Question

1. simplify.

a) \\( \sqrt{24} \\)
b) \\( \sqrt{x^8} \\)
c) \\( \sqrt{50m^3} \\)
d) \\( \sqrt{6a^3} \cdot \sqrt{12a^7} \\)

1. are the triangles

Explanation:

Part a)

Step1: Factor 24 into perfect square and other

We know that \(24 = 4\times6\), where \(4\) is a perfect square.

Step2: Use square - root property \(\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\)

\(\sqrt{24}=\sqrt{4\times6}=\sqrt{4}\times\sqrt{6}\)
Since \(\sqrt{4} = 2\), we have \(\sqrt{24}=2\sqrt{6}\)

Part b)

Step1: Recall the property of exponents and square - roots

For \(\sqrt{x^{n}}\), when \(n\) is even, \(\sqrt{x^{n}}=x^{\frac{n}{2}}\) (assuming \(x\geq0\)). Here \(n = 8\)

Step2: Calculate the exponent

\(\sqrt{x^{8}}=x^{\frac{8}{2}}=x^{4}\)

Part c)

Step1: Factor \(50m^{3}\)

We can write \(50m^{3}=25m^{2}\times2m\), where \(25m^{2}\) is a perfect square (\(25m^{2}=(5m)^{2}\))

Step2: Use square - root property

\(\sqrt{50m^{3}}=\sqrt{25m^{2}\times2m}=\sqrt{25m^{2}}\times\sqrt{2m}\)
Since \(\sqrt{25m^{2}} = 5m\) (assuming \(m\geq0\)), we get \(\sqrt{50m^{3}} = 5m\sqrt{2m}\)

Part d)

Step1: Use the property \(\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}\)

\(\sqrt{6a^{3}}\cdot\sqrt{12a^{7}}=\sqrt{(6a^{3})\times(12a^{7})}\)

Step2: Multiply the coefficients and add the exponents of like bases

First, multiply the coefficients: \(6\times12 = 72\)
Then, for the variable \(a\), using the rule \(a^{m}\cdot a^{n}=a^{m + n}\), we have \(a^{3}\cdot a^{7}=a^{3 + 7}=a^{10}\)
So, \((6a^{3})\times(12a^{7})=72a^{10}\)

Step3: Simplify the square - root

\(\sqrt{72a^{10}}\). Factor \(72a^{10}\) as \(36a^{10}\times2\), where \(36a^{10}=(6a^{5})^{2}\)
\(\sqrt{72a^{10}}=\sqrt{36a^{10}\times2}=\sqrt{36a^{10}}\times\sqrt{2}=6a^{5}\sqrt{2}\)

Answer:

s:
a) \(\boldsymbol{2\sqrt{6}}\)

b) \(\boldsymbol{x^{4}}\)

c) \(\boldsymbol{5m\sqrt{2m}}\)

d) \(\boldsymbol{6a^{5}\sqrt{2}}\)