algebra 2
unit 5 part 2 bonus
name
solve each equation. remember to check for extraneous solutions.
date
1) $14 = 5 + \sqrt{11b - 7}$
2) $\sqrt{4 - 3n} = n$
solve each equation.
3) $(x - 25)^{\frac{3}{2}} - 2 = 727$
4) $3 + 2(3v - 15)^{\frac{3}{2}} = 57$

Question

Accepted Answer

### Problem 1: $ 14 = 5 + \sqrt{11b - 7} $
# Explanation:
## Step 1: Isolate the square root
Subtract 5 from both sides:
$ 14 - 5 = \sqrt{11b - 7} $
$ 9 = \sqrt{11b - 7} $

## Step 2: Eliminate the square root
Square both sides:
$ 9^2 = (\sqrt{11b - 7})^2 $
$ 81 = 11b - 7 $

## Step 3: Solve for $ b $
Add 7 to both sides:
$ 81 + 7 = 11b $
$ 88 = 11b $
Divide by 11:
$ b = \frac{88}{11} = 8 $

## Step 4: Check for extraneous solutions
Substitute $ b = 8 $ into the original equation:
Left side: $ 14 $
Right side: $ 5 + \sqrt{11(8) - 7} = 5 + \sqrt{88 - 7} = 5 + \sqrt{81} = 5 + 9 = 14 $
Solution is valid.

# Answer: $ b = 8 $

### Problem 2: $ \sqrt{4 - 3n} = n $
# Explanation:
## Step 1: Eliminate the square root
Square both sides:
$ (\sqrt{4 - 3n})^2 = n^2 $
$ 4 - 3n = n^2 $

## Step 2: Rearrange into quadratic form
$ n^2 + 3n - 4 = 0 $

## Step 3: Solve the quadratic equation
Factor:
$ (n + 4)(n - 1) = 0 $
Solutions: $ n = -4 $ or $ n = 1 $

## Step 4: Check for extraneous solutions
- For $ n = -4 $:
  Left side: $ \sqrt{4 - 3(-4)} = \sqrt{16} = 4 $
  Right side: $ -4 $
  $ 4
eq -4 $, so $ n = -4 $ is extraneous.

- For $ n = 1 $:
  Left side: $ \sqrt{4 - 3(1)} = \sqrt{1} = 1 $
  Right side: $ 1 $
  Solution is valid.

# Answer: $ n = 1 $

### Problem 3: $ (x - 25)^{\frac{3}{2}} - 2 = 727 $
# Explanation:
## Step 1: Isolate the rational exponent term
Add 2 to both sides:
$ (x - 25)^{\frac{3}{2}} = 727 + 2 $
$ (x - 25)^{\frac{3}{2}} = 729 $

## Step 2: Rewrite the exponent
Recall $ a^{\frac{3}{2}} = (\sqrt{a})^3 $, so:
$ (\sqrt{x - 25})^3 = 729 $

## Step 3: Take the cube root of both sides
$ \sqrt{x - 25} = \sqrt[3]{729} $
$ \sqrt{x - 25} = 9 $

## Step 4: Eliminate the square root
Square both sides:
$ x - 25 = 9^2 $
$ x - 25 = 81 $

## Step 5: Solve for $ x $
Add 25 to both sides:
$ x = 81 + 25 = 106 $

## Step 6: Check (optional, since exponent $ \frac{3}{2} $ requires $ x - 25 \geq 0 $, and $ x = 106 $ satisfies this)
Substitute $ x = 106 $:
$ (106 - 25)^{\frac{3}{2}} - 2 = 81^{\frac{3}{2}} - 2 = (9)^3 - 2 = 729 - 2 = 727 $, which matches.

# Answer: $ x = 106 $

### Problem 4: $ 3 + 2(3v - 15)^{\frac{3}{2}} = 57 $
# Explanation:
## Step 1: Isolate the rational exponent term
Subtract 3 from both sides:
$ 2(3v - 15)^{\frac{3}{2}} = 57 - 3 $
$ 2(3v - 15)^{\frac{3}{2}} = 54 $

Divide by 2:
$ (3v - 15)^{\frac{3}{2}} = 27 $

## Step 2: Rewrite the exponent
$ (\sqrt{3v - 15})^3 = 27 $

## Step 3: Take the cube root of both sides
$ \sqrt{3v - 15} = \sqrt[3]{27} $
$ \sqrt{3v - 15} = 3 $

## Step 4: Eliminate the square root
Square both sides:
$ 3v - 15 = 3^2 $
$ 3v - 15 = 9 $

## Step 5: Solve for $ v $
Add 15 to both sides:
$ 3v = 9 + 15 $
$ 3v = 24 $

Divide by 3:
$ v = \frac{24}{3} = 8 $

## Step 6: Check (optional, since $ 3v - 15 \geq 0 $ when $ v \geq 5 $, and $ v = 8 $ satisfies this)
Substitute $ v = 8 $:
$ 3 + 2(3(8) - 15)^{\frac{3}{2}} = 3 + 2(24 - 15)^{\frac{3}{2}} = 3 + 2(9)^{\frac{3}{2}} = 3 + 2(27) = 3 + 54 = 57 $, which matches.

# Answer: $ v = 8 $

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QUESTION IMAGE

Question

Explanation:

Problem 1: \( 14 = 5 + \sqrt{11b - 7} \)

Step 1: Isolate the square root

Step 2: Eliminate the square root

Step 3: Solve for \( b \)

Step 4: Check for extraneous solutions

Step 1: Eliminate the square root

Step 2: Rearrange into quadratic form

Step 3: Solve the quadratic equation

Step 4: Check for extraneous solutions

Step 1: Isolate the rational exponent term

Step 2: Rewrite the exponent

Step 3: Take the cube root of both sides

Step 4: Eliminate the square root

Step 5: Solve for \( x \)

Step 6: Check (optional, since exponent \( \frac{3}{2} \) requires \( x - 25 \geq 0 \), and \( x = 106 \) satisfies this)

Answer:

Problem 2: \( \sqrt{4 - 3n} = n \)