How do you solve #x^2<=4x-2# using a sign chart?

1 Answer

Answer:

#2-sqrt2 <= x <= 2+sqrt2#

Explanation:

We have #x^2<=4x-2# i.e. #x^2-4x+2<=0# and using the quadratic formula is

#x=(4+-sqrt(4^2-4xx1xx2))/2=2+-sqrt2#

and our inequality is therefore

#(x-2+sqrt2)(x-2-sqrt2)<=0#

From this, we know that the product #(x-2+sqrt2)(x-2-sqrt2)# is negative or equal to #0#. It is apparent that sign of binomials #(x-2+sqrt2)# and #x-2-sqrt2# will change around the values #2-sqrt2# and #2+sqrt2# respectively. In a sign chart, we divide the real number line using these values, i.e. below #2-sqrt2#, between #2-sqrt2# and #2+sqrt2# and above #2+sqrt2# and see how the sign of #x^2-4x+2# changes.

Sign Chart

#color(white)(XXXXXXXXXXX)2-sqrt2color(white)(XXXXX)2+sqrt2#

#(x-2+sqrt2)color(white)(X)-ive color(white)(XXXX)+ive color(white)(XXXX)+ive#

#(x-2-sqrt2)color(white)(X)-ive color(white)(XXXX)-ive color(white)(XXXX)+ive#

#(x^2-4x+2)color(white)(XX)+ive color(white)(XXX)-ive color(white)(XXXX)+ive#

It is observed that #x^2-4x+2 <= 0# when either #x >= 2-sqrt2# or #x <= 2+sqrt2# i.e. #x# lies between #2-sqrt2# and #2+sqrt2# including these numbers or #2-sqrt2 <= x <= 2+sqrt2#, which is the solution for the inequality.

In interval form solution is #[2-sqrt2,2+sqrt2]#

graph{x^2-4x+2 [-2.976, 7.024, -2.5, 2.5]}