Question

Sadie simplified the expression StartRoot 54 a Superscript 7 b cubed EndRoot, where a greater-than-or-equal-to 0, as shown colon StartRoot 54 a Superscript 7 baseline b cubed EndRoot = StartRoot 3 squared times 6 times a squared times a Superscript 5 Baseline times b squared times b EndRoot = 3 a b StartRoot 6 a Superscript 5 Baseline b EndRoot Describe the error Sadie made, and explain how to find the correct answer.\

Answer 1

Sadie did not factor a to the 7th power using the largest perfect power.

Instead of a squared times a to the 5th power, the factors should be a to the 6th times a to the 1st power.

In Sadie’s final answer, the exponent on a should have been 3.

Step-by-step explanation: EDGE 2021:)

Answer 2

Sadie's error is " she made error in step 2
`=√(3^2* 6* a^2* a^5* b^2* b)` where
`a\geqslant 0`"

Because she made error in splitting the powers to simplify the square root

Therefore the correct answer for Sadie's expression is
`3ab√(6ab)` where
`a\geqslant 0`

Explanation:

Given that " Sadie simplified the expression StartRoot 54 a Superscript 7 b cubed EndRoot, where a greater-than-or-equal-to 0, "

It can be written as
`√(54a^7b^3)` where
`a\geqslant 0`

The given expression is
`√(54a^7b^3)` where
`a\geqslant 0`

To find Sadie's error and explain the correct answer :

Sadie's steps are

`√(54a^7b^3)` where
`a\geqslant 0`

`=√(3^2* 6* a^2* a^5* b^2* b)`

`=3ab√(6a^5b)`

`√(54a^7b^3)=3ab√(6a^5b)` where
`a\geqslant 0`

Now corrected steps are

`√(54a^7b^3)` where
`a≥0`

`=\sqrt{(9* 6)(a^(6+1))(b^(2+1))`

`=\sqrt{(3^2* 6)(a^6.a^1)(b^2.b^1)` (by using the identity
`a^(m+n)=a^m.a^n`

`=\sqrt{3^2* 6* ((a^3)^2.a)(b^2.b)` (by using the identity
`a^(mn)=(a^m)^n` )

`=3ab√(6ab)`

Therefore
`√(54a^7b^3)=3ab√(6ab)` where
`a\geqslant 0`

The correct answer is
`3ab√(6ab)` where
`a\geqslant 0`

Sadie's error is " she made error in step 2
`=√(3^2* 6* a^2* a^5* b^2* b)` " where
`a\geqslant 0`

Because she made error in splitting the powers to simplify the square root

Therefore the correct answer for Sadie's expression is
`3ab√(6ab)` where
`a\geqslant 0`