QuanticaPrealgebraOpen in the courseIn 8.2 you estimated \(\sqrt{12} \approx 3.5\). This lesson rewrites it exactly. The number \(12 = 4 \times 3\) hides a perfect square, and \(\sqrt{ab} = \sqrt{a}\sqrt{b}\) lets you pull \(\sqrt{4} = 2\) out front, giving \(2\sqrt{3}\).
Problem
Compute \(\sqrt{4} \times \sqrt{25}\) and compare it to \(\sqrt{4 \times 25}\). Do they agree? What is \(\sqrt{100}\)?
Show a hint
- A perfect square is a number whose root is whole. Ask what whole number times itself gives \(100\).
- Since \(10 \times 10 = 100\), the root of \(100\) is \(10\). Notice it matches \(\sqrt{4} \times \sqrt{25} = 2 \times 5\).
Show the full solution
\(\sqrt{4} \times \sqrt{25} = 2 \times 5 = 10\), and \(\sqrt{4 \times 25} = \sqrt{100} = 10\), since \(10 \times 10 = 100\). Both routes give \(\boxed{10}\). Rooting a product and multiplying the roots always agree, and that is the rule the rest of this lesson runs on.
Problem
Simplify \(\sqrt{12}\) using \(12 = 4 \times 3\). Split into \(\sqrt{4} \times \sqrt{3}\), then root the perfect-square piece. What whole number comes out in front of \(\sqrt{3}\)?
Show a hint
- Split \(12\) as \(4 \times 3\), so \(\sqrt{12} = \sqrt{4} \times \sqrt{3}\). The \(\sqrt{3}\) part cannot be tidied, but \(\sqrt{4}\) can.
- Since \(\sqrt{4} = 2\), the root becomes \(2 \times \sqrt{3}\), written \(2\sqrt{3}\). The whole number in front is \(2\).
Show the full solution
Since \(12 = 4 \times 3\), $$\sqrt{12} = \sqrt{4} \times \sqrt{3} = 2\sqrt{3}.$$ The whole number in front is \(\boxed{2}\). The \(3\) has no perfect-square factor, so it stays under the radical. Squaring back gives \((2\sqrt{3})^2 = 4 \times 3 = 12\).
Problem
Simplify \(\sqrt{72}\). The largest perfect square dividing \(72\) is \(36\), since \(72 = 36 \times 2\). What whole number appears in front of \(\sqrt{2}\)?
Show a hint
- Write \(72\) as \(36 \times 2\), so \(\sqrt{72} = \sqrt{36} \times \sqrt{2}\).
- Since \(\sqrt{36} = 6\), the simplest form is \(6\sqrt{2}\). The number in front is \(6\).
Show the full solution
Since \(72 = 36 \times 2\), $$\sqrt{72} = \sqrt{36} \times \sqrt{2} = 6\sqrt{2}.$$ The whole number in front is \(\boxed{6}\). The leftover \(2\) is square-free, so \(6\sqrt{2}\) is simplest, and \((6\sqrt{2})^2 = 36 \times 2 = 72\) confirms it.
Problem
Simplify \(\sqrt{48}\). The largest perfect square dividing \(48\) is \(16\), since \(48 = 16 \times 3\). What number remains under the radical?
Show a hint
- Write \(48 = 16 \times 3\), so \(\sqrt{48} = \sqrt{16} \times \sqrt{3}\). Which piece stays under the root?
- Since \(\sqrt{16} = 4\), the simplest form is \(4\sqrt{3}\). The number left under the radical is \(3\).
Show the full solution
Since \(48 = 16 \times 3\), $$\sqrt{48} = \sqrt{16} \times \sqrt{3} = 4\sqrt{3}.$$ The number left under the radical is \(\boxed{3}\). The \(4\) comes out front and the \(3\) stays put, because \(3\) has no perfect-square factor bigger than \(1\).
Problem
Simplify \(\sqrt{98}\). Since \(98 = 49 \times 2\), split the root. What whole number is in front of \(\sqrt{2}\)?
Show a hint
- The perfect squares to test are \(4, 9, 16, 25, 36, 49\). Which one divides \(98\)? Try \(49\).
- Since \(98 = 49 \times 2\), we get \(\sqrt{98} = \sqrt{49} \times \sqrt{2} = 7\sqrt{2}\). The number in front is \(7\).
Show the full solution
Since \(98 = 49 \times 2\), $$\sqrt{98} = \sqrt{49} \times \sqrt{2} = 7\sqrt{2}.$$ The whole number in front is \(\boxed{7}\). To find that factor, run through the squares \(4, 9, 16, 25, 36, 49\) and keep the largest one that divides the radicand.
Problem
Factor \(30 = 2 \times 3 \times 5\). Every prime appears just once, so no repeated factor forms a perfect square. What is the largest perfect square dividing \(30\)?
Show a hint
- A perfect square needs a repeated factor, like \(2 \times 2\) or \(3 \times 3\). In \(30 = 2 \times 3 \times 5\), is any factor repeated?
- None of \(4, 9, 16, 25\) divides \(30\). The only perfect square that does is \(1\), so \(\sqrt{30}\) is already in simplest form.
Show the full solution
Factor \(30 = 2 \times 3 \times 5\). Every prime shows up once, so there is no repeated prime to build a square from, and none of \(4\), \(9\), \(16\), or \(25\) divides \(30\). The largest perfect-square factor is \(\boxed{1}\). Pulling out \(\sqrt{1} = 1\) changes nothing, so \(\sqrt{30}\) is already in simplest form.
Problem
Simplify \(\sqrt{10} \times \sqrt{5}\). Combine under one radical to get \(\sqrt{50}\). Then pull out the perfect-square factor. What whole number is in front of \(\sqrt{2}\)?
