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!