Why is -9^2+81 = 0? Shouldn’t it be 162?

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I just read an article on MSN about a tricky mathematical problem that apparently most people get wrong. The premise of the article is that, while (-9)^(2) \+ 81 would solve as 162, the lack of braces causes us to do the exponentiation first (PEMDAS) resulting in -81 + 81 = 0, which seems very incorrect as -9 isn’t an operation, it’s a literal. (PEMDAS doesn’t talk about a number’s sign.)

Stating a term of -9 doesn’t mean “take 9 and negate it”, it means “take the literal which is 9 less than 0”. Unless there’s some historical shorthand I’m missing here or something? Does -9 represent a shorthand for 0 – 9? Most programming languages treat a negative literal as a literal and not an operand. (In languages where you can override the – operand, specifying a number with a – in front of it still takes the literal without calling the operand override function.)

So, I guess my question is this: Does -9^(2) evaluate as (-9)^(2) = 81 or as -(9^(2)) = -81, and why?

Here’s the article in question.

[https://www.msn.com/en-us/lifestyle/parenting/tricky-equation-confuses-public-mathematical-misconception-leads-to-widespread-error/ar-BB1hD1mJ?rc=1&ocid=winp1taskbar&cvid=4a3ad03d63b84eb1d76097a4d6c147d3&ei=22](https://www.msn.com/en-us/lifestyle/parenting/tricky-equation-confuses-public-mathematical-misconception-leads-to-widespread-error/ar-BB1hD1mJ?rc=1&ocid=winp1taskbar&cvid=4a3ad03d63b84eb1d76097a4d6c147d3&ei=22)

Who’s right (me or the article) and why?

In: Mathematics

4 Answers

Anonymous 0 Comments

When you encounter exponentials, you always take only the first thing before it. So -9^2 indeed means “take nine, raise it to the power of two, then negate”, as number and its negation are two things.

If you want to raise to the power whole -9, then you have to write it (-9)^2. In this case, negation is also raised.

This is actually popular problem and many matematicians are educated wrong way. But simple equation is enough to explain:

a – b^2 = 0

By logic, a has to be greater than zero, as b^2 cannot be negative (ignoring complex numbers). If we use 9 as b and raise whole -9 to the power of two, we get:

a + 81 = 0 => a = -81.

Contradiction with previous statement, thus we made an error.

Other example is when we rearrange equation to a = b^2. That way we get rid of negation and we see that 9 and -81 are not solutions.

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