Question

Solve the inequality plASE ?

Answer 1

`x > -7`

Explanation:

First considering `x ne -5`

`sqrt(x^2+x-6) +3x+13 > x+5` or

`sqrt(x^2+x-6) > -(2x+8)` or

`-sqrt(x^2+x-6) < 2x+8`

now squaring both sides

`x^2+x-6 < (2x+8)^2` or

`3x^2+31x+70 > 0` and then

`{x > -7} uu {x < -10/3}`

but after checking, the feasible solution is

`x > - 7`

NOTE

The squaring operation introduces extraneous additional solutions.

Answer 2

Assumption: this is `((sqrt(x^2+x-6))+(3x+13))/((x+5))>1`

Note that this solution set `color(red)(" EXCLUDES "x=-5`

`-7.59 < x <3.07` as an approximate answer

`color(white)("d")-(32+2sqrt(46))/6 < x < + (-32+2sqrt(46))/6` as an exact answer

Explanation:

I am using brackets to group 'things' at the moment.

Multiply both sides by `(x+5)` giving

`color(green)(((sqrt(x^2+x-6))+(3x+13))/((x+5)) xxcolor(red)((x+5))color(white)("dd")>color(white)("dd")1color(red)(xx(x+5))`

`color(green)([(sqrt(x^2+x-6))+(3x+13)] xxcolor(red)((x+5))/((x+5))color(white)("dd")>color(white)("dd")color(red)((x+5)))`

But `(x+5)/(x+5) =1`

`color(green)([(sqrt(x^2+x-6))+(3x+13)] xxcolor(white)("dd")1color(white)("ddddd")>color(white)("dd")color(red)((x+5)))`

`color(green)((sqrt(x^2+x-6))+(3x+13) color(white)("dddddddddddd")>color(white)("dd")(x+5))`

Subtract `(3x+13)` from both sides

`color(green)(sqrt(x^2+x-6)color(white)("ddd") >color(white)("ddd") (x+5)-(3x+13))`

but `-(3x+13)` is the same as `-3x-13`

`color(green)(sqrt(x^2+x-6)color(white)("ddd") >color(white)("ddd") x+5-3x-13)`

`color(green)(sqrt(x^2+x-6)color(white)("ddd") >color(white)("ddd") -2x-8)`

Square both sides

`color(green)(x^2+x-6 > (-2x-8)^2)`

`color(green)(x^2+x-6 > +4x^2+32x+64)`

Subtract `x^2+x-6` from both sides

`color(green)(0>3x^2+32x+70)`

Using `ax^2+bx+c -> x=(-b+-sqrt(b^2-4ac))/(2a)`
where `a=3; b=32 and c=70` giving:

`x=(-32+-sqrt(32^2-4(3)(70)))/(2(3))`

`x=(-32+-sqrt(184))/6`

`x=(-32+-sqrt(2^2xx46))/6 = (-32+-2sqrt(46))/6`

`x~~3.07 and x~~-7.59` 2 to decimal places

But this is an inequality and these are the extremes of the domain (input `->x` values) giving:

`-7.59 < x <3.07` as an approximate answer

`color(white)("d")-(32+2sqrt(46))/6 < x < + (-32+2sqrt(46))/6` as an exact answer
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Looking back at the original inequality

`((sqrt(x^2+x-6))+(3x+13))/((x+5))>1`

This is undefined when the denominator becomes 0. So `x=-5` is 'not allowed'