082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

1.6 Solution of inequalities 37
Case(ii)
x−2<0andsox<2.
x+2>0andsox>−2.
Here we havex < 2 andx > −2. This is usually written as −2 < x < 2. Thus all
values ofxbetween −2 and 2 will ensure thatx 2 < 4.
Insummary,x 2 < 4 when −2 <x<2.
The previous example illustratesageneral rule.
Ifx 2 >kthenx> √ korx<− √ k.
Ifx 2 <kthen − √ k<x< √ k.
Example1.37 Solve the following inequalities:
(a)x 2 +x−6>0
(b)x 2 +8x+1<0
Solution (a) x 2 +x−6> 0
(x−2)(x+3)>0
For the product (x −2)(x +3) tobepositive requires either
(i) x−2>0andx+3>0
or
(ii) x−2<0andx+3<0.
Case(i)
x−2>0andsox>2.
x+3>0andsox>−3.
Bothofthese inequalitiesare satisfied only whenx > 2.
Case(ii)
x−2<0andsox<2.
x+3<0andsox<−3.
Bothofthese inequalitiesare satisfied only whenx < −3.
Insummary,x 2 +x−6 > 0 when eitherx > 2 orx < −3.
(b) The quadratic expression x 2 + 8x + 1 does not factorize and so the technique of
completingthe squareisused.
Hence
x 2 +8x+1=(x+4) 2 −15
(x+4) 2 −15<0
(x+4) 2 <15