Question

solving quadratic inequalities

Answer 1

Answer:

$\boxed{\sf x < -3\quad \mathrm{or}\quad \:x > -1}$

Step-by-step explanation:

$\sf x^2+4x+3 > \:0$

In order to solve inequality, we need to factor the left hand side. we can use the transformation $ax^2+bx+c=a(x-x_1)(x-x_2)$ to factor quadratic polynomials. where  x(1) & x(2)  are the solutions of the quadratic equation  ax²+bx+c=0 .

$\sf x^2+4x+3=0$

Quadratic formula:-

$\boxed{\sf x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}}$

$\sf a=1\\b=4\\c=3$

$\sf \cfrac{-4\pm \sqrt{4^2-4\times \:1\times \:3}}{2\times \:1}$ ← Calculate

$\sf \sqrt{4^2-4\times \:1\times \:3}=\boxed{2}$

$\sf \cfrac{-4\pm \:2}{2\times \:1}$

Now, let's Separate the solutions,

$\sf x_1=\cfrac{-4+2}{2\times \:1},\:x_2=\cfrac{-4-2}{2\times \:1}$

Do the calculations,

$\sf x_1=\cfrac{-4+2}{2\times \:1}=\boxed{-1}$

$\sf x_2=\cfrac{-4-2}{2\times \:1}=\boxed{-3}$

$\boxed{\sf x < -1\quad \mathrm{or}\quad \:x > -3}$