Incremental distance

Algebra Level 2

The distance between two numbers -2 and 7 on a number line is $7 - (-2) = 9,$ as shown above.

Now, we have three numbers $a, a^2, a^3$ on a number line, as shown below. If the distance between $a$ and $a^2$ is 6, what is the distance between $a^2$ and $a^3?$

The answer is 18.

4 solutions

Luis Ortiz Montero
Jan 31, 2017

I ended up with 2 answers (see EDIT)

$a^2 - a = 6$

$a^2 - a -6 = 0$

$a=3 , a= -2$

$a^2 = 9 , a^2 = 4$

$a^3 = 27 , a^3 =-8$

$a^3 - a^2 = 18 , |a^3 - a^2| = 12$

EDIT:

The only valid answer is a= 18, as the last picture clearly shows that $a^2$ must be between $a$ and $a^3$ . All credit to @Pi Han Goh for clearing the mistake made in this solution in the comment section.

See the report section. It explains why the answer cannot be 12.

Pi Han Goh - 4 years, 4 months ago

Excuse me but what/where is the "report section"? In the question did you explicitly state that $a$ is non-negative? Or did you mean to state by the image?

A Former Brilliant Member - 4 years, 4 months ago

Click the "dot dot dot" button below the "view wiki" button, then click "report section".

The reason why $a = -2$ cannot be a solution is because it doesn't satisfy $a<a^2<a^3$ that was provided in the number line.

Pi Han Goh - 4 years, 4 months ago

@Pi Han Goh – yeah, the last picture limits "a" to be strictly 18, thanks for clearing that. it is my fault

Luis Ortiz Montero - 4 years, 4 months ago

Zach Abueg
Jan 30, 2017

Given $a^2 - a = 6$ , find $a^3 - a^2$ .

$a^2 - a = a(a - 1) = 6 \Longrightarrow a = 3$

$a^3 - a^2 = a(a^2 - a) = 3 \times 6 = 18$

Must end with 2 solutions because of the nature of a quadratic. I don't believe there is anything given in the question to sway towards only one answer.

Reuben Joshua - 4 years, 4 months ago

Right, a quadratic equation has two roots. In this case, the two solutions for $(a, a^2, a^3)$ are $(3, 9, 27)$ and $(-2, 4, -8)$ . Note that number line shows that $a < a^2 < a^3$ . The second solution doesn't satisfy this criteria since $4$ is not between $-2$ and $-8$ .

Pranshu Gaba - 4 years, 4 months ago

The distance between $a$ and $b$ is better represented by $|a - b|$ , the absolute value of $a - b$ , because we want the distance to be positive. So it would be more accurate to start with the equation $|a^2-a| = 6$ , and find $|a^3 - a^2|$ .

However, the number line shows that $a < a^2 < a^3$ , so $|a^2 -a |$ is equal to $a^2-a$ and $|a^3 - a^2|$ is equal to $a^3 - a^2$ , so we can do away with the absolute value signs.

Pranshu Gaba - 4 years, 4 months ago

Jack Ceroni
Feb 4, 2017

We can see from the given information that we must find what the value of $a$ is. We are first given that the distance between $a$ and $a^{2}$ is $6$ . We can set up the following equation to solve for $a$ :

$a^{2}$ - $a$ = $6$

By subtracting $6$ , we can get the following equation:

$a^{2}$ - $a$ - $6$ = $0$

We can factor this equation, and get the following results for $a$ :

$a$ = $3$ , $a$ = $-2$

Since $a^{3}$ is evidently higher on the number line than $a^{2}$ , and if $a$ were a negative number, then $a^{3}$ would be negative, while $a^{2}$ would be positive, $a^{2}$ would be higher on the number line than $a^{3}$ , and from the diagram, this cannot be the case, therefore $a$ must equal to $3$ . This would mean that we can find the distance between $a^{3}$ and $a^{2}$ to be $18$ .

Osman Ates
Feb 4, 2017

a can be either -2 or 3 but considering the figure and number line, a<a^2<a^3 so a=3, answer=18

Incremental distance

The answer is 18.

4 solutions

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