Find x when 1, √2, x, x² have range 4

[alextrainer]

List contains 1, square root of 2, x and x squared and the list range is 4.

I guessed 2 but it is not correct.

 

[Greg]

Hi alextrainer. Would you please explain what a "list range" is?

 

[MarkFL]

List contains 1, square root of 2, x and x squared and the list range is 4.

I guessed 2 but it is not correct.

I am going to assume that a "list range" is the difference between the largest and smallest elements in the list. So, if we assume then that we are going to have $x^2$ as the largest member of the list (since we will require $1<x$), where $1$ is the smallest, then we need:

$$x^2-1=4$$

Can you solve this where $1<x$? Is $\sqrt{2}<x^2$?

Another solution arises from assuming $x$ will be the smallest member of the list and $x^2$ will be the largest:

$$x^2-x=4$$

Can you find this solution?

 

[alextrainer]

You got the problem correctly stated?

x squared - x = 4

so not an integer

without just randomly trying fractions, what other strategy?

x squared = 4 - x? does not help

 

[MarkFL]

Let's first look at:

$$x^2-1=4$$

Add $1$ to both sides:

$$x^2=5$$

Hence, taking the root where $1<x$, we obtain:

$$x=\sqrt{5}$$

And so the list contains (ordered from smallest to largest):

$$\{1,\sqrt{2},\sqrt{5},5\}$$

We can see the "list range" is:

$$5-1=4$$

Next, let's look at:

$$x^2-x=4$$

Arrange in standard form:

$$x^2-x-4=0$$

Using the quadratic formula and taking the root such that $x<1$, we obtain:

$$x=\frac{1-\sqrt{17}}{2}\implies x^2=\frac{9-\sqrt{17}}{2}$$

And so the list contains (ordered from smallest to largest):

$$\left\{\frac{1-\sqrt{17}}{2},1,\sqrt{2},\frac{9-\sqrt{17}}{2}\right\}$$

We can see the "list range" is:

$$\frac{9-\sqrt{17}}{2}-\frac{1-\sqrt{17}}{2}=4$$