Znajdź takie x i y , aby ciąg ( 8 , x , y ) był arytmetyczny , a ciąg ( 8, x , y+2 ) geometryczny .
Proszę o pomoc .
x - 8 = y - x
x / 8 = (y + 2) / x
2x - 8 = y
x² = 8(y + 2)
y = 2x - 8
x² = 8(2x - 8 + 2)
x² = 8(2x - 6)
x² = 16x - 48
x² - 16x + 48 = 0
Δ = 256 - 192 = 64
√Δ = 8
x1 = (16 - 8) / 2 = 8 / 2 = 4
x2 = (16 + 8) / 2 = 24 / 2 = 12
x1 = 4
y1 = 2x - 8 = 2 * 4 - 8 = 8 - 8 = 0
lub
x2 = 12
y2 = 2x - 8 = 24 - 8 = 16
odp. x = 4 i y = 0 lub x = 12 i y = 16
{x=(8+y)/2 ---- z wlasnosci c.a
{x² = 8*(y+2) ---- z wlasnosci c.g
{ 2x=8+y
{ x²= 8y +16
{y=2x-8
{x²= 8(2x-8) +16
x²=16x - 64 +16
x²-16x+48=0
Δ=b²-4ac=256 - 192=64 √Δ=8
x1=(-b-√Δ)/2a=(16-8)/2=4
x2=(-b+√Δ)/2a=(16+8)/2=12
y1= 2x1-8 ∨ y2=2x2-8
y1=2*4 -8 y2=2*12-8
y1=0 y2=16
{x=4 lub x=12
{y=0 y=16
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x - 8 = y - x
x / 8 = (y + 2) / x
2x - 8 = y
x² = 8(y + 2)
y = 2x - 8
x² = 8(2x - 8 + 2)
x² = 8(2x - 6)
x² = 16x - 48
x² - 16x + 48 = 0
Δ = 256 - 192 = 64
√Δ = 8
x1 = (16 - 8) / 2 = 8 / 2 = 4
x2 = (16 + 8) / 2 = 24 / 2 = 12
x1 = 4
y1 = 2x - 8 = 2 * 4 - 8 = 8 - 8 = 0
lub
x2 = 12
y2 = 2x - 8 = 24 - 8 = 16
odp. x = 4 i y = 0 lub x = 12 i y = 16
{x=(8+y)/2 ---- z wlasnosci c.a
{x² = 8*(y+2) ---- z wlasnosci c.g
{ 2x=8+y
{ x²= 8y +16
{y=2x-8
{x²= 8(2x-8) +16
x²=16x - 64 +16
x²-16x+48=0
Δ=b²-4ac=256 - 192=64 √Δ=8
x1=(-b-√Δ)/2a=(16-8)/2=4
x2=(-b+√Δ)/2a=(16+8)/2=12
y1= 2x1-8 ∨ y2=2x2-8
y1=2*4 -8 y2=2*12-8
y1=0 y2=16
{x=4 lub x=12
{y=0 y=16