Question

what is the solution to the following equation?
$2\sqrt3{3x + 1} - 11 = -3$
select one:
○ a. $x = 5$
○ b. $x = 9$
○ c. $x = 21$
○ d. $x = \frac{11}{3}$
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what is the solution to the following equation?
$\sqrt{a + 9} = -3 - a$
select one:
○ a. $a = 0$
○ b. $a = -5$
○ c. $a = \\{0, 5\\}$
○ d. $a = \\{-5, 0\\}$

Explanation:

Step1: Isolate the cube root term

$2\sqrt[3]{3x+1} = -3 + 11$
$2\sqrt[3]{3x+1} = 8$

Step2: Solve for the cube root

$\sqrt[3]{3x+1} = \frac{8}{2}$
$\sqrt[3]{3x+1} = 4$

Step3: Eliminate the cube root

$(\sqrt[3]{3x+1})^3 = 4^3$
$3x+1 = 64$

Step4: Solve for $x$

$3x = 64 - 1$
$3x = 63$
$x = \frac{63}{3} = 21$

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Step1: Square both sides to eliminate root

$(\sqrt{a+9})^2 = (-3-a)^2$
$a+9 = a^2 + 6a + 9$

Step2: Rearrange to quadratic equation

$0 = a^2 + 6a + 9 - a - 9$
$a^2 + 5a = 0$

Step3: Factor and solve for $a$

$a(a+5) = 0$
$a=0$ or $a=-5$

Step4: Verify solutions in original equation

For $a=0$: $\sqrt{0+9} = -3-0$ → $3=-3$ (false, discard)
For $a=-5$: $\sqrt{-5+9} = -3-(-5)$ → $2=2$ (true)

Answer:

First equation: c. $x = 21$
Second equation: b. $a = -5$