Solve the equation by making an appropriate substitution.

x minus 8 StartRoot x EndRoot plus 15x−8x+15equals=0

We are given the equation:

\[
x - 8\sqrt{x} + 15 = 0
\]

To solve this equation, we can make a substitution to simplify the expression involving the square root. Let:

\[
u = \sqrt{x}
\]

Then, squaring both sides:

\[
u^2 = x
\]

Substituting \(u\) into the original equation:

\[
u^2 - 8u + 15 = 0
\]

This is now a quadratic equation in terms of \(u\). To solve it, we use the quadratic formula:

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

For the equation \(u^2 - 8u + 15 = 0\), we have \(a = 1\), \(b = -8\), and \(c = 15\). Plugging these values into the quadratic formula:

\[
u = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(15)}}{2(1)}
\]
\[
u = \frac{8 \pm \sqrt{64 - 60}}{2}
\]
\[
u = \frac{8 \pm \sqrt{4}}{2}
\]
\[
u = \frac{8 \pm 2}{2}
\]

Thus, we have two possible solutions for \(u\):

\[
u = \frac{8 + 2}{2} = 5 \quad \text{or} \quad u = \frac{8 - 2}{2} = 3
\]

Now, recall that \(u = \sqrt{x}\), so we square both solutions to find \(x\):

\[
u = 5 \implies \sqrt{x} = 5 \implies x = 25
\]
\[
u = 3 \implies \sqrt{x} = 3 \implies x = 9
\]

Thus, the solutions to the original equation are:

\[
x = 25 \quad \text{or} \quad x = 9
\]

Would you like further details or have any questions?

Here are 5 related questions for further exploration:
1. How do you solve quadratic equations using factoring?
2. What is the importance of the discriminant in the quadratic formula?
3. How do you verify the solution of a square root equation?
4. What are some common strategies for solving non-linear equations?
5. Can you apply substitution methods to solve more complex equations?

**Tip**: Always check both solutions in the original equation to ensure they satisfy it, especially when dealing with square roots.

Solve the equation by making an appropriate substitution.
x - 8√x + 15 = 0

Learn how to solve the equation x - 8√x + 15 = 0 using substitution and the quadratic formula. This problem covers key algebraic concepts like solving square root equations and applying the quadratic formula. Ideal for high school students in Grades 9-12.

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