Solve 2+x > 15/x ...and a few questions on simplifying...?

1. 2 + x > 15/x

Case 1: if x>0.
2x + x^2 > 15
x^2 + 2x + 1 > 15 + 1
(x + 1)^2 > 16
x + 1 > +/- 4
x > +/- 4 - 1
x> +4 - 1
x> 3 and
x> - 4 - 1
x> - 5
so x > 3 ANSWER

Case 2: if x < 0:
2x + x^2 < 15
x^2 + 2x + 1 < 15
(x + 1)^2 < 15
( x + 1)^2 < 15 - 1
(x + 1)^2 < 14
x + 1 < +/-sqrt(14)
x < +/- sqrt(14) - 1
x < + sqrt(14) - 1 and
x < - sqrt(14) - 1
so, x < - sqrt(14) - 1 ANSWER

2. Can 8-y be simplified to -(8-y)? No.
-(8-y) is -8 + y which is NOT equal to 8-y.

3. Can 8-y / 3y be simplified any further? (I assume it is
(8-y)/(3y)). It is simplified already. But, of course, you
can also write it this way: 8/3y - y/3y = 8/3y - 1/3. This
one isn't any simpler than the original.

4. Also, Solve b^4 + b^2 -20 = 0 would be +- 2 and
+- sqrt5 (i)?
Yes, you're right.

teddy boy

all of us understand that the denominator won't be able to be 0, as a results of fact branch by way of 0 is undefined. We additionally understand that (2+x) won't be able to be unfavorable, as a results of fact then a million/(2+x) may be under 0. this implies (2+x) would desire to be greater beneficial than 0, so x may be any variety greater beneficial than -2.

Question 1
x² + 2x - 15 > 0
(x + 5)(x - 3) > 0

(x + 5) > 0 and (x - 3) > 0
OR
(x + 5) < 0 and (x - 3) < 0

x > - 5 and x > 3
x > 3
OR
x < - 5 and x < 3
x < - 5

{ x : x < - 5 , x > 3 , x Є R }

Question 2
NO !
- ( 8 - y ) = - 8 + y

Question 3
No !

Question 4
(b² + 5)(b² - 4) = 0
b = ± i √5 , b = ± 2
We agree on this one !

1)
2 + x > 15/x
x(2 + x) > 15
2x + x^2 > 15
x^2 + 2x - 15 > 0
(x + 5)(x - 3) > 0

x + 5 > 0
x > -5

x - 3 > 0
x > 3

-5 < x < 3

= = = = = = = =

2)
8 - y ≠ -(8 - y)

= = = = = = = =

3)
(8 - y)/3y
= 8/3y - y/3y
= 8/3y - 1/3

= = = = = = = =

4)
b^4 + b^2 - 20 = 0
(b^2 + 5)(b^2 - 4) = 0
(b^2 + 5)(b + 2)(b - 2) = 0

b^2 = 5
b = ±√5

b + 2 = 0
b = -2

b - 2 = 0
b = 2

∴ b = ±√5 , ±2

1) 2+x > 15/x
multiply by x both sides
x(2+x)>15
x^2+2x-15>0
(x-3)(x+5)>0
x-3>0 -> x>3
x+5>0 -> x>-5

2) Can 8-y be simplified to -(8-y)?
it can be simplified to -(8+y)

3) Can 8-y / 3y be simplified any further?
no

1) To simplify, try to get all the terms onto one side of the '>' (so you end up with an equation > 0) .. you should then be able to factorise the equation...

2) No. First having brackets is NOT simpler :-)
Second -(8-y) = -8+y (or y-8) which is NOT the same as 8-y

3) No (unless you mean 8 - (y/3y) in which case it's 8 - 1/3 = 7 2/3rds)

4) correct

Solve 2+x > 15/x ...and a few questions on simplifying...? 2+x > 15/x
so 2x +x^2 > 15
so x^2 + 2x - 15 > 0
so (x+5)(x-3) > 0
so x > 5 or x < 3

Can 8-y be simplified to -(8-y)?
No

Can 8-y / 3y be simplified any further?
Not really. Depends if you mean
Can (8-y) / 3y be simplified any further?
or Can 8-(y / 3y) be simplified any further?

Also, Solve b^4 + b^2 -20 = 0
would be +- 2 and +- sqrt5 (i)?

Put x=b^2
The x^2 + x - 20 = 0
or (x+5)(x-4)=0
so x=-5 or x=+4
so b=sqrt(-5) or +2 or -2
.