10.1: Simplify Radicals (2024)

If \(a\), \(b\) are any two positive real numbers, then \[\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\nonumber\] In general, if \(a\), \(b\) are any two positive real numbers, then \[\sqrt[n]{ab}=\sqrt[n]{a}\cdot\sqrt[n]{b},\nonumber\] where \(n\) is a positive integer and \(n\geq 2\).

Simplify: \(\sqrt{75}\)

Solution

We can apply the product rule for radicals to simplify this number. We need to find the largest factor of \(75\) that is a perfect square (since we have a square root) and rewrite the radicand as a product of this perfect square and its other factor. The largest factor of radicand \(75\) that is a perfect square is \(25\).

\[\begin{array}{rl}\sqrt{75}&\text{Rewrite radicand as a product of }25\text{ and }3 \\ \sqrt{25\cdot 3}&\text{Apply product rule for radicals} \\ \sqrt{25}\cdot\sqrt{3}&\text{Simplify each square root} \\ 5\cdot\sqrt{3}&\text{Rewrite} \\ 5\sqrt{3}&\text{Simplified expression}\end{array}\nonumber\]

If the radicand is not a perfect square, we leave as is; hence, we left \(\sqrt{3}\) as is.

Simplify: \(\sqrt{72}\)

Solution

We can apply the product rule for radicals to simplify this number. We need to find the largest factor of \(72\) that is a perfect square (since we have a square root) and rewrite the radicand as a product of this perfect square and its other factor. The largest factor of radicand \(72\) that is a perfect square is \(36\).

\[\begin{array}{rl}\sqrt{72}&\text{Rewrite radicand as a product of }36\text{ and }2 \\ \sqrt{36\cdot 2}&\text{Apply product rule for radicals} \\ \sqrt{36}\cdot\sqrt{2}&\text{Simplify each square root} \\ 6\cdot\sqrt{2}&\text{Rewrite} \\ 6\sqrt{2}&\text{Simplified expression}\end{array}\nonumber\]

If the radicand is not a perfect square, we leave as is; hence, we left \(\sqrt{2}\) as is.

Simplify: \(5\sqrt{63}\)

Solution

We can apply the product rule for radicals to simplify this number and multiply coefficients in the last steps. We need to find the largest factor of \(63\) that is a perfect square (since we have a square root) and rewrite the radicand as a product of this perfect square and its other factor. The largest factor of radicand \(63\) that is a perfect square is \(9\).

\[\begin{array}{rl}5\sqrt{63}&\text{Rewrite radicand as a product of }9\text{ and }7 \\ 5\sqrt{9\cdot 7}&\text{Apply product rule for radicals} \\ 5\cdot\sqrt{9}\cdot\sqrt{7}&\text{Simplify each square root} \\ 5\cdot 3\cdot\sqrt{7}&\text{Rewrite and simplify coefficients} \\ 15\sqrt{7}&\text{Simplified expression}\end{array}\nonumber\]

If the radicand is not a perfect square, we leave as is; hence, we left \(\sqrt{7}\) as is.

Let \(a\) be the base, and \(m\) and \(n\) be real real numbers. Then

\[a^{\dfrac{m}{n}}=\sqrt[n]{a^m}=\left(\sqrt[n]{a}\right)^m\nonumber\]

The denominator of a rational exponent is the root on the radical and vice versa

Rewrite each radical with its corresponding rational exponent.

1. \((\sqrt[5]{x})^3\)
2. \((\sqrt[6]{3x})^5\)
3. \(\dfrac{1}{\left(\sqrt[7]{a}\right)^3}\)
4. \(\dfrac{1}{\left(\sqrt[3]{xy}\right)^2}\)

Solution

1. For the expression \((\sqrt[5]{x})^3\), we see the root is \(5\). This means that the denominator of the rational exponent is \(5\). Hence, the numerator is the exponent \(3\): \((\sqrt[5]{x})^3=x^{\dfrac{3}{5}}\).
2. For the expression \((\sqrt[6]{3x})^5\), we see the root is \(6\). This means that the denominator of the rational exponent is \(6\). Hence, the numerator is the exponent \(5\): \((\sqrt[6]{3x})^5=(3x)^{\dfrac{5}{6}}\).
3. For the expression \(\dfrac{1}{(\sqrt[7]{a})^3}\), we see the root is \(7\). This means that the denominator of the rational exponent is \(7\). Hence, the numerator is the exponent \(3\). Furthermore, since the expression with the radical is in the denominator, we can rewrite the expression using a negative exponent: \(\dfrac{1}{(\sqrt[7]{a})^3}=(a)^{-\dfrac{3}{7}}\).
4. For the expression \(\dfrac{1}{(\sqrt[3]{xy})^2}\), we see the root is \(3\). This means that the denominator of the rational exponent is \(3\). Hence, the numerator is the exponent \(2\). Furthermore, since the expression with the radical is in the denominator, we can rewrite the expression using a negative exponent: \(\dfrac{1}{(\sqrt[3]{xy})^2}=(xy)^{-\dfrac{2}{3}}\).

