Question

i can simplify radical expressions into reduced radical form.
a. $\sqrt{625x^6y^{18}}$
b. $\sqrt3{27x^6y^{12}}$
c. $\sqrt{60x^{15}y^3}$
d. $\sqrt3{320x^8y^{18}}$
e. $\sqrt{5x^8} \cdot \sqrt4{10x^7}$
f. $\sqrt3{2x^5} \cdot \sqrt9{6x^5}$
g. $\sqrt{6x^5} \cdot \sqrt3{2x^4}$
h. $\frac{\sqrt{10y}}{\sqrt{2x^5}}$
i. $\sqrt{\frac{5x^7}{6y^7}}$
j. $\sqrt3{\frac{3y^{10}}{25x^4}}$

Explanation:

Step1: Simplify \(\sqrt{625x^{6}y^{18}}\)

We know that \(\sqrt{a^2}=a\) for \(a\geq0\), and \(\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\) (\(a\geq0,b\geq0\)). First, factor the numbers and variables inside the square root:
\(625 = 25^2\), \(x^{6}=(x^{3})^{2}\), \(y^{18}=(y^{9})^{2}\).
So \(\sqrt{625x^{6}y^{18}}=\sqrt{25^{2}\cdot(x^{3})^{2}\cdot(y^{9})^{2}}\)
Using the property \(\sqrt{abc}=\sqrt{a}\cdot\sqrt{b}\cdot\sqrt{c}\), we get:
\(\sqrt{25^{2}}\cdot\sqrt{(x^{3})^{2}}\cdot\sqrt{(y^{9})^{2}}\)
Since \(\sqrt{a^{2}} = a\) for non - negative \(a\), this simplifies to \(25x^{3}y^{9}\)

Step2: Simplify \(\sqrt[3]{27x^{6}y^{12}}\)

We know that \(\sqrt[3]{a^{3}}=a\) and \(\sqrt[3]{abc}=\sqrt[3]{a}\cdot\sqrt[3]{b}\cdot\sqrt[3]{c}\) (\(a,b,c\) real numbers).
Factor the terms inside the cube root: \(27 = 3^{3}\), \(x^{6}=(x^{2})^{3}\), \(y^{12}=(y^{4})^{3}\)
So \(\sqrt[3]{27x^{6}y^{12}}=\sqrt[3]{3^{3}\cdot(x^{2})^{3}\cdot(y^{4})^{3}}\)
Using the property of cube roots, we have \(\sqrt[3]{3^{3}}\cdot\sqrt[3]{(x^{2})^{3}}\cdot\sqrt[3]{(y^{4})^{3}}=3x^{2}y^{4}\)

Step3: Simplify \(\sqrt{60x^{15}y^{3}}\)

First, factor \(60 = 4\times15\), \(x^{15}=x^{14}\cdot x=(x^{7})^{2}\cdot x\), \(y^{3}=y^{2}\cdot y\)
So \(\sqrt{60x^{15}y^{3}}=\sqrt{4\times15\times(x^{7})^{2}\cdot x\times y^{2}\cdot y}\)
Using the property \(\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\), we get:
\(\sqrt{4}\cdot\sqrt{(x^{7})^{2}}\cdot\sqrt{y^{2}}\cdot\sqrt{15xy}\)
Since \(\sqrt{4} = 2\), \(\sqrt{(x^{7})^{2}}=x^{7}\), \(\sqrt{y^{2}}=y\) (assuming \(x\geq0,y\geq0\)), this simplifies to \(2x^{7}y\sqrt{15xy}\)

Step4: Simplify \(\sqrt[3]{320x^{8}y^{18}}\)

Factor \(320=64\times5 = 4^{3}\times5\), \(x^{8}=x^{6}\cdot x^{2}=(x^{2})^{3}\cdot x^{2}\), \(y^{18}=(y^{6})^{3}\)
So \(\sqrt[3]{320x^{8}y^{18}}=\sqrt[3]{4^{3}\times5\times(x^{2})^{3}\cdot x^{2}\cdot(y^{6})^{3}}\)
Using the property of cube roots \(\sqrt[3]{abc}=\sqrt[3]{a}\cdot\sqrt[3]{b}\cdot\sqrt[3]{c}\), we get:
\(\sqrt[3]{4^{3}}\cdot\sqrt[3]{(x^{2})^{3}}\cdot\sqrt[3]{(y^{6})^{3}}\cdot\sqrt[3]{5x^{2}}\)
Simplifying the cube - root of perfect cubes: \(4x^{2}y^{6}\sqrt[3]{5x^{2}}\)

