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:

$$\frac{9-\sqrt{17}}{2}-\frac{1-\sqrt{17}}{2}=4$$
 
I have been insisting to my statistics students that for probabilities, the rule is the number of significant figures is the number of digits past the leading zeros or leading nines. For example to give 4 significant figures for a probability: 0.000001234 and 0.99999991234 are the correct number of decimal places. That way the complementary probability can also be given to the same significant figures ( 0.999998766 and 0.00000008766 respectively). More generally if you have a value that...

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