Question

The graph below represents the solution set of which inequality?

Answer 1

Answer:

option: B ($x^2+2x-8$) 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:

$x^2-2x-8$

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

$x^2-4x+2x-8$

⇒  $x(x-4)+2(x-4)=0$

⇒ $(x+2)(x-4)$

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:

$x+2>0$ and $x-4$

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

so we have the region as:

(-2,4)

Case 2:

$x+2$ and $x-4>0$

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:

$x^2+2x-8$

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

$x^2+4x-2x-8$

⇒ $x(x+4)-2(x+4)$

⇒ $(x-2)(x+4)$

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:

$x-2>0$ and $x+4$

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

Hence, we do not get a common region.

Case 2:

$x-2$ and $x+4>0$

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)

$x^2-2x-8>0$

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

$x^2-4x+2x-8>0$

⇒ $x(x-4)+2(x-4)>0$

⇒ $(x-4)(x+2)>0$

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:

$x+2>0$ and $x-4>0$

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

Hence, the common region is (4,∞)

Case 2:

$x+2$ and $x-4$

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)

$x^2+2x-8>0$

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

$x^2+4x-2x-8>0$

⇒ $x(x+4)-2(x+4)>0$

⇒ $(x-2)(X+4)>0$

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:

$x-2$ and $x+4$

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

Hence, the common region is: (-∞,-4)

Case 2:

$x-2>0$ and $x+4>0$

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.