Step5: Simplify \(\sqrt{5x^{8}}\cdot\sqrt[4]{10x^{7}}\)

First, rewrite the radicals with fractional exponents. Recall that \(\sqrt[n]{a}=a^{\frac{1}{n}}\) and \(a^{m}\cdot a^{n}=a^{m + n}\)
\(\sqrt{5x^{8}}=(5x^{8})^{\frac{1}{2}}=5^{\frac{1}{2}}x^{4}\) (since \((x^{8})^{\frac{1}{2}}=x^{4}\))
\(\sqrt[4]{10x^{7}}=(10x^{7})^{\frac{1}{4}}=10^{\frac{1}{4}}x^{\frac{7}{4}}\)
Now multiply them together: \(5^{\frac{1}{2}}\times10^{\frac{1}{4}}\times x^{4+\frac{7}{4}}\)
Simplify the exponents of \(x\): \(4+\frac{7}{4}=\frac{16 + 7}{4}=\frac{23}{4}=5+\frac{3}{4}\)
\(5^{\frac{1}{2}}=5^{\frac{2}{4}}\), so \(5^{\frac{2}{4}}\times10^{\frac{1}{4}}=(25\times10)^{\frac{1}{4}}=250^{\frac{1}{4}}\)
And \(x^{\frac{23}{4}}=x^{5+\frac{3}{4}}=x^{5}x^{\frac{3}{4}}\)
So \(\sqrt{5x^{8}}\cdot\sqrt[4]{10x^{7}}=250^{\frac{1}{4}}x^{5}x^{\frac{3}{4}}=x^{5}\sqrt[4]{250x^{3}}\) (rewriting back to radical form)

Step6: Simplify \(\sqrt[3]{2x^{5}}\cdot\sqrt[9]{6x^{5}}\)

Rewrite with fractional exponents: \((2x^{5})^{\frac{1}{3}}\cdot(6x^{5})^{\frac{1}{9}}\)
Using the property \((ab)^{n}=a^{n}b^{n}\), we get \(2^{\frac{1}{3}}x^{\frac{5}{3}}\cdot6^{\frac{1}{9}}x^{\frac{5}{9}}\)
Multiply the coefficients and the variables separately:
For coefficients: \(2^{\frac{1}{3}}\cdot6^{\frac{1}{9}}=2^{\frac{3}{9}}\cdot6^{\frac{1}{9}}=(8\times6)^{\frac{1}{9}} = 48^{\frac{1}{9}}\)
For variables: \(x^{\frac{5}{3}+\frac{5}{9}}=x^{\frac{15 + 5}{9}}=x^{\frac{20}{9}}=x^{2+\frac{2}{9}}=x^{2}x^{\frac{2}{9}}\)
So \(\sqrt[3]{2x^{5}}\cdot\sqrt[9]{6x^{5}}=x^{2}\sqrt[9]{48x^{2}}\) (rewriting back to radical form)

Step7: Simplify \(\frac{\sqrt{10y}}{\sqrt{2x^{5}}}\)

Rationalize the denominator by multiplying the numerator and denominator by \(\sqrt{2x^{5}}\):
\(\frac{\sqrt{10y}\cdot\sqrt{2x^{5}}}{\sqrt{2x^{5}}\cdot\sqrt{2x^{5}}}=\frac{\sqrt{20x^{5}y}}{2x^{5}}\)
Factor \(20 = 4\times5\), \(x^{5}=x^{4}\cdot x=(x^{2})^{2}\cdot x\)
\(\sqrt{20x^{5}y}=\sqrt{4\times5\times(x^{2})^{2}\cdot x\cdot y}=2x^{2}\sqrt{5xy}\)
So \(\frac{2x^{2}\sqrt{5xy}}{2x^{5}}=\frac{\sqrt{5xy}}{x^{3}}\)

Step8: Simplify \(\sqrt{\frac{5x^{7}}{6y^{7}}}\)

