Znajdź pierwiastki podanych trójmianów: a) y = x² - 3x - 4 = 0 Δ = (-3)² -4*1*(-4) = 9 + 16 = 25 √Δ = √25 = 5 x1 = ( 3 -5): 2*1 = (-2) : 2 = -1 x2 = (3 + 5) : 2*1 = 8 : 2 = 4 b) y = -2x² + 2x + 24 = 0 /:(-2) x² -x - 12 = 0 Δ = (-1)² - 4*1*(-12) = 1 + 48 = 49 √Δ = √ 49 = 7 x1 = ( 1 -7) : 2*1 = (-6) :2 = -3 x2 = ( 1 + 7) :2*1 = 8 : 2 = 4 c) y = -8x² - 52x - 60 -8x² - 52x - 60 = 0 ?:(-4) 2x² + 13x + 15 = 0 Δ = 13² -4*2*15= 169 - 120 = 49 √Δ=√49 = 7 x1 = (-13- 7): 2*2 = (-20) : 4 = -5 x2 = (-13 +7) : 2*2 = (-6) : 4 = - 6/4 = - 3/2 d) y = -2√2x² + 9x - 3√3 -2√2x² + 9x - 3√3 = 0 a = -2√2 b = 9 c = -3√3 Δ = 9² - 4*(-2√2)*(-3√3) = 81 - 24*√6 trudno obliczyć √Δ, gdzieś musi być bład w zadaniu e) y = -(5x - 1)² - (9 - 2x)² + 3x + 82 -(5x - 1)² - (9 - 2x)² + 3x + 82 = 0 -(25x² -10x +1) -(81 - 36x + 4x²) + 3x + 82 = 0 -25x² + 10x -1 -81 +36x -4x² + 3x + 82 = 0 -19x² + 49x -82 + 82 = 0 -19x² + 49 x = 0 - x( 19x - 49) = 0 -x = 0 lub 19x - 49 = 0 x = 0 lub 19x = 49 x= 0 lub x = 49/19
a) y = x² - 3x - 4 Δ=b²-4ac Δ=-3²-4×1×-4 Δ=9+16 Δ=25 x₁=-b-√Δ /2a x₂=-b+√Δ /2a x₁=3-5 /2 x₂=3+5 /2 x₁=-1 x₂=4 b) y = -2x² + 2x + 24 Δ=b²-4ac Δ=2²-4×-2×24 Δ=4+192 Δ=196 x₁=-b-√Δ /2a x₂=-b+√Δ /2a x₁=-2-14 /-4 x₂=-2+14 /-4 x₁=4 x₂=-3 c) y = -8x² - 52x - 60 Δ=b²-4ac Δ=-52²-4×-8×-60 Δ=2704-1920 Δ=784 x₁=-b-√Δ /2a x₂=-b+√Δ /2a x₁=52-28 /-16 x₂=52+28 /-16 x₁=-1,5 x₂=-5 d) y = -2√2x² + 9x - 3√3 Δ=b²-4ac Δ=9²-4×-√8×-√27 Δ=81-4√216 Δ=22,2 (w przybliżeniu) x₁=-b-√Δ /2a x₂=-b+√Δ /2a x₁=-9-4,7 /5,7 x₂=-9+4,7 /5,7 x₁=-2,4 x₂=-0,75 e) y = -(5x - 1)² - (9 - 2x)² + 3x + 82 y = -(25x²-10x+1) - (4x²-36x+81) + 3x + 82 y = -25x²+10x-1-4x²+36x-81+3x+82 y = -29x²+49x Δ=b²-4ac Δ=49²-4×-29×0 Δ=2401-0 Δ=2401 x₁=-b-√Δ /2a x₂=-b+√Δ /2a x₁=-49-49 /-58 x₂=-49+49 /-58 x₁=-1,5 x₂=1,7 http://www.fizykon.org/wzory/wzory_matem_trojmian_kwadratowy.htm
