I just had a simple question my professor used the substition rule to prove squareroot(16-r^2). Would it have been easier to simpilfy root(16) and then solve? So, it would be 4-r.
Thank you very much!!!!
2007-01-30 13:40:42 · 5 answers · asked by Johnny O 1 in Science & Mathematics ➔ Mathematics
Square-Root is not a linear function. Linearity is a property by which a function, say f, satisfies the following:
f(x+y) = f(x) + f(y)
Hence:
(!= means "not-equal-to")
sqrt(16-r^2) != sqrt(16)-sqrt(r^2) = 4-r
2007-01-30 14:25:02 · answer #1 · answered by mohrahit 1 · 0⤊ 0⤋
16 - r^2 is a square difference because :
16 - r^2 = (4 - r) (4 + r) = 0
solve : 4- r = 0 => r = 4
4 + r = 0 => r = -4
2007-01-30 21:50:16 · answer #2 · answered by frank 7 · 0⤊ 0⤋
I think you mean solving Squareroot(16-r^2) {let us call squareroot, SR}, not prroving. I think it is easier to simplify SR(16-r^2)=SR[(4-r)(4+r)]=SR(4-r).SR(4+r)=0
So the result is: r=+4 and r= -4
2007-01-31 06:33:48 · answer #3 · answered by Newton 1 · 0⤊ 0⤋
It is not possible to do that because you need to a have a product. In this particular case you have a sum. First, you need to do the sum and only then you extract the square root.
2007-01-30 21:46:21 · answer #4 · answered by cloud_strife 2 · 0⤊ 0⤋
it cannot be simplified that way...it is true only when r=0
because
16-r^2= (4+r)(4-r)
2007-01-31 02:24:31 · answer #5 · answered by grandpa 4 · 0⤊ 0⤋