Step-by-step explanation:
First let solve the inequality
[tex]5 {x}^{2} + 2x - 3 < 0[/tex]
Factor by grouping
[tex]5 {x}^{2} + 5x - 3x - 3 < 0[/tex]
[tex]5x(x + 1) - 3(x + 1)[/tex]
So the factor are
[tex](5x - 3)(x + 1)[/tex]
So the factor are
[tex]x = \frac{3}{5} [/tex]
and
[tex]x = - 1[/tex]
Solutions to a quadratic can be represented by a absolute value equation because remeber quadratics
creates 2 roots and/or double roots.
The inequality
[tex] |x - b| < c[/tex]
works as
b is the midpoint between 2 roots. And c is the
[tex] |x + b| = c[/tex]
We know that the midpoint between both roots is-1/5.
so
[tex] |x - ( - \frac{1}{5} )| < c[/tex]
[tex] |x + \frac{1}{5} | < c[/tex]
Let use roots 3/5
[tex] | \frac{3}{5} + \frac{1}{5} | = \frac{4}{5} [/tex]
-1 works as well.
[tex] | - 1 + \frac{1}{5} | = | - \frac{4}{5} | = \frac{4}{5} [/tex]
So the absolute value equation is
[tex] |x + \frac{1}{5} | < \frac{4}{5} [/tex]
