a)
b)
4x+5 > x²
0 > x²-4x-5 ⇔ x²-4x-5 < 0
Δ=(-4)²-4·1·(-5)=16+20=36 , √Δ=√36=6
x1=(4-6)/2
x1=-1
x2=(4+6)/2
x2=5
x∈(-1,5)
6x²-x > 12
6x²-x-12 > 0
Δ=(-1)²-4·6·12=1+288=289, √Δ=√289=17
x1=(1-17)/12
x1=-16/12
x=-4/3
x2=(1+17)/12
x2=18/12
x2=3/2
x∈(-∞,-4/3)∪(3/2,∞)
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a)
b)
a)
4x+5 > x²
0 > x²-4x-5 ⇔ x²-4x-5 < 0
Δ=(-4)²-4·1·(-5)=16+20=36 , √Δ=√36=6
x1=(4-6)/2
x1=-1
x2=(4+6)/2
x2=5
x∈(-1,5)
b)
6x²-x > 12
6x²-x-12 > 0
Δ=(-1)²-4·6·12=1+288=289, √Δ=√289=17
x1=(1-17)/12
x1=-16/12
x=-4/3
x2=(1+17)/12
x2=18/12
x2=3/2
x∈(-∞,-4/3)∪(3/2,∞)