How to solve (x^2 - 4x +3)(x^2 -3)<0?

1 Answer
Jul 10, 2015

x epsilon (-sqrt(3), -1)xε(3,1)
or
x epsilon (sqrt(3),3)xε(3,3)

Explanation:

If (x^2-4x+3)(x^2-3)<0(x24x+3)(x23)<0
then one (but not both) of the terms must be less than zero.

Case 1: (x^2-4x+3 < 0)(x24x+3<0) and (x^2-3 > 0)(x23>0)
color(white)("XXXX")XXXXx^2-4x+3 < 0x24x+3<0
color(white)("XXXX")XXXX(x-3)(x-1) < 0(x3)(x1)<0
color(white)("XXXX")XXXXagain one of the terms must be less than zero (and the other greater)
color(white)("XXXX")XXXXSince x-3 < x-1x3<x1
color(white)("XXXX")XXXXcolor(white)("XXXX")XXXXx-3 < 0x3<0 and x-1 > 0x1>0
color(white)("XXXX")XXXXcolor(white)("XXXX")XXXXx < 3x<3 and x > 1x>1
color(white)("XXXX")XXXXcolor(white)("XXXX")XXXXx epsilon (1,3)xε(1,3)

and
color(white)("XXXX")XXXXx^2-3>0x23>0
color(white)("XXXX")XXXXrarr x^2 > 3x2>3
color(white)("XXXX")XXXXrArr x > sqrt(3)x>3

rArr sqrt(3) < x < 33<x<3

Case 2: (x^2-4x+3 > 0)(x24x+3>0) and (x^2-3 < 0)(x23<0)
color(white)("XXXX")XXXXx^2-4x+3 > 0x24x+3>0
color(white)("XXXX")XXXX(x-3)(x-1) > 0(x3)(x1)>0
color(white)("XXXX")XXXXboth terms must be negative or both terms must be positive
color(white)("XXXX")XXXXand since x-3 < x-1x3<x1
color(white)("XXXX")XXXXcolor(white)("XXXX")XXXXx-3 > 0x3>0 or x-1 < 0x1<0
color(white)("XXXX")XXXXcolor(white)("XXXX")XXXXrArr x >3x>3 or x < -1x<1

and
color(white)("XXXX")XXXXx^2-3 < 0x23<0
color(white)("XXXX")XXXXrArr x^2 < 3x2<3
color(white)("XXXX")XXXXrArr -sqrt(3) < x < sqrt(3)3<x<3

rArr -sqrt(3) < x < -13<x<1

Combining: Case 1 or Case 2
-sqrt(3) < x < -13<x<1
or
sqrt(3) < x < 33<x<3