Question

show that if abs(x+2)<2 then abs((x+2)(x+5))<4

Answer 1

let t=x+2 so we have

|t|<2 and we want to prove |t(t+3)|<4

Answer 2

right?

Answer 3

I never tried it that way.
I tried to show that since x+<1 yields x<-1
and that x+5<4 yields x<-1
and then I figured I needed to solve the expression (x+2)(x+5)<(x+5)
x^2+7x+10<x+5
x^2+6x+5<0
(x+5)(x+1)<0 and since they both yield values less than 4 I figured that was true, but it was wrong

Answer 4

emm...ignore my first reply :) lets do it with x

Answer 5

Hahaha okie dokie then

Answer 6

|x+2|<2
so
-2<x+2<2
-4<x<0
right?

Answer 7

Hmmm...so somehow that needs to translate to being <4?

Answer 8

Wait, it is supposed to be
\[\left| x+2 \right|<1\]

Answer 9

Oops sorry for the confusion...haha

Answer 10

now

\[(x+2)(x+5)=x^2+7x+10=x^2+7x+\frac{49}{4}-\frac{49}{4}+10=(x+\frac{7}{2})^2-\frac{9}{4}\]and we have -4<x<0 so\[-\frac{1}{2}<x+\frac{7}{2}<\frac{7}{2}\]\[\frac{1}{4}<(x+\frac{7}{2})^2<\frac{49}{4}\]\[\frac{1}{4}-\frac{9}{4}<(x+\frac{7}{2})^2-\frac{9}{4}<\frac{49}{4}-\frac{9}{4}\]\[-2<(x+\frac{7}{2})^2-\frac{9}{4}<4\]so\[|(x+\frac{7}{2})^2-\frac{9}{4}|<4\]