ZAD.1 A/ 2x² -3x-1=0 Δ= b² - 4ac = (-3)² - 4*2*(-1) = 9 +8=17 √Δ = √17 ≈4,12 x₁ = (-b-√Δ)/2a = (3-4,12)/4 = -1,12/4 = -0,28 x₂ = (-b+√Δ)/2a = (3+4,12)/4 = 7,12/4 = 1,78 B/ 5x(x+2)=60+5x 5x² + 10x = 60 + 5x 5x² + 10x - 60 - 5x = 0 5x² + 5x - 60 = 0 Δ= b² - 4ac = 5² - 4*5*(-60) = 25 + 1200 =125 √Δ = √1225 = 35 x₁ = (-b-√Δ)/2a = (-5 - 35)/10 = -40/10 = -4 x₂ = (-b+√Δ)/2a = (-5 + 35)/10 = 30/10 = 3 ZAD.2. A/ x²+ 5x+5 > 0 Δ= b² - 4ac = 5² - 4*1*5 = 25 -20 = 5 √Δ = √5 ≈ 2,24 x₁ = (-b-√Δ)/2a = (-5 – 2,24)/2 = -7,24/2 = -3,62 x₂ = (-b+√Δ)/2a = (-5+ 2,24)/2 = -2,76/2 = -1,38 a > 0 więc ramiona paraboli skierowane do góry x€ (-∞; -3,62 ) U( -1,38 ; + ∞) B/ : 2x(x-4) ≤ 2x+28 2x² -8x -2x -28 ≤ 0 2x² -10x -28≤ 0 Δ= b² - 4ac = (-10)² - 4*2*(-28) =100 +224 = 324 √Δ= √334 = 18 x₁ = (-b-√Δ)/2a = (-10 – 18)/4 = -28/4 = -7 x₂ = (-b+√Δ)/2a = = (-10 +18)/4 = 8/4 = 2 a > 0 więc ramiona paraboli skierowane do góry x € < 2,-7>
zad.1 a) 2x²-3x -1=0 4x-3x-1=0 4x-3x=0+1 x=1 b) 5x·(x+2)=60+5x 5x²+10x=60+5x 25x+10x-5x=60 30x=60/÷30 x=2 pierwsze zrobiłam, drugie coś nie moge xD
