#x^2 - 4 <= 4x + 1#
First, let's move everything to the left side to make this quadratic inequality equal to 0.
#x^2 - 4 - 4x - 1 <= 0#
We know that #-4# and #-1# can simplify #=> -4-1 = -5#. So now our inequality looks like this:
#x^2 - 5 - 4x <= 0#
Now let's put this equation in descending order, or from highest exponent to lowest/no exponent.
#x^2 - 4x - 5 <= 0#
What we do now is called factoring.
We need to find 2 numbers that multiply to -5 AND add up to -4. Are there 2 numbers that do so? YES! #-5# and #1# multiply to -5 and add up to -4.
So now we do this:
#(x-5)(x+1)# <= 0
Then we set each of the things inside the parenthesis to equal 0, like this:
#x - 5 <= 0# and #x + 1 <= 0#
#x <= 5# and #x <= -1#
Since we know that #x <= 5#, if we plug in #5# back into the equation #(x-5)(x+1)# to check our work:
#(5-5)(5+1) >= 0#
#(0)(6) >= 0# (any number times 0 is 0), so we know that 5 works.
Now let's try a number less than 5, like 4.
#(4-5)(4+1) <=0#?
#(-1)(5) <= 0#?
#-5 <= 0# Yes! So we know that #x <= 5# is true.
Now we try the other answer, #x <= -1#. Let's plug back #-1# into the equation and see if that works:
#(-1-5)(-1+1) <= 0#?
#(-6)(0) <= 0# (any number times 0 is 0), so we know that -1 works.
Now let's try a number less than -1, like -2.
#(-2-5)(-2+1) <= 0#?
#(-7)(-1) <= 0#?
#(7) cancel(<=)0# No! It doesn't work.
Since we know that #x <= 5# is TRUE and only #x = -1# works in #x <= -1#, then we know that only #-1# to #5# works!
If you write it in interval notation, you can say #x = [-1, 5]# or #-1 <= x <= 5#.
