Hi, I have experience this in one of Knewton's practice problems which is stated as a valid square root: sqrt(a^2) = -a, if a<0 I don't understand. If a is negative, say a = -1 then (a)^2 is positive. And the square root per definition has positive result, no? Grateful for clarification

This has more to do with how a square root function is defined...so this result is by definition of a square root function sqrt(a^2) = mod of a = a for all a>= 0 and -a for all a<0

bkw wrote:Hi, I have experience this in one of Knewton's practice problems which is stated as a valid square root: sqrt(a^2) = -a, if a<0 I don't understand. If a is negative, say a = -1 then (a)^2 is positive. And the square root per definition has positive result, no? Grateful for clarification Both you and Knewton's statement are correct: If a = -1, then √a² = √(-1)² = √1 = 1. -a = -(-1) = 1. Thus, when a<0, √a² = -a.

mk101 wrote:This has more to do with how a square root function is defined...so this result is by definition of a square root function sqrt(a^2) = mod of a = a for all a>= 0 and -a for all a<0 what do you mean by "mod", modulo %?

GMATGuruNY wrote:bkw wrote:Hi, I have experience this in one of Knewton's practice problems which is stated as a valid square root: sqrt(a^2) = -a, if a<0 I don't understand. If a is negative, say a = -1 then (a)^2 is positive. And the square root per definition has positive result, no? Grateful for clarification Both you and Knewton's statement are correct: If a = -1, then √a² = √(-1)² = √1 = 1. -a = -(-1) = 1. Thus, when a<0, √a² = -a. Gotcha, Thanks!

bkw wrote:mk101 wrote:This has more to do with how a square root function is defined...so this result is by definition of a square root function sqrt(a^2) = mod of a = a for all a>= 0 and -a for all a =0 and -a for a<0 ....