I'm a bit stumped on this one.
I'm either missing something or there's an error in the solution.
I'm hoping someone can help me figure out which.

question 16 of 18 in the set called "radical actions"

The Questions is written [ with" representing powers] (7"1/2 - 6/(5"1/2) + 3 × 5"1/2) (71/2 + 6/(5"1/2) – 3 × 5"1/2) =

but the solution that follows analyses the questions as if the line up of the signs in the parentheses match. They don't. With two differing signs within, + & -, in different orders isn't this disqualified for regular factoring method a-la-quadratic?

the solution breaks it down like this:

(71/2 – [6/(51/2) – 3(51/2)]) × (71/2 + [6/(51/2) – 3(51/2)] )

Can you just do that on the left? Put parentheses around the 2nd and 3rd inner term and switch the sign on the inside but not on the outside?

I'm confused... maybe the extra numbers are throwing me off.... Thanks for any thoughts on other ways to look at this.

Heres the rest of the solution...

Notice the two terms : 71/2 and [6/(51/2) – 3(51/2)].

The expression above is the difference of these two terms, taken to the second power, and can be rewritten in the following way:

(71/2 )2 – (6/(51/2) – 3(51/2))2

***** BUT this wouldn't be true the sign in the original problem is still a plus sign... right? anyone?

Well, that is the right. this means taking the factor (-1) from both of the terms 6/(51/2) and 3(51/2). So the previous expression
- 6/(5"1/2) + 3 × 5"1/2 is equivalent to
(-1) * 6/(5"1/2) + (-1) * (- 3 × 5"1/2) - notice the minus sign before 3 now.

and the next step is to take the factor (-1) outside and group the other terms 6/(51/2) and 3(51/2) inside the brackets.


Hope this was helpful.

Caveman is right. To see it another way, you might work the steps backwards to see that (7^1/2 – [6/(5^1/2) – 3(5^1/2)]) becomes (7"1/2 - 6/(5^1/2) + 3 × 5^1/2) when you distribute the negative sign over the terms inside the [...].

More importantly, though, is recognizing why you should bother to do this at all. As you point out, the convenient thing is the repeated terms, while the inconvenient things are the different signs and the ugliness of the terms themselves.

On my paper, my procedure is always the following:
1. Replace the repeated expressions with temporary variables
2. Simplify the expression using the temporary variables,
3. Replace the temporary variables with the original expressions.

I'll model that here, using a slightly different approach from the official solution--one that doesn’t require the little sign trick you questioned.

Original problem: (7^1/2 - 6/(5^1/2) + 3 × 5^1/2) (7^1/2 + 6/(5^1/2) – 3 × 5^1/2) =

That’s of the form: (a - b + c)(a + b - c) = ?
BTW: Here's where you can also see that the sign manipulation was OK: (a - b + c) does equal (a - (b - c))
Where:
a = 7^1/2
b = 6/(5^1/2)
c = 3(5^1/2)

Work to simplify the expression with a, b, and c:
(a - b + c)(a + b - c) =
(a)(a + b - c) + (- b)(a + b - c) + (c)(a + b - c) =
(a^2 + ab - ac) + (-ab - b^2 + bc) + (ac + bc - c^2) =
a^2 - b^2 + 2bc - c^2
(Notice that many terms cancelled out…this would have been a nightmare with the original expression.)

Finally, replace the variables with the original expressions:
a^2 - b^2 + 2bc - c^2 =
7 - 36/5 + (2)(6/(5^1/2))(3(5^1/2)) - 9(5) =
7 - 36/5 + 36 - 45
-2 - 36/5 =
-10/5 - 36/5 =
-46/5

_________________
Emily Sledge
Instructor
ManhattanGMAT