Show a hint
- Combine first, \(\sqrt{10} \times \sqrt{5} = \sqrt{50}\). Then look for the largest perfect square dividing \(50\).
- Since \(50 = 25 \times 2\), we get \(\sqrt{50} = \sqrt{25} \times \sqrt{2} = 5\sqrt{2}\). The number in front is \(5\).
Show the full solution
Combine the roots, \(\sqrt{10} \times \sqrt{5} = \sqrt{50}\). The largest perfect square dividing \(50\) is \(25\), so $$\sqrt{50} = \sqrt{25} \times \sqrt{2} = 5\sqrt{2}.$$ The whole number in front is \(\boxed{5}\). Combining first often exposes a square factor that neither root showed on its own.
Problem
Simplify \(\sqrt{507}\) to the form \(a\sqrt{3}\). What is \(a\)?
Show a hint
- The number \(507\) is odd, so \(4, 16, 36, 64, 100\) cannot divide it. Try odd perfect squares like \(9\), \(25\), \(49\), \(121\), \(169\).
- Since \(507 = 169 \times 3\) and \(\sqrt{169} = 13\), the simplest form is \(13\sqrt{3}\). The number in front is \(13\).
Show the full solution
Since \(507 = 169 \times 3\) and \(169 = 13^2\), $$\sqrt{507} = \sqrt{169} \times \sqrt{3} = 13\sqrt{3},$$ so \(a = \boxed{13}\). Because \(507\) is odd, only odd squares can divide it, which shortens the search to \(9, 25, 49, 121, 169\).
Practice these ideas
Practice
Simplify \(\sqrt{8}\). The largest perfect square dividing \(8\) is \(4\), since \(8 = 4 \times 2\). Written as a whole number times \(\sqrt{2}\), what is the whole number in front?
Show the solution
Since \(8 = 4 \times 2\), $$\sqrt{8} = \sqrt{4} \times \sqrt{2} = 2\sqrt{2}.$$ The whole number in front is \(\boxed{2}\).
Practice
Simplify \(\sqrt{75}\). Since \(75 = 25 \times 3\) and \(25\) is a perfect square, the root tidies into a whole number times \(\sqrt{3}\). What is the whole number in front?
Show the solution
Since \(75 = 25 \times 3\), $$\sqrt{75} = \sqrt{25} \times \sqrt{3} = 5\sqrt{3}.$$ The whole number in front is \(\boxed{5}\).
Practice
Simplify \(\sqrt{44}\). The largest perfect square dividing \(44\) is \(4\), since \(44 = 4 \times 11\). What number is left under the radical?
Show the solution
Since \(44 = 4 \times 11\), $$\sqrt{44} = \sqrt{4} \times \sqrt{11} = 2\sqrt{11}.$$ The \(11\) is square-free and stays under the radical, so the number left under it is \(\boxed{11}\).
Practice
Simplify \(\sqrt{128}\). The largest perfect square dividing \(128\) is \(64\), since \(128 = 64 \times 2\). Written as a whole number times \(\sqrt{2}\), what is the whole number in front?
Show the solution
Since \(128 = 64 \times 2\), $$\sqrt{128} = \sqrt{64} \times \sqrt{2} = 8\sqrt{2}.$$ The whole number in front is \(\boxed{8}\).
Practice
Combine \(\sqrt{3} \times \sqrt{12}\) under one radical and evaluate. What whole number is the result?
Show the solution
Join the roots, \(\sqrt{3} \times \sqrt{12} = \sqrt{3 \times 12} = \sqrt{36}\). Since \(36\) is a perfect square, \(\sqrt{36} = 6\), a clean whole number. So the value is \(\boxed{6}\).
Practice
Factor \(60 = 2 \times 2 \times 3 \times 5\). The repeated prime forms a perfect square. What is the largest perfect square dividing \(60\)?
Show the solution
Factor \(60 = 2 \times 2 \times 3 \times 5\). The only repeated prime is \(2\), giving the perfect square \(2 \times 2 = 4\). No larger perfect square divides \(60\), since \(3\) and \(5\) appear only once. So the largest perfect-square factor is \(\boxed{4}\). (It would simplify \(\sqrt{60}\) to \(2\sqrt{15}\).)
Practice
Decide whether \(\sqrt{33}\) can be simplified. Factor \(33\) completely, then find the largest perfect square that divides it. What is that largest perfect square?
Show the solution
Factor \(33 = 3 \times 11\). Each prime appears once, so there is no repeated factor to form a perfect square. The largest perfect square dividing \(33\) is \(\boxed{1}\), which means \(\sqrt{33}\) is already in simplest form.
Practice
Combine \(\sqrt{15} \times \sqrt{6}\) under one radical to get \(\sqrt{90}\). Since \(90 = 9 \times 10\), simplify. What whole number appears in front of \(\sqrt{10}\)?
Show the solution
Join the roots, \(\sqrt{15} \times \sqrt{6} = \sqrt{15 \times 6} = \sqrt{90}\). The largest perfect square dividing \(90\) is \(9\), and \(90 = 9 \times 10\), so $$\sqrt{90} = \sqrt{9} \times \sqrt{10} = 3\sqrt{10}.$$ The whole number in front is \(\boxed{3}\).
Practice
Simplify \(\sqrt{675}\). Since \(675 = 225 \times 3\) and \(\sqrt{225} = 15\), what whole number is in front of \(\sqrt{3}\)?
Show the solution
Since \(675 = 225 \times 3\) and \(225 = 15^2\), $$\sqrt{675} = \sqrt{225} \times \sqrt{3} = 15\sqrt{3}.$$ The whole number in front is \(\boxed{15}\). Squaring back, \((15\sqrt{3})^2 = 225 \times 3 = 675\).