3x(x+3)=(x+3) ^2 3x²+9x=x²+6x+9 3x²-x²+9x-6x-9=0 2x²+3x-9=0 Δ=b²-4ac = 9+72=81 √Δ=9 x1 = (-b-√Δ)/2a = (-3-9)/4 = -12/4 = -3 x2 = (-b+√Δ)/2a = (-3+9)/4 = 6/4 = 1,5 2 2x^2+ 7x > lub rowne 4 2x²+7x-4 ≥0 Δ = 49+32=81 √Δ = 9 x1 = (-7-9)/4 = -4 x2 = (-7+9)/4 = 2/4 = 0,5 przy x² jest znak + czyli x∈(-∞-4> V <0,5;∞) 3 y= 1/4 x^2 - 2x +4 Δ = 4-4=0 Δ=0 x = -b/2a = 2/0,5 = 4 czyli funkcja ma jeden podwójny pierwiastek y = a(x-x₀)² = 1/4(x-4)²
[latex]3x(x+3)=(x+3) ^2\ 3x^2+9x=x^2+6x+9\ 2x^2+3x-9=0\ \ Delta = b^2-4ac=3^2-4cdot 2 cdot (-9)=81\ \ x_{1}=frac{-b-sqrt{Delta}}{2a}=frac{-3-9}{4}=-3\ x_{1}=frac{-b+sqrt{Delta}}{2a}=frac{-3+9}{4}=frac{3}{2}\[/latex] [latex]2x^2+ 7x ge 4\ 2x^2+7x-4ge 0\ \ Delta=7^2-4cdot 2 cdot (-4)=49+32=81\ \ x_{1}=frac{-7-9}{4}=-4\ x_{2}=frac{-7+9}{4}=frac{1}{2}\ \ x in (-infty, -4> cup
