Question

Can someone please explain to me how to solve these kinds of problems?

Answer 1

1)
`10\sqrt{10`

3)
`15`

5)
`9\sqrt3`

7)
`35\sqrt7`

Explanation:

We can simplify these values by factoring the value underneath the square root sign.

1) We can factor 1000 as:

`1000 = 10 \cdot 10 \cdot 10`

From this factorization, we can see that there is a pair of 10s. So, we can represent this as one 10 outside the square root.

`√(1000)`

`= √(10 \cdot 10 \cdot 10)`

`= \boxed{10√(10)}`

3) We can factor 225 as:

`225 = 5 \cdot 5 \cdot 3 \cdot 3`

We can see that there is a pair of 5s and a pair of 3s. So, we have one 5 and one 3 outside the square root.

`√(225)`

`= √(5\cdot 5 \cdot 3 \cdot 3)`

`= 5 \cdot 3 √(1)`

`= \boxed{15}`

5) We can factor 243 as:

`243 = 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3`

We can see that there are two pairs of 3s. So, there will be two 3s outside the square root.

`\sqrt{243`

`= \sqrt{3 \cdot 3 \cdot3 \cdot 3\cdot 3`

`= 3 \cdot 3\sqrt3`

`=\boxed{9\sqrt3}`

7) We can factor 175 as:

`175 = 5 \cdot 5 \cdot 7`

We can see that there is a pair of 5s. So, there will be one 5 outside the square root multiplied by the 7 that is already there.

`7\sqrt{175`

`= 7√( 5 \cdot 5 \cdot 7)`

`= 7 \cdot 5\sqrt7`

`= \boxed{35\sqrt7}`