Question

solve the inequality. use interval notion to state the solution (x+3)^2>2(x^2+7)

Answer 1

Answer:

we conclude that:

$x^2+6x+9>2x^2+14\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:1<x<5\:\\ \:\mathrm{Interval\:Notation:}&\:\left(1,\:5\right)\end{bmatrix}$

Hence, (1, 5) is the solution in interval notation.

Please also check the attached graph.

Step-by-step explanation:

Given the inequality expression

$\left(x+3\right)^2>\:2\left(x^2+7\right)$

as

(x + 3)² = x² + 6x + 9

2(x² + 7) = 2x² + 14

so

$\:x^2+6x+9\:>\:2x^2+14$

rewriting in the standard form

$-x^2+6x-5>0$

Factor -x² + 6x - 5:  - (x - 1) (x - 5)

$-\left(x-1\right)\left(x-5\right)>0$

Multiply both sides by -1 (reverse the inequality)

$\left(-\left(x-1\right)\left(x-5\right)\right)\left(-1\right)<0\cdot \left(-1\right)$

Simplify

$\left(x-1\right)\left(x-5\right)<0$

so

$1<x<5$

Therefore, we conclude that:

$x^2+6x+9>2x^2+14\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:1<x<5\:\\ \:\mathrm{Interval\:Notation:}&\:\left(1,\:5\right)\end{bmatrix}$

Hence, (1, 5) is the solution in interval notation.

Please also check the attached graph.