How do you solve the inequality #(x-1)(x-4) <2(x-4)#?

#x^2 - x - 4x + 4 < 2x - 8#

You must put everything to one side of the inequality so that the other side is 0. You must then solve like a regular quadratic equation.

#x^2 - 7x + 12 = 0#

#(x - 4)(x - 3) = 0#

#x = 4, x = 3#

Now you must select test points, one smaller than the smaller value of x, one between the two values of x, and one larger than the largest value of x.

Let's say our test points are 2, 3.5 and 5.

Plug them into the original inequality.

#(2)^2 - 7(2) + 12 < 0#

#2 < 0#

This doesn't not satisfy the inequality

#(3.5)^2 - 7(3.5) + 12 < 0#

#-0.25 < 0#

This satisfies our inequality, so the solution is #3 < x < 4#

Practice exercises

Solve the following inequality.

a). #(2x + 3)(4x - 1) < (x - 3)(3x + 5)#

b). #(2x + 5)/(2x) > (3x - 2)/(6x - 11)#

#1#. Expand the left side of the inequality.

#(color(red)xcolor(blue)(-1))(color(orange)(x)color(green)(-4))<2(x-4)#

#(color(red)x)(color(orange)x)color(red)(+x)(color(green)(-4))color(blue)(-1)(color(orange)x)color(blue)(-1)(color(green)(-4))<2(x-4)#

#x^2-4x-x+4<2(x-4)#

#x^2-5x+4<2(x-4)#

#2#. Expand the right side of the inequality.

#x^2-5x+4<2x-8#

#3#. Bring all the terms to one side. In this case, the terms will be brought to the left side. Note that in inequalities, adding and subtracting is the same as in equalities.

#x^2-7x+12<0#

#4#. Factor the quadratic inequality. Note that in inequalities, factoring is the same as in equalities.

#(x-4)(x-3)<0#

#5#. In order for the two factors multiplied together to be less than #0#, one of the factors must be negative. In other words, a positive number multiplied by a negative number will always produce a negative number less than #0#. Thus the possible solutions are:

Set #1#: #color(white)(XXX)(x-4)>0color(white)(XXXXXX)(x-3)<0#

#color(white)(XXXXXXXXXXXXXXX)color(purple)("or")#

Set #2#: #color(white)(XXX)(x-4)<0color(white)(XXXXXX)(x-3)>0#

#6#. Solve for #x# in the possible solution sets.

Set #1#: #color(white)(XXX)x>4color(white)(XXXXXX)x<3#

#color(white)(XXXXXXXXXXX)color(purple)("or")#

Set #2#: #color(white)(XXX)x<4color(white)(XXXXXX)x>3#

#7#. For set #1#, if you substitute #color(brown)x# as any number, you will notice that no number satisfies the conditions. For example, if you used the number, #color(brown)2#, #color(brown)2# cannot be greater than #4# and less than #3# at the same time. Thus, set #1# is not a solution of the inequality.

Set #1#: #color(white)(XXX)color(brown)2>4color(white)(XXXXXXX)color(brown)2<3#

#8#. For set #2#, if you substitute #color(turquoise)x# as #color(turquoise)(3.5)#, you will notice that this number does satisfy the conditions. Thus, set #2# is a solution of the inequality.

Set #2#: #color(white)(XXX)color(turquoise)(3.5)<4color(white)(XXXXXX)color(turquoise)3.5>3#

#9#. Rewrite the solution as an algebraic inequality that satisfies the conditions.

#3##<##x<4#

#:.#, the solution is #3##<##x<4#.