Question

solve the following inequality 1< 3x^2-7x+8≤2​

Answer 1

Answer:

1≤

x

2

+1

3x

2

−7x+8

≤2

x

2

+1

3x

2

−7x+8−2(x

2

+1)

≤0

(x

2

+1>0)

3x

2

−7x+8−2x

2

−2≤0

x

2

−7x+6≤0

x

2

−6x−x+6≤0

(x−1)(x−6)≤0

xε(1,6)

(ii)1≤

x

2

+1

3x

2

−7x+8

x

2

+1

3x

2

−7x+8−(x

2

+1)

≥0

3x

2

−7x+8−x

2

−x≥0

2x

2

−7x+7≥0

D=b

2

−4ac=49−(4×2×7)<0

hencexεallrealnumbers

from(i)&(ii)

xε(1,6)