Question

For which interval does this quadratic inequality hold true? 6x − x2 − 8 > 0

a. (1, 3)
b. (2, 4)
c. (3, 5) D(4, 6)

Answer 1

-x²+6x-8>0 a=-1 , b=6 , c=-8
Δ = b² - 4.a.c
Δ = 6² - 4 . -1 . -8
Δ = 36 - 32
Δ = 4

x = (-b +- √Δ) / 2a

x' = (-6 + √4) / 2(-1) x'' = (-6 - √4)/ 2(-1)
x' = -4 / -2 x'' = -8 / -2
x' = 2 x'' = 4


2>x>4 ou (2,4)

Answer 2

`6x-x^2-8 \ \textgreater \ 0 \\ -x^2+6x-8\ \textgreater \ 0 \\ -x^2+2x+4x-8\ \textgreater \ 0 \\ -x(x-2)+4(x-2)\ \textgreater \ 0 \\ (-x+4)(x-2)\ \textgreater \ 0 \\ \\ \hbox{the zeros:} \\ (-x+4)(x-2)=0 \\ -x+4=0 \ \lor \ x-2=0 \\ x=4 \ \lor \ x=2`

The coefficient of x is negative, so the parabola opens downwards. The values greater than 0 are between the zeros.
The solution set for the inequality is:

`x \in (2,4)`

The answer is B.