Solve the equation with rational exponents 3(2𝑥𝑥 + 3)2 3⁄ + 5 = 17. Show all work.
2. Solve the quadratic like equation 𝑥𝑥 + 4√𝑥𝑥 − 21 = 0 by making an appropriate

Let's solve each problem step by step.

### 1. Solve the equation with rational exponents:

The equation is:

\[
3(2x + 3)^{\frac{2}{3}} + 5 = 17
\]

#### Step 1: Isolate the term with the exponent

First, subtract 5 from both sides of the equation:

\[
3(2x + 3)^{\frac{2}{3}} = 17 - 5
\]

\[
3(2x + 3)^{\frac{2}{3}} = 12
\]

#### Step 2: Divide both sides by 3

\[
(2x + 3)^{\frac{2}{3}} = \frac{12}{3}
\]

\[
(2x + 3)^{\frac{2}{3}} = 4
\]

#### Step 3: Eliminate the fractional exponent by raising both sides to the reciprocal power (which is \(\frac{3}{2}\))

\[
\left( (2x + 3)^{\frac{2}{3}} \right)^{\frac{3}{2}} = 4^{\frac{3}{2}}
\]

This simplifies to:

\[
2x + 3 = (4^{\frac{3}{2}})
\]

Since \( 4^{\frac{3}{2}} = (2^2)^{\frac{3}{2}} = 2^3 = 8 \), we get:

\[
2x + 3 = 8
\]

#### Step 4: Solve for \( x \)

Now, subtract 3 from both sides:

\[
2x = 8 - 3
\]

\[
2x = 5
\]

Finally, divide both sides by 2:

\[
x = \frac{5}{2}
\]

So, the solution to the first equation is:

\[
x = \frac{5}{2}
\]

---

### 2. Solve the quadratic-like equation:

The equation is:

\[
x + 4\sqrt{x} - 21 = 0
\]

#### Step 1: Make an appropriate substitution

Let \( u = \sqrt{x} \), so \( u^2 = x \). Substituting this into the equation, we get:

\[
u^2 + 4u - 21 = 0
\]

#### Step 2: Solve the quadratic equation

Now solve the quadratic equation \( u^2 + 4u - 21 = 0 \) using the quadratic formula:

\[
u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \( a = 1 \), \( b = 4 \), and \( c = -21 \). Plugging these values into the quadratic formula:

\[
u = \frac{-4 \pm \sqrt{4^2 - 4(1)(-21)}}{2(1)}
\]

\[
u = \frac{-4 \pm \sqrt{16 + 84}}{2}
\]

\[
u = \frac{-4 \pm \sqrt{100}}{2}
\]

\[
u = \frac{-4 \pm 10}{2}
\]

Now, solve for the two possible values of \( u \):

\[
u = \frac{-4 + 10}{2} = \frac{6}{2} = 3
\]

\[
u = \frac{-4 - 10}{2} = \frac{-14}{2} = -7
\]

#### Step 3: Substitute back for \( \sqrt{x} \)

Now, substitute \( u = \sqrt{x} \):

- If \( u = 3 \), then \( \sqrt{x} = 3 \), so \( x = 3^2 = 9 \).
- If \( u = -7 \), then \( \sqrt{x} = -7 \), which is not possible because the square root of a number cannot be negative.

So, the only valid solution is:

\[
x = 9
\]

---

### Final Solutions:

1. \( x = \frac{5}{2} \) for the first equation.
2. \( x = 9 \) for the second equation.

Would you like more details or clarification on any step?

---

Here are five related questions to further explore these concepts:

1. What happens if the exponent in the first equation is \( \frac{3}{4} \) instead of \( \frac{2}{3} \)?
2. How would the solution change if we added a coefficient in front of \( \sqrt{x} \) in the second equation?
3. Can the quadratic formula be used if the discriminant is negative?
4. How would you solve \( (3x - 2)^{\frac{3}{2}} = 27 \)?
5. What is the significance of rational exponents in algebra, and how do they relate to roots?

---

**Tip:** When solving equations involving rational exponents, converting them into integer exponents by raising both sides to the reciprocal power simplifies the problem considerably.

Solve the equation with rational exponents 3(2𝑥 + 3)^{2/3} + 5 = 17. Solve the quadratic-like equation 𝑥 + 4√𝑥 − 21 = 0 by making an appropriate substitution.

Learn how to solve the equation 3(2x + 3)^{2/3} + 5 = 17 involving rational exponents and the quadratic-like equation x + 4√x - 21 = 0 using substitution. This tutorial includes key algebraic concepts such as the properties of exponents and the quadratic formula. Suitable for students in Grades 9-11.

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