How do you insert parentheses to make #4*2-2^2 ÷9+2=6# true?

2 Answers
May 15, 2016

Answer:

#(4*2-2)^2 -: 9 + 2 = 6#

or better:

#((4*2-2)^2 -: 9)+2 = 6#

Explanation:

PEMDAS is a mnemonic for the following conventions of order of operations:

P Parentheses.

E Exponents.

MD Multiplication and Division - evaluated left to right.

AS Addition and Subtraction - evaluated left to right.

Note that multiplication and division have the same priority and addition and subtraction have the same priority.

Using PEMDAS (or BODMAS, etc.) it is sufficient to add just one pair of parentheses:

#(4*2-2)^2 -: 9 + 2#

#= (8-2)^2 -: 9 + 2#

#= 6^2 -: 9 + 2#

#= 36 -: 9 + 2#

#= 4 + 2#

#=6#

Further considerations

PEMDAS (or BODMAS, etc) are intended to help clarify the meaning of expressions involving arithmetic operations, but sometimes lead to confusion instead.

For example, what is the intended meaning of the following expression?

#6-:2(1+2)#

PEMDAS says that the division (#-:#) should be performed before the multiplication by juxtaposition, so the interpretation is like:

#(6-:2)(1+2) = 9#

Those unfamiliar with PEMDAS or having slightly varied rules would probably interpret it more like:

#6-:(2(1+2)) = 1#

It would be better to add more parentheses to the expression to make the intended meaning clear.

Since the current question does not specify PEMDAS, and does not specify that only one pair of parentheses are allowed, it is probably better to add a second pair of parentheses to make it clear that you want the division performed before adding #2#:

#((4*2-2)^2 -: 9) + 2 = 6#

Historically the obelus #-:# symbol has sometimes meant that you should divide the whole expression to the left of the obelus by the whole expression on the right. That would not work for us, since it would result in something more like:

#(4*2-2)^2-: (9 + 2) = 36/11#

May 15, 2016

Answer:

#(4⋅2−2)^2÷9+2=6#

Explanation:

To insert parenthesis appropriately so that following is always true.

#4⋅2−2^2÷9+2=6#

Step 1. From PEMDAS we know that #+ and -# operators are processed at the end. Therefore we can work backwards. Independent terms associated with #+# operator can be taken to RHS of the equation. We get
#4⋅2−2^2÷9=6-2#
#4⋅2−2^2÷9=4#, must also be true.

Step 2. Inspection reveals that exponent operator will get precedence over the #÷ and -# operators for last two terms. If we execute #2^2# first and then execute #÷# operator, we obtain a fraction. This is not acceptable. Only way to avoid this situation is to place a right parenthesis before the exponent.
#4⋅2−2)^2÷9=4#

Step 3. Use brute force to place left parenthesis before #4# or take the independent term #9# to RHS
#4⋅2−2)^2=4cdot 9#
#4⋅2−2)^2=36#

Step 4. We have only two places to put the left parenthesis.
(a) #4⋅(2−2)^2!=36#, failure.
(b) #(4⋅2−2)^2=36#, success.

Final result
#(4⋅2−2)^2÷9+2=6#