Solving Hard Inequalities: 1/(x+4)>x-4 & 1/(x+7)>x-7

[alewisGB]

I am looking at questions like
1/(x+4)>x-4
or 1/(x+7)>x-7
I have no idea how to solve them,
I have simplified to (-x^2+50) /(7+x)
however I don't think it is correct and I don't know what do do from there.

 

[tiny-tim]

hi alewisGB!

(try using the X² button just above the Reply box )

obviously, you need to get rid of the fraction!

in an ordinary equation, you'd just multiply both sides by the denominator

but that might be negative (which would change the > to <),

so, instead, multiply both sides by the square of the denominator …

what do you get? ​

 

[alewisGB]

for:
1/(x+2)>x-2
Step 1: x+2>x³+2x²-4x-8
Step 2: 0>x³+2x²-5x-10
Step 3: 0>(x+2)(x²-5)
Step 4: if 0=(x+2)(x²-5)
x = -2 or +√5 or -√5
Step 5: Unsure, do input -2.01 and -1.99 and see which one is true? (same for +or-√5) or do I do something else

 

[tiny-tim]

hi alewisGB!

this way is easier …

1/(x+2)>x-2
Step 1: x+2>(x+2)(x²-4)
Step 3: 0>(x+2)(x²-5)
Step 4: 0>(x+√5)(x+2)(x-√5)

… and you now have this in the form 0 > (x-a)(x-b)(x-c) with a<b<c,

sooo … ? ​

 

[alewisGB]

I have got the answer of -√5 > x and 5 > x > -2
I have not done further maths and although inequalities were covered they were not nearly this complex
I have not seen the rule "0 > (x-a)(x-b)(x-c) with a<b<c"
So I am not sure what that means
Thank you for your help :)

 

[tiny-tim]

I have not seen the rule "0 > (x-a)(x-b)(x-c) with a<b<c"
So I am not sure what that means

oh, that's easy …

the RHS is the product of three numbers,

and it can only be negative if one or all three of them are negative,

which means either x < a or b < x < c ​

 

[alewisGB]

Thank you ever so much, you have no idea how grateful I am. This will help me in now but also in the future :)