p^2+2pq+q^2=1
p= .49
solve for q
2007-05-12 14:17:41 · 6 answers · asked by tc 1 in Science & Mathematics ➔ Mathematics
p^2+2pq+q^2 = (p+q)^2
therefore
(p+q)^2 = 1
p+q = 1 or -1
q = 0.51 or -1.49
2007-05-12 14:22:04 · answer #1 · answered by ong_joce 2 · 0⤊ 0⤋
p^2+2pq+q^2=1
(p+q)^2 = 1
if p = 0.49 then q = 0. 51
2007-05-12 21:22:56 · answer #2 · answered by nelaq 4 · 0⤊ 0⤋
Note that p² + 2pq + q² = (p+q)².
So, taking square roots, we have
p + q = 1
.49 + q = 1
q = .51
or
p + q = -1
.49 + q = -1
q = -1.49.
2007-05-12 21:24:29 · answer #3 · answered by steiner1745 7 · 0⤊ 0⤋
(p + q)² = 1
(p + q) = ± 1
If p + q = 1
0.49 + q = 1
q = 0.51
If p + q = - 1
0.49 + q = - 1
q = - 1.49
2007-05-13 10:03:04 · answer #4 · answered by Como 7 · 0⤊ 0⤋
p^2+2pq+q^2=1
p= .49
0.49^2+2*0.49q+q^2=1
q^2+0.98q-0.7599=0
q=(-0.98屉(0.98^2+4*0.7599))/2
q=(-0.98屉(0.9604+3.0396))/2
q=(-0.98屉4)/2
q=(-0.98±2)/2
q=0.51
q=-1.49
2007-05-12 21:48:12 · answer #5 · answered by yupchagee 7 · 0⤊ 0⤋
ya what they said
2007-05-12 21:31:28 · answer #6 · answered by Martion.Alex Martion. 2 · 0⤊ 0⤋