Solve the following equation by making an appropriate substitution.

4 x minus 13 x Superscript 1 divided by 2 Baseline plus 10 equals 0

The equation is:

\[
4x - 13x^{\frac{1}{2}} + 10 = 0
\]

### Step 1: Substitution
Let \( u = x^{\frac{1}{2}} \), or equivalently, \( u^2 = x \).

Substituting this into the equation, we replace \( x^{\frac{1}{2}} \) with \( u \) and \( x \) with \( u^2 \). The equation becomes:

\[
4u^2 - 13u + 10 = 0
\]

### Step 2: Solve the quadratic equation
Now solve the quadratic equation \( 4u^2 - 13u + 10 = 0 \).

We can use the quadratic formula:

\[
u = \frac{-(-13) \pm \sqrt{(-13)^2 - 4(4)(10)}}{2(4)}
\]

Simplifying the discriminant:

\[
u = \frac{13 \pm \sqrt{169 - 160}}{8}
\]

\[
u = \frac{13 \pm \sqrt{9}}{8}
\]

\[
u = \frac{13 \pm 3}{8}
\]

This gives two solutions for \( u \):

\[
u = \frac{13 + 3}{8} = \frac{16}{8} = 2
\]
\[
u = \frac{13 - 3}{8} = \frac{10}{8} = \frac{5}{4}
\]

### Step 3: Back-substitute for \( u \)
Recall that \( u = x^{\frac{1}{2}} \), so:

- If \( u = 2 \), then \( x^{\frac{1}{2}} = 2 \), which means \( x = 2^2 = 4 \).
- If \( u = \frac{5}{4} \), then \( x^{\frac{1}{2}} = \frac{5}{4} \), which means \( x = \left( \frac{5}{4} \right)^2 = \frac{25}{16} \).

### Final Answer:
The solutions to the equation are:
\[
x = 4 \quad \text{or} \quad x = \frac{25}{16}
\]

Would you like further details on any part of this solution?

### Related Questions:
1. How do you solve a general quadratic equation using substitution?
2. What is the process for finding the square root of a fraction?
3. Can this method be applied to solve higher-degree polynomials?
4. What is the importance of checking solutions after substitution?
5. How can you confirm if a quadratic equation is factorable?

**Tip:** When using substitutions in equations, always remember to convert back to the original variable at the end!

Solve the following equation by making an appropriate substitution. 4x - 13x^(1/2) + 10 = 0

Learn how to solve the equation 4x - 13x^(1/2) + 10 using substitution. This step-by-step solution introduces substitution for square roots and uses the quadratic formula to find the solution. Ideal for high school students (Grades 9-12) studying algebra and quadratic equations.

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