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60 Square Root Simplified

What is the 60 square root simplified, and how can \(\sqrt{60}\) be rewritten in simplest radical form using prime factorization and perfect-square extraction?

Subject: Math Algebra Chapter: Algebraic Expressions and Polynomials Topic: Rationalizing Algebraic Expressions Answer included
60 square root simplified simplify radicals simplest radical form square root simplification perfect square factor prime factorization extracting squares radical expression
Accepted answer Answer included

Goal

The task “60 square root simplified” means rewriting \(\sqrt{60}\) so that no perfect-square factor remains inside the radical. The final answer should be in simplest radical form.

Key property

If \(a \ge 0\) and \(b \ge 0\), then \[ \sqrt{a\cdot b} = \sqrt{a}\cdot\sqrt{b}. \] In particular, if \(a\) is a perfect square, \(\sqrt{a}\) becomes an integer and can be moved outside the radical.

Method 1: Factor out the largest perfect square

Find a perfect-square factor of \(60\). Since \(60 = 4\cdot 15\) and \(4\) is a perfect square:

  1. Rewrite using a perfect square: \[ \sqrt{60} = \sqrt{4\cdot 15}. \]
  2. Split the radical: \[ \sqrt{4\cdot 15} = \sqrt{4}\cdot\sqrt{15}. \]
  3. Evaluate \(\sqrt{4}\): \[ \sqrt{4}\cdot\sqrt{15} = 2\cdot\sqrt{15}. \]

60 square root simplified: \(\;\sqrt{60} = 2\sqrt{15}\)

Method 2: Prime factorization and pairing

Prime factorization makes the perfect-square structure explicit:

\[ 60 = 2\cdot 2\cdot 3\cdot 5 = 2^2\cdot 3\cdot 5. \]

Use \(\sqrt{2^2\cdot 3\cdot 5} = \sqrt{2^2}\cdot\sqrt{3\cdot 5}\):

\[ \sqrt{60} = \sqrt{2^2\cdot 3\cdot 5} = 2\cdot\sqrt{15}. \]

Quick verification

Squaring the simplified expression should reproduce the original radicand:

\[ (2\sqrt{15})^2 = 2^2\cdot(\sqrt{15})^2 = 4\cdot 15 = 60. \]

Common pitfalls

  • Stopping too early: \(\sqrt{60}=\sqrt{4\cdot 15}\) is not yet simplified until the perfect square \(4\) is extracted.
  • Incorrect splitting: \(\sqrt{a+b}\ne \sqrt{a}+\sqrt{b}\) in general; only products split: \(\sqrt{a\cdot b}=\sqrt{a}\cdot\sqrt{b}\).
  • Leaving a square inside: after simplification, the number inside the radical must have no perfect-square factor greater than \(1\).

Visualization: factor tree and the “paired” square

Factor tree for 60 (pair the primes to extract perfect squares) 60 6 10 2 3 2 5 The pair \(2\cdot 2 = 2^2\) becomes \(2\) outside: \(\sqrt{60}=2\sqrt{15}\).
Prime factors of \(60\) are \(2,2,3,5\). The paired \(2\cdot 2\) forms a perfect square, allowing extraction of \(2\) from \(\sqrt{60}\).
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