2x^2 +10x +47 = 0
solve by completeing the square.
2007-02-22 14:04:41 · 3 answers · asked by Joshua V 2 in Science & Mathematics ➔ Mathematics
divide all by 2
x^2 + 5x +23.5 = 0
subtract 23.5 from both sides
add 6.25 to both because 5/2^2 = 6.25
simplify to
(x + 5/2)^2
(sorry to mathassignment, but (5/2)^2 = 25/4, not 5/4)
x= -5/2
2007-02-22 14:11:24 · answer #1 · answered by lifelesswonder 4 · 0⤊ 0⤋
2x^2 +10x +47 = 0
2x^2+10x = -47
divide by 2
x^2 + 5x = -23.5
(x+5/2)^2 = -23.5 +5/4
x+5 = sqrt(-22.25)
x=+-4.72i - 5
2007-02-22 22:09:58 · answer #2 · answered by leo 6 · 0⤊ 0⤋
first move the 47 to the other side
2x^2+10x=-47
then divide by 2
x^2+5x=-23.5
then add half of 5 squared to each side (6.25)
x^2+5x+6.25=-17.25
then factor it
(x+2.5)^2=-17.25
then square root each side so
x+2.5=the sqaure root of (-17.25)
then solve for x so
x=-2.5(+or-)i[square root of 17.25]
2007-02-22 22:14:01 · answer #3 · answered by L_Bizzy25 2 · 0⤊ 0⤋