MHB Guess the Mystery Number in this List of 4!

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The discussion revolves around solving for a mystery number in a list containing 1, the square root of 2, x, and x squared, with a specified list range of 4. Participants explore the definition of "list range" as the difference between the largest and smallest elements. Two equations are derived: x squared minus 1 equals 4 and x squared minus x equals 4, leading to different values for x. The first solution yields x as the square root of 5, while the second involves a quadratic formula resulting in a complex expression. Ultimately, both approaches confirm that the list range remains consistent at 4.
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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.
 
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Hi alextrainer. Would you please explain what a "list range" is?
 
alextrainer said:
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?
 
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
 
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:

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