Question

1. Express the solution of, x^2 + 6x + 5 ≥ 0, using interval notation.

2. Express the solution of, x^3 + 7x^2 < x + 7, using interval notation.

3. A ball is thrown straight up from the top of a tower that is 280 ft high with an initial velocity of 48 ft/s. The height of the object can be modeled by the equation s(t) = -16t^2 + 48t + 280.
Determine the time(s) the ball is lower than the building. Express your answer in interval notation. In two or more sentences, describe your solution method.

Answer 1

Solution: (1)The interval notation of
`x^2+6x+5\geq 0` is
`(-\infty ,5]\cup [-1,\infty )`.

Step-by-step explanation:

`x^2+6x+5\geq 0`

`x^2+5x+x+5\geq 0\\x(x+5)+1(x+5)\geq 0\\(x+5)(x+1)\geq 0`

The product of two expression is positive if and only if both are positive or both are negative.

case 1:
`(x+5)\geq 0` and
`(x+1)\geq 0`

Then
`x\geq -5` and
`x\geq -1`

From above conditions
`x\geq -1`.

case 2:
`(x+5)\leq 0` and case 2:
`(x+1)\leq 0`

Then
`x\leq -5` and
`x\leq -1`

From above conditions
`x\leq -5`.

By case 1 and 2 it is conclude that the interval notation of
`x^2+6x+5\geq 0` is
`(-\infty ,5]\cup [-1,\infty )`.

Solution: (2) The interval notation of
`x^3+7x^2<x+7` is
`(-\infty ,-7)\cup (-1,1 )`.

Step-by-step explanation:

`x^3+7x^2<x+7`

`x^3+7x^2-x-7<0\\x^2(x+7)-1(x+7)<0\\(x+7)(x^2-1)<0\\(x+7)(x-1)(x+1)<0`

The relative equation of the above inequality is
`(x+7)(x-1)(x+1)=0`

Equate each factor equal to 0. The zeros of above equation is -7,-1 and 1. These points divides the number line in 4 parts or intervals. The any value from those interval to check the inequality. If the test point satisfies the equality then that interval is the solution of the inequality.

By the above process the interval notation of
`x^3+7x^2<x+7` is
`(-\infty ,-7)\cup (-1,1 )`.

Solution: (3) The ball is lower than the building for
`t>3`.

Step-by-step explanation:

The height of the ball is given by the equation
`s(t)=-16t^2+48t+280` and the height of the building is 280ft.

The ball is lower than the building if the height of the ball is less than 280ft, i.e.,
`s(t)<280`.

`-16t^2+48t+280<280\\16t(-t+3)<0`

The time is always positive, so the first factor 16t is always positive. Since first factor is positive and the product of both factors is less than zero, therefore the second factor (-t+3) is always less than zero.

`-t+3<0\\-t<-3\\t>3`

Therefore, the ball is lower than the building for
`t>3`.