Rewrite as \(\frac{\sqrt{5x^{7}}}{\sqrt{6y^{7}}}\). Factor \(x^{7}=x^{6}\cdot x=(x^{3})^{2}\cdot x\), \(y^{7}=y^{6}\cdot y=(y^{3})^{2}\cdot y\)
\(\sqrt{5x^{7}}=x^{3}\sqrt{5x}\), \(\sqrt{6y^{7}}=y^{3}\sqrt{6y}\)
Rationalize the denominator by multiplying numerator and denominator by \(\sqrt{6y}\):
\(\frac{x^{3}\sqrt{5x}\cdot\sqrt{6y}}{y^{3}\sqrt{6y}\cdot\sqrt{6y}}=\frac{x^{3}\sqrt{30xy}}{y^{3}\times6y}=\frac{x^{3}\sqrt{30xy}}{6y^{4}}\)

Step9: Simplify \(\sqrt[3]{\frac{3y^{10}}{25x^{4}}}\)

Rewrite as \(\frac{\sqrt[3]{3y^{10}}}{\sqrt[3]{25x^{4}}}\). Factor \(y^{10}=y^{9}\cdot y=(y^{3})^{3}\cdot y\), \(x^{4}=x^{3}\cdot x\)
\(\sqrt[3]{3y^{10}}=y^{3}\sqrt[3]{3y}\), \(\sqrt[3]{25x^{4}}=x\sqrt[3]{25x}\)
Rationalize the denominator by multiplying numerator and denominator by \(\sqrt[3]{25^{2}x^{2}}\):
\(\frac{y^{3}\sqrt[3]{3y}\cdot\sqrt[3]{625x^{2}}}{x\sqrt[3]{25x}\cdot\sqrt[3]{625x^{2}}}=\frac{y^{3}\sqrt[3]{1875x^{2}y}}{x\sqrt[3]{15625x^{3}}}\)
Since \(\sqrt[3]{15625x^{3}} = 25x\)
\(\frac{y^{3}\sqrt[3]{1875x^{2}y}}{25x^{2}}=\frac{y^{3}\sqrt[3]{125\times15x^{2}y}}{25x^{2}}=\frac{5y^{3}\sqrt[3]{15x^{2}y}}{25x^{2}}=\frac{y^{3}\sqrt[3]{15x^{2}y}}{5x^{2}}\)

Answer:

a. \(25x^{3}y^{9}\)

b. \(3x^{2}y^{4}\)

c. \(2x^{7}y\sqrt{15xy}\)

d. \(4x^{2}y^{6}\sqrt[3]{5x^{2}}\)

e. \(x^{5}\sqrt[4]{250x^{3}}\)

f. \(x^{2}\sqrt[9]{48x^{2}}\)

g. \(25x^{3}y^{9}\) (Wait, no, for g: \(\sqrt{6x^{5}}\cdot\sqrt[3]{2x^{4}}\). Rewrite with exponents: \((6x^{5})^{\frac{1}{2}}\cdot(2x^{4})^{\frac{1}{3}}=6^{\frac{1}{2}}x^{\frac{5}{2}}\cdot2^{\frac{1}{3}}x^{\frac{4}{3}}\)
\(6^{\frac{1}{2}}=6^{\frac{3}{6}}\), \(2^{\frac{1}{3}}=2^{\frac{2}{6}}\), so coefficient: \((6^{3}\times2^{2})^{\frac{1}{6}}=(216\times4)^{\frac{1}{6}} = 864^{\frac{1}{6}}\)
Exponent of \(x\): \(\frac{5}{2}+\frac{4}{3}=\frac{15 + 8}{6}=\frac{23}{6}=3+\frac{5}{6}\), \(x^{3}x^{\frac{5}{6}}\)
So \(\sqrt{6x^{5}}\cdot\sqrt[3]{2x^{4}}=x^{3}\sqrt[6]{864x^{5}}\))

h. \(\frac{\sqrt{5xy}}{x^{3}}\)

i. \(\frac{x^{3}\sqrt{30xy}}{6y^{4}}\)

j. \(\frac{y^{3}\sqrt[3]{15x^{2}y}}{5x^{2}}\)

(Note: The original hand - written answers in the image have some errors. The above are the correct simplification steps and results for each radical expression.)