Rewrite each expression in its equivalent radical form.

1. \(a^{\dfrac{5}{3}}\)
2. \((2mn)^{\dfrac{2}{7}}\)
3. \(x^{-\dfrac{4}{5}}\)
4. \((xy)^{-\dfrac{2}{9}}\)

Solution

1. From the definition, we know that the denominator of the rational exponent is the root making the numerator the power: \(a^{\dfrac{5}{3}}=\sqrt[3]{a^5}\) or \((\sqrt[3]{a})^5\).
2. From the definition, we know that the denominator of the rational exponent is the root making the numerator the power: \((2mn)^{\dfrac{2}{7}}=\sqrt[7]{(2mn)^2}\) or \((\sqrt[7]{2mn})^2\).
3. From the definition, we know that the denominator of the rational exponent is the root making the numerator the power: \(x^{−\dfrac{4}{5}} = (\sqrt[5]{x})^{-4}\). Notice that the expression still contains a negative exponent. Hence, we need to reciprocate the radical to rewrite the expression with only positive exponents: \[x^{-\dfrac{4}{5}}=\dfrac{1}{(\sqrt[5]{x})^4}\nonumber\]
4. From the definition, we know that the denominator of the rational exponent is the root making the numerator the power: \((xy)^{−\dfrac{2}{9}} = (\sqrt[9]{x})^{−2}\). Notice that the expression still contains a negative exponent. Hence, we need to reciprocate the radical to rewrite the expression with only positive exponents: \[(xy)^{-\dfrac{2}{9}}=\dfrac{1}{(\sqrt[9]{xy})^2}\nonumber\]

Nicole Oresme, a Mathematician born in Normandy was the first to use rational exponents. He used the notation \(\dfrac{1}{3} • 9^{p}\) to represent \(9^{\dfrac{1}{3}}\). However, his notation went largely unnoticed

Evaluate \(27^{-\dfrac{4}{3}}\).

Solution

We first rewrite the expression with only positive exponents, then evaluate the exponen

\[\begin{array}{rl}27^{-\dfrac{4}{3}}&\text{Rewrite the expression with positive exponents} \\ \dfrac{1}{27^{\dfrac{4}{3}}}&\text{Rewrite in radical form} \\ \dfrac{1}{(\sqrt[3]{27})^4}&\text{Evaluate radical }\sqrt[3]{27}=3 \\ \dfrac{1}{(3)^4}&\text{Evaluate exponent }3^4=81 \\ \dfrac{1}{81}&\text{Result} \end{array}\nonumber\]

Thus, \(27^{−\dfrac{4}{3}} = \dfrac{1}{81}\). This result should emphasize the fact that negative exponents means reciprocals, and not negative numbers.

Simplify: \(\sqrt{x^6 y^5}\)

Solution

We can apply the product rule for radicals to simplify by rewriting the variable’s exponent and rewrite the exponents so that one of the exponents is the largest even number.

\[\begin{array}{rl}\sqrt{x^6y^5}&\text{Rewrite radicand} \\ \sqrt{x^6\cdot y^4\cdot y^1}&\text{Apply product rule for radicals} \\ \sqrt{x^6}\cdot\sqrt{y^4}\cdot\sqrt{y}&\text{Simplify each square root} \\ x^3\cdot y^2\cdot\sqrt{y}&\text{Rewrite and simplify coefficients} \\ x^3y^2\sqrt{y}&\text{Simplified expression}\end{array}\nonumber\]

Notice that \((x^3)^2\) and \((y^2)^2=y^4\); hence, we extract the perfect squares of the variables and leave the \(\sqrt{y}\) as is.

Recall, when taking a square root of a number, the radicand must be greater than or equal to zero. So, when we are applying the square root to variables, the variables must also be greater than or equal to zero.

Notice, we are essentially dividing the exponents on the variables by two and the factor that remains in the radicand has exponent \(1\).

