Question

Solve the inequality and express your answer in interval notation x^2+12x+7<0

Answer 1

(-11.385, -0.615)

Explanation:

Inequation: x^2+12x+7<0

Equation: x^2+12x+7=0

Use the quadratic formula to get the roots of the equation.

`(-b \pm √(b^2 - 4* a * c))/(2* a)`

`(-12 \pm √(12^2 - 4* 1 * 7))/(2* 1)`

`(-12 \pm 10.77)/(2)`

Root 1:

`(-12 + 10.77)/(2)`

-0.615

Root 2:

`(-12 - 10.77)/(2)`

-11.385

Then, the inequality can be expressed as

(x + 0.615)*(x + 11.385)<0

There are 2 ways to satisfy it:

x + 0.615 < 0 and x + 11.385 > 0

or

x + 0.615 > 0 and x + 11.385 < 0

For the first case:

x + 0.615 < 0 and x + 11.385 > 0

x < -0.615 and x > -11.385

In interval notation (-11.385, -0.615)

For the second case:

x + 0.615 > 0 and x + 11.385 < 0

x > -0.615 and x < -11.385

Which is impossible to satisfy. Therefore, the solution is (-11.385, -0.615)

Answer 2

x∈
`(-6-√(29),-6+√(29))`

Explanation:

The given equation is:

`x^(2)+12x+7<0`

Using the quadratic formula, we get

`x=\frac{-(12){\pm}\sqrt{(12)^(2)-4{*}1{*}7}}{2(1)}`

=
`\frac{-12{\pm}√(144-28)}{2}`

=
`\frac{-12{\pm}√(116)}{2}`

=
`\frac{-12{\pm}2√(29)}{2}`

=
`\frac{-6{\pm}√(29)}{1}`

Thus, x∈
`(-6-√(29),-6+√(29))`