Generally, sqr(x^2)= the absolute value of x (lxl)
What about sqr(x^4)? is it gonna be just x^2 or lx^2l?
2006-12-02 05:13:37 · 4 answers · asked by 7 in Science & Mathematics ➔ Mathematics
sqrt(x^4) is just going to be x^2, because saying |x^2| would be redundant (there's no way x^2 could be a negative number within the realm of real numbers).
2006-12-02 05:16:12 · answer #1 · answered by Puggy 7 · 0⤊ 0⤋
Yahoo Answers is such a crap precisely because incompetent people like to jump in. If you are "not sure, but you think you know", please, don't answer.
Gopal is right. The solution to sqrt(x^4) is a two-item set {x^2,-x^2}. By the same, token the solution to sqrt(x^2) is not |x| but +/-x.
2006-12-02 06:00:32 · answer #2 · answered by Angelique 1 · 0⤊ 0⤋
gopal ur an idiot. sorry.
yes, sqrt(x^4) = x^2 for all cases:
examples
x=-2
sqrt(16) = 4 ==true
x=1+i
sqrt(-4) = 2i ==true
2006-12-02 05:22:10 · answer #3 · answered by MooseBoys 6 · 0⤊ 0⤋
sq.rt of x^2=+/-x
sq.rt of x^4=+/-x^2
2006-12-02 05:16:45 · answer #4 · answered by raj 7 · 0⤊ 0⤋