct (or possibly incorrect if you made a mistake in (a))? 3. two of the following values of $x$ are solutions to the equation: $x^2 + 4x - 12 = 10x + 4$. determine which they are and provide a justification for your answer. $x = -2$ $x = -5$ $x = 6$ $x = 8$ common core algebra ii, unit #1 – essential algebra concepts – lesson #6 emathinstruction, red hook, ny 12571, © 2015

Explanation:

Problem 1

Step1: Substitute x=-8

$7(-8)+18 = -56+18$

Step2: Substitute x=3

$3(3)^2-2(3)+5 = 27-6+5$

Step3: Substitute x=-5

$(-5)^3+5(-5)^2-4(-5)-20 = -125+125+20-20$

Step4: Substitute x=1, compute absolute value

$|(1)^2-2(1)-8| = |1-2-8|$

Step5: Substitute x=2

$\frac{5(2)-3}{4(2)^2+5} = \frac{10-3}{16+5}$

Step6: Substitute x=-5

$\sqrt{\frac{4-(-5)}{-5+9}} = \sqrt{\frac{9}{4}}$

(a) Step1: Isolate x terms

$6x-4x = 9+3$

(a) Step2: Simplify to solve x

$2x=12 \implies x=\frac{12}{2}$

(b) Step1: Substitute x=6 into left side

$6(6)-3 = 36-3$

(b) Step2: Substitute x=6 into right side

$4(6)+9 = 24+9$

(c) Step1: State equality check rule

A solution makes both sides equal

Step1: Rearrange the equation

$x^2+4x-12-10x-4=0 \implies x^2-6x-16=0$

Step2: Test x=-2

$(-2)^2-6(-2)-16 = 4+12-16$

Step3: Test x=-5

$(-5)^2-6(-5)-16 = 25+30-16$

Step4: Test x=6

$(6)^2-6(6)-16 = 36-36-16$

Step5: Test x=8

$(8)^2-6(8)-16 = 64-48-16$

Answer:

(a) $-38$
(b) $26$
(c) $0$
(d) $9$
(e) $\frac{7}{21} = \frac{1}{3}$
(f) $\frac{3}{2}$

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Problem 2