Answer:
x∈[tex](-6-\sqrt{29},-6+\sqrt{29})[/tex]
Step-by-step explanation:
The given equation is:
[tex]x^{2}+12x+7<0[/tex]
Using the quadratic formula, we get
[tex]x=\frac{-(12){\pm}\sqrt{(12)^{2}-4{\times}1{\times}7}}{2(1)}[/tex]
=[tex]\frac{-12{\pm}\sqrt{144-28}}{2}[/tex]
=[tex]\frac{-12{\pm}\sqrt{116}}{2}[/tex]
=[tex]\frac{-12{\pm}2\sqrt{29}}{2}[/tex]
=[tex]\frac{-6{\pm}\sqrt{29}}{1}[/tex]
Thus, x∈[tex](-6-\sqrt{29},-6+\sqrt{29})[/tex]
Answer:
(-11.385, -0.615)
Step-by-step explanation:
Inequation: x^2+12x+7<0
Equation: x^2+12x+7=0
Use the quadratic formula to get the roots of the equation.
[tex]\frac{-b \pm \sqrt{b^2 - 4\times a \times c}}{2\times a}[/tex]
[tex]\frac{-12 \pm \sqrt{12^2 - 4\times 1 \times 7}}{2\times 1}[/tex]
[tex]\frac{-12 \pm 10.77}{2}[/tex]
Root 1:
[tex]\frac{-12 + 10.77}{2}[/tex]
-0.615
Root 2:
[tex]\frac{-12 - 10.77}{2}[/tex]
-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)
