i have looked everywhere but cant find out how to solve these last two.
#1 involves the principal of zero products
0=t(t+3)
what is the solution for t?
#2 multiply
(p-(1/4))x(p+(1/4))
2006-11-03 01:41:16 · 8 answers · asked by sean_mchugh6 3 in Science & Mathematics ➔ Mathematics
#1
t=0 or -3
#2
p^2 - (1/4)^2 which is p^2 - 1/16
2006-11-03 01:46:30 · answer #1 · answered by 6 · 2⤊ 0⤋
1). There are 2 solutions: Either
t = 0 or t + 3 = 0, t = -3
2). p^2 - 1/16. (Sum and difference rule for 2 squares)
2006-11-03 06:19:52 · answer #2 · answered by steiner1745 7 · 0⤊ 0⤋
Like all quadratics, #1 has two solutions: 0 and -3
2006-11-03 01:52:01 · answer #3 · answered by JJ 7 · 0⤊ 0⤋
1 set each factor equal to zero and solve
t=0 and t+3=0= -3
2 this is the called the difference of prefect squares
so you could square the first term p and square the last term so you get (p2-(1/16))
2006-11-03 01:49:10 · answer #4 · answered by Jdicu812 1 · 0⤊ 0⤋
1.t=0 or -3
2.p^2-(1/16)
2006-11-03 01:49:35 · answer #5 · answered by raj 7 · 0⤊ 0⤋
#1 = -3
#2 = p^2-(1/16)
(someone better double check that)
2006-11-03 01:46:21 · answer #6 · answered by ? 3 · 0⤊ 1⤋
#1 is -3, I think, pretty sure
2006-11-03 01:50:10 · answer #7 · answered by liz 1 · 0⤊ 0⤋
number 1 is impossible and dunno bout 2
2006-11-03 01:44:09 · answer #8 · answered by Jamie S 1 · 0⤊ 2⤋