Respuesta :

Answer:

option: B ([tex]x^2+2x-8<0[/tex]) is correct.

Step-by-step explanation:

We are given the solution set as seen from the graph as:

(-4,2)

1)

On solving the first inequality we have:

[tex]x^2-2x-8<0[/tex]

On using the method of splitting the middle term we have:

[tex]x^2-4x+2x-8<0[/tex]

⇒  [tex]x(x-4)+2(x-4)=0[/tex]

⇒ [tex](x+2)(x-4)<0[/tex]

And we know that the product of two quantities are negative if either one of them is negative so we have two cases:

case 1:

[tex]x+2>0[/tex] and [tex]x-4<0[/tex]

i.e. x>-2 and x<4

so we have the region as:

(-2,4)

Case 2:

[tex]x+2<0[/tex] and [tex]x-4>0[/tex]

i.e. x<-2 and x>4

Hence, we did not get a common region.

Hence from both the cases we did not get the required region.

Hence, option 1 is incorrect.

2)

We are given the second inequality as:

[tex]x^2+2x-8<0[/tex]

On using the method of splitting the middle term we have:

[tex]x^2+4x-2x-8<0[/tex]

⇒ [tex]x(x+4)-2(x+4)<0[/tex]

⇒ [tex](x-2)(x+4)<0[/tex]

And we know that the product of two quantities are negative if either one of them is negative so we have two cases:

case 1:

[tex]x-2>0[/tex] and [tex]x+4<0[/tex]

i.e. x>2 and x<-4

Hence, we do not get a common region.

Case 2:

[tex]x-2<0[/tex] and [tex]x+4>0[/tex]

i.e. x<2 and x>-4

Hence the common region is (-4,2) which is same as the given option.

Hence, option B is correct.

3)

[tex]x^2-2x-8>0[/tex]

On using the method of splitting the middle term we have:

[tex]x^2-4x+2x-8>0[/tex]

⇒ [tex]x(x-4)+2(x-4)>0[/tex]

⇒ [tex](x-4)(x+2)>0[/tex]

And we know that the product of two quantities are positive if either both of them are negative or both of them are positive so we have two cases:

Case 1:

[tex]x+2>0[/tex] and [tex]x-4>0[/tex]

i.e. x>-2 and x>4

Hence, the common region is (4,∞)

Case 2:

[tex]x+2<0[/tex] and [tex]x-4<0[/tex]

i.e. x<-2 and x<4

Hence, the common region is: (-∞,-2)

Hence, from both the cases we did not get the desired answer.

Hence, option C is incorrect.

4)

[tex]x^2+2x-8>0[/tex]

On using the method of splitting the middle term we have:

[tex]x^2+4x-2x-8>0[/tex]

⇒ [tex]x(x+4)-2(x+4)>0[/tex]

⇒ [tex](x-2)(X+4)>0[/tex]

And we know that the product of two quantities are positive if either both of them are negative or both of them are positive so we have two cases:

Case 1:

[tex]x-2<0[/tex] and [tex]x+4<0[/tex]

i.e. x<2 and x<-4

Hence, the common region is: (-∞,-4)

Case 2:

[tex]x-2>0[/tex] and [tex]x+4>0[/tex]

i.e. x>2 and x>-4.

Hence, the common region is: (2,∞)

Hence from both the case we do not have the desired region.

Hence, option D is incorrect.




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