Answer:
Question A)
[tex]=\sqrt{6}x[/tex]
Question B)
[tex]=3a\sqrt{a}[/tex]
Question C)
[tex]=5\sqrt{2}b^2[/tex]
Step-by-step explanation:
A)
We are given:
[tex]\sqrt{6x^2}\, \text{ where } x\geq 0[/tex]
We can rewrite the expression:
[tex]=\sqrt{6}\cdot \sqrt{x^2}[/tex]
The square root and square will cancel each other out. Thus:
[tex]=\sqrt{6}x[/tex]
B)
We are given:
[tex]\sqrt{9a^3}[/tex]
Rewrite:
[tex]=\sqrt{9}\cdot \sqrt{a^3}[/tex]
Note that the square root of 9 is simply 3. We can also factor the second part:
[tex]=3\cdot \sqrt{a^2\cdot a}[/tex]
Rewriting:
[tex]=3\cdot\sqrt{a^2}\cdot\sqrt{a}[/tex]
Simplify:
[tex]=3a\sqrt{a}[/tex]
C)
We are given:
[tex]\sqrt{50b^4}[/tex]
Rewrite. Note that 50 = 25(2):
[tex]=\sqrt{25}\cdot \sqrt{2}\cdot \sqrt{b^4}[/tex]
Simplify. We can rewrite the factor as:
[tex]=5\cdot \sqrt{2}\cdot \sqrt{(b^2)^2}[/tex]
The square and square root will cancel out. Thus:
[tex]=5\sqrt{2}b^2[/tex]
