Question

enhanced algebra: concepts and connections - unit 3
georgia’s k-12 mathematics standards
name:
date:
formative assessment
solve and explain the reasoning behind the solutions. hint: some may have multiple solutions.
a. $a^2 = 6$
b. $b^3 = 7$
c. $c^2 = 8$
d. $d^2 = 7$
e. $e^3 = -8$
f. $f^2 = 9$
g. $g^3 = 11$

Explanation:

Part A: \( a^2 = 6 \)

Step1: Take square root of both sides

To solve for \( a \), we take the square root of both sides of the equation \( a^2 = 6 \). Remember that when taking the square root of a number, we get both a positive and a negative solution because \( (\pm x)^2 = x^2 \).
\[
a = \pm\sqrt{6}
\]

Step2: Simplify (if needed)

\( \sqrt{6} \) cannot be simplified further into a whole number or a simpler radical, so the solutions are \( a = \sqrt{6} \) and \( a = -\sqrt{6} \).

Part B: \( b^3 = 7 \)

Step1: Take cube root of both sides

To solve for \( b \), we take the cube root of both sides of the equation \( b^3 = 7 \). Unlike square roots, cube roots of real numbers have only one real solution (since a positive number cubed is positive, a negative number cubed is negative, and zero cubed is zero).
\[
b = \sqrt[3]{7}
\]

Part C: \( c^2 = 8 \)

Step1: Take square root of both sides

To solve for \( c \), we take the square root of both sides of the equation \( c^2 = 8 \). We can simplify \( \sqrt{8} \) as \( \sqrt{4 \times 2} = \sqrt{4} \times \sqrt{2} = 2\sqrt{2} \). Since squaring a negative gives a positive, we have two solutions.
\[
c = \pm\sqrt{8} = \pm 2\sqrt{2}
\]

Part D: \( d^2 = 7 \)

Answer:

s:

• A: \( a = \pm\sqrt{6} \)
• B: \( b = \sqrt[3]{7} \)
• C: \( c = \pm 2\sqrt{2} \)
• D: \( d = \pm\sqrt{7} \)
• E: \( e = -2 \)
• F: \( f = \pm 3 \)
• G: \( g = \sqrt[3]{11} \)