2007-01-03 18:22:00 · 7 answers · asked by Lost in math 1 in Science & Mathematics ➔ Mathematics
Solve,
(t - 4)^2 - 64 = 0
Add 64 to both sides,
(t - 4)^2 - 64 + (64) = 0 + (64)
(t - 4)^2 = 64
Square-root both sides
sqr( (t - 4)^2 ) = sqr( 64 )
(t - 4) = 8 or (t - 4) = - 8
t = 8 + 4 or t = - 8 + 4
t = 12 or t = - 4
2007-01-03 20:10:23 · answer #1 · answered by ideaquest 7 · 0⤊ 0⤋
So we want (t-4)^2 = 64.
Since 8^2 and (-8)^2 both equal 64, we have
(t-4) = 8 or (t-4) = -8.
This gives t=12 or t=-4.
2007-01-03 18:27:21 · answer #2 · answered by GeorgeMKLam 1 · 0⤊ 0⤋
This is a difference of squares where a^2-b^2=(a+b)(a-b):
(t-4)^2 - 8^2 = (t-4 +8)(t-4 -8) since a=t-4 and b=8
=(t+4)(t-12)=0
t+4=0 so t=-4
t-12=0 so t=12
2007-01-03 18:26:21 · answer #3 · answered by Professor Maddie 4 · 0⤊ 0⤋
(t-4)^2-64=0
(t-4)^2=64
(t-4)=16
Hence t=16-4 = 12
Solved and simplified
2007-01-03 18:26:47 · answer #4 · answered by ? 2 · 0⤊ 2⤋
(t-4)^2-64=0 (add 64 to both sides)
(t-4)^2 = 64 (sqrt both sides)
t-4=plus or minus 8 (add 4 to both sides)
t= 12 or -4
2007-01-03 18:27:55 · answer #5 · answered by Tommy R 2 · 0⤊ 0⤋
(t-4)^2-64=0
Add 64 to each side
(t-4)^2=64
Take the square root of each side
t-4=8 and t-4=-8
Add 4 to each side
t=12 and t=-4
2007-01-03 18:27:00 · answer #6 · answered by unhrdof 3 · 1⤊ 0⤋
(t-4)^2 = 64
t-4 = sqrt(64)
t-4 = +/- 8
t = 12 or -4
2007-01-03 18:26:53 · answer #7 · answered by hznfrst 6 · 0⤊ 1⤋