Simplify: \(-5\sqrt{18x^4y^6z^{10}}\). Assume all variables are positive.

Solution

We can apply the product rule for radicals to simplify by rewriting the variable’s exponent and rewrite the exponents so that one of the exponents is the largest even number.

\[\begin{array}{rl}-5\sqrt{18x^4y^6z^{10}}&\text{Rewrite radicand} \\ -5\cdot\sqrt{9\cdot 2\cdot x^4\cdot y^6\cdot x^{10}}&\text{Apply product rule for radicals} \\ -5\cdot\sqrt{9}\cdot\sqrt{2}\cdot\sqrt{x^4}\cdot\sqrt{y^6}\cdot\sqrt{z^{10}}&\text{Simplify each square root} \\ -5\cdot 3\cdot\sqrt{2}\cdot x^2\cdot y^3\cdot z^5 &\text{Rewrite and simplify coefficients} \\ -15x^2y^3z^5\sqrt{2}&\text{Simplified expression}\end{array}\nonumber\]

Simplify: \(\sqrt{20x^5y^9z^6}\). Assume all variables are positive.

Solution

We can apply the product rule for radicals to simplify by rewriting the variable’s exponent and rewrite the exponents so that one of the exponents is the largest even number.

\[\begin{array}{rl}\sqrt{20x^5y^9z^6}&\text{Rewrite radicand} \\ \sqrt{4\cdot 5\cdot x^4\cdot x\cdot y^8\cdot y\cdot z^6}&\text{Apply product rule for radicals} \\ \sqrt{4}\cdot\sqrt{5}\cdot\sqrt{x^4}\cdot\sqrt{x}\cdot\sqrt{y^8}\cdot\sqrt{y}\cdot\sqrt{z^6}&\text{Simplify each square root} \\ 2\cdot\sqrt{5}\cdot x^2\cdot\sqrt{x}\cdot y^4\cdot\sqrt{y}\cdot z^3&\text{Rewrite and simplify coefficients} \\ 2x^2y^4z^3\sqrt{5xy}&\text{Simplified expression}\end{array}\nonumber\]

\(\sqrt{245}\)

\(\sqrt{36}\)

\(\sqrt{12}\)

\(3\sqrt{12}\)

\(6\sqrt{128}\)

\(-8\sqrt{392}\)

\(\sqrt{192n}\)

\(\sqrt{196v^2}\)

\(\sqrt{252x^2}\)

\(-\sqrt{100k^4}\)

\(-7\sqrt{64x^4}\)

\(-5\sqrt{36m}\)

\(\sqrt{45x^2y^2}\)

\(\sqrt{16x^3y^3}\)

\(\sqrt{320x^4y^4}\)

\(6\sqrt{80xy^2}\)

\(5\sqrt{245x^2y^3}\)

\(-2\sqrt{180u^3v}\)

\(-8\sqrt{180x^4y^2z^4}\)

\(2\sqrt{80hj^4k}\)

\(-4\sqrt{54mnp^2}\)

\(\sqrt{125}\)

\(\sqrt{196}\)

\(\sqrt{338}\)

\(5\sqrt{32}\)

\(7\sqrt{128}\)

\(-7\sqrt{63}\)

\(\sqrt{343b}\)

\(\sqrt{100n^3}\)

\(\sqrt{200a^3}\)

\(-4\sqrt{175p^4}\)

\(-2\sqrt{128n}\)

\(8\sqrt{112p^2}\)

\(\sqrt{72a^3b^4}\)

\(\sqrt{512a^4b^2}\)

\(\sqrt{512m^4n^3}\)

\(8\sqrt{98mn}\)

\(2\sqrt{72x^2y^2}\)

\(-5\sqrt{72x^3y^4}\)

\(6\sqrt{50a^4bc^2}\)

\(-\sqrt{32xy^2z^3}\)

\(-8\sqrt{32m^2p^4q}\)

\(m^{\dfrac{3}{5}}\)

\((7x)^{\dfrac{3}{2}}\)

\((10r)^{-\dfrac{3}{4}}\)

\((6b)^{-\dfrac{4}{3}}\)

\(\dfrac{1}{(\sqrt{6x})^3}\)

\(\dfrac{1}{(\sqrt[4]{n})^7}\)

\(\sqrt{v}\)

\(\sqrt{5a}\)

\(8^{\dfrac{2}{3}}\)

\(4^{\dfrac{3}{2}}\)

\(16^{\dfrac{1}{4}}\)

\(100^{-\dfrac{3}{2}}\)