Question

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

Answer 1

`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]}