Verified Answer :

  • Combine the radicals into one: \(\frac{1}{2} \sqrt{\frac{50x^4y^2z}{5}}\).
  • Simplify the fraction inside the square root: \(\frac{1}{2} \sqrt{10x^4y^2z}\).
  • Extract the square roots of the variable powers \(\sqrt{x^4} = x^2\) and \(\sqrt{y^2} = |y|\).
  • State the final simplified form: \(\boxed{\frac{x^2 |y| \sqrt{10 z}}{2}}\).

Explanation

  1. Analyze the Expression and Properties

To simplify the expression \(\frac{\sqrt{50 x^4 y^2 z}}{2 \sqrt{5}}\), we first examine the structure of the expression. We have a quotient of two square roots, and we can utilize the property of radicals that states \(\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}\) for \(a, b > 0\). This allows us to combine the terms under a single square root. We also assume that \(x, y, z\) are such that the expression is well-defined, specifically \(z \geq 0\).

  1. Combine and Simplify the Radicand

Using the property \(\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}\), we can rewrite the expression as:
\[\frac{\sqrt{50 x^4 y^2 z}}{2 \sqrt{5}} = \frac{1}{2} \sqrt{\frac{50 x^4 y^2 z}{5}}\]
Next, we simplify the numerical fraction inside the radical:
\[\frac{50}{5} = 10\]
So, the expression simplifies to:
\[\frac{1}{2} \sqrt{10 x^4 y^2 z}\]

  1. Extract Perfect Squares and Form Final Expression

Now we can extract the perfect square variables from inside the square root. We know that for any real numbers \(x\) and \(y\), \(\sqrt{x^4} = x^2\) and \(\sqrt{y^2} = |y|\). Applying these properties, we pull these terms out of the square root:
\[\sqrt{10 x^4 y^2 z} = \sqrt{x^4} \cdot \sqrt{y^2} \cdot \sqrt{10 z} = x^2 |y| \sqrt{10 z}\]
Substituting this back into our expression, we get:
\[\frac{1}{2} \cdot x^2 |y| \sqrt{10 z}\]
This can also be written as:
\[\frac{x^2 |y| \sqrt{10 z}}{2}\]

Examples

Simplifying radicals is a fundamental skill used in physics and engineering, particularly when dealing with formulas like the distance formula in coordinate geometry or the magnitude of a vector. For example, when calculating the velocity of an object moving in two dimensions, you might end up with an expression like \(\sqrt{v_x^2 + v_y^2}\). Simplifying these expressions is crucial for performing further calculations efficiently and maintaining precision in scientific modeling. Mastering these algebraic manipulations helps streamline complex equations in many technical fields.