x\sqrt{x} - 7x + 10\sqrt{x} = 0.

["# Solve the Equation: ( x\sqrt{x} - 7x + 10\sqrt{x} = 0 )", "Understanding and solving equations involving both polynomial terms and square roots can be challenging, but techniques like substitution simplify the process. The equation:", "[\nx\sqrt{x} - 7x + 10\sqrt{x} = 0\n]", "is a classic example where substitution helps rewrite the equation into a more manageable form. In this article, we’ll guide you step-by-step through solving this equation, explain key mathematical concepts, and highlight practical insights to help you master similar problems.", "---", "## Understanding the Equation", "First, observe the presence of (\sqrt{x}) alongside (x) and (x\sqrt{x}). Since (\sqrt{x}) is equivalent to (x^{1/2}), and (x = x^{1}), the equation contains terms with fractional powers. This motivates us to substitute to simplify the variable exponents.", "---", "## Step 1: Let’s Use Substitution", "Let ( y = \sqrt{x} ). Then,\nSince ( x = (\sqrt{x})^2 = y^2 ),\nand ( x\sqrt{x} = y^2 \cdot y = y^3 ).", "Substituting into the original equation:", "[\ny^3 - 7y^2 + 10y = 0\n]", "Now we have a cubic equation in ( y ), which is easier to handle.", "---", "## Step 2: Factor the Substituted Equation", "Factor out the common term ( y ):", "[\ny(y^2 - 7y + 10) = 0\n]", "This gives two factors:\n1. ( y = 0 )\n2. ( y^2 - 7y + 10 = 0 )", "---", "## Step 3: Solve Each Factor", "### Solving ( y = 0 )", "[\ny = 0 \implies \sqrt{x} = 0 \implies x = 0\n]", "### Solving ( y^2 - 7y + 10 = 0 )", "Factor the quadratic:", "[\ny^2 - 7y + 10 = (y - 2)(y - 5) = 0\n]", "So,", "[\ny = 2 \quad \ ext{or} \quad y = 5\n]", "Recall ( y = \sqrt{x} ), so:", "- ( \sqrt{x} = 2 \implies x = 4 )\n- ( \sqrt{x} = 5 \implies x = 25 )", "---", "## Step 4: Verify All Solutions", "Check each candidate in the original equation ( x\sqrt{x} - 7x + 10\sqrt{x} = 0 ):", "- For ( x = 0 ):\n ( 0 - 0 + 0 = 0 \quad \checkmark )", "- For ( x = 4 ):\n ( 4 \cdot 2 - 7 \cdot 4 + 10 \cdot 2 = 8 - 28 + 20 = 0 \quad \checkmark )", "- For ( x = 25 ):\n ( 25 \cdot 5 - 7 \cdot 25 + 10 \cdot 5 = 125 - 175 + 50 = 0 \quad \checkmark )", "All solutions are valid.", "---", "## Practical Takeaways", "- Substitution is powerful for equations with mixed variables and fractional powers. Here, letting ( y = \sqrt{x} ) transformed the equation from mixed-degree terms into a solvable cubic.\n- Always verify solutions in the original equation, especially with radicals, to avoid extraneous roots.\n- Recognizing perfect squares and factoring quadratics quickly simplifies otherwise complex problems.", "---", "## Advanced Tip: Graphical and Numerical Insights", "Graphing ( f(x) = x\sqrt{x} - 7x + 10\sqrt{x} ) reveals the function crosses zero at ( x = 0 ), ( x = 4 ), and ( x = 25 )—confirming our algebraic solutions. In real-world applications, such equations model growth patterns where power-law relationships appear with square-root modifiers.", "---", "## Summary", "The equation ( x\sqrt{x} - 7x + 10\sqrt{x} = 0 ) is solved elegantly using substitution and algebraic factoring. By letting ( y = \sqrt{x} ), we reduce the problem to solving a cubic equation, then back-substitute to find:", "[\n\boxed{x = 0,\ 4,\ \ ext{or}\ 25}\n]", "Mastering this method equips you to tackle a variety of radical and polynomial equations—key skills in algebra and applied mathematics.", "---", "Keywords: solve ( x\sqrt{x} - 7x + 10\sqrt{x} = 0 ), substitution method, radical equations, solving polynomial radicals, ( y = \sqrt{x} ) substitution, step-by-step equation solving, algebraic manipulation, verifying roots.", "---", "If you found this guide helpful, explore more about substitution techniques and radical equations—your mastery of algebra grows with every problem solved!"]