Answer:
[tex]\sqrt{3}\cdot\sqrt{6}=3\cdot\sqrt{2}[/tex]
Step-by-step explanation:
Radicals
We have the expression:
[tex]\sqrt{3}\cdot\sqrt{6}[/tex]
Since 6=2*3, it can be rewritten as:
[tex]\sqrt{3}\cdot\sqrt{6}=\sqrt{3}\cdot\sqrt{2}\cdot\sqrt{3}[/tex]
Joining like radicals:
[tex]\sqrt{3}\cdot\sqrt{6}=\sqrt{3}\cdot\sqrt{3}\cdot\sqrt{2}[/tex]
[tex]\sqrt{3}\cdot\sqrt{6}=\sqrt{9}\cdot\sqrt{2}[/tex]
[tex]\boxed{\sqrt{3}\cdot\sqrt{6}=3\cdot\sqrt{2}}[/tex]
This is the required form with b=2
The integer b for [tex]\rm b\sqrt{2}[/tex] such as [tex]\rm \sqrt{3}\times \sqrt{6} = b\sqrt{2}[/tex] holds good is 3 .
According to the property of square root for any positive integer x
[tex]\rm \sqrt{x} \times \sqrt{x} = x....(1)[/tex]
According to the given condition
[tex]\rm \sqrt{3}\times \sqrt{6} = b\sqrt{2}[/tex]
We can simplify the left hand side of the equation as follows
[tex]\sqrt{3}\times \sqrt{6}[/tex]
[tex]= \sqrt{3}\times \sqrt{2\times 3 }[/tex]
[tex]= 3\sqrt{2}[/tex] ( By using equation (1))
So on comparing with right hand side of equation (1) we get
[tex]\rm 3\sqrt{2} = b\sqrt{2}\\hence \; on\; comparing\\b = 3[/tex]
The integer b for [tex]\rm b\sqrt{2}[/tex] such as [tex]\rm \sqrt{3}\times \sqrt{6} = b\sqrt{2}[/tex] holds good is 3 .
For more information please refer to the link below
https://brainly.com/question/17309565
