+0

Problem Solving Strategy: Make a Model

0
95
2

The sum of a number and its cube, is 68. What is the number?

Guest Feb 23, 2017

#2
+18777
+15

The sum of a number and its cube, is 68.

What is the number?

$$\begin{array}{|rcll|} \hline n+n^3 &=& 68 \\ \underbrace{n}_{=4}\cdot \underbrace{(1+n^2)}_{=17} &=& 4\cdot 17 \\\\ 1+n^2&=&17\\ n^2 &=& 16 \\ n &=& 4 \checkmark \\ \hline \end{array}$$

The number is 4

heureka  Feb 24, 2017
Sort:

#1
+5

Let the number =N

N + N^3 = 68, solve for N

Solve for N:
N^3 + N = 68

Subtract 68 from both sides:
N^3 + N - 68 = 0

The left hand side factors into a product with two terms:
(N - 4) (N^2 + 4 N + 17) = 0

Split into two equations:
N - 4 = 0 or N^2 + 4 N + 17 = 0

N = 4 or N^2 + 4 N + 17 = 0

Subtract 17 from both sides:
N = 4 or N^2 + 4 N = -17

N = 4 or N^2 + 4 N + 4 = -13

Write the left hand side as a square:
N = 4 or (N + 2)^2 = -13

Take the square root of both sides:
N = 4 or N + 2 = i sqrt(13) or N + 2 = -i sqrt(13)

Subtract 2 from both sides:
N = 4 or N = (0 + 1 i) sqrt(13) - 2 or N + 2 = -i sqrt(13)

Subtract 2 from both sides:
Answer: |N = 4               {or N = (0 + 1 i) sqrt(13) - 2 or N = (0 - i) sqrt(13) - 2} Discard these

Guest Feb 23, 2017
#2
+18777
+15

The sum of a number and its cube, is 68.

What is the number?

$$\begin{array}{|rcll|} \hline n+n^3 &=& 68 \\ \underbrace{n}_{=4}\cdot \underbrace{(1+n^2)}_{=17} &=& 4\cdot 17 \\\\ 1+n^2&=&17\\ n^2 &=& 16 \\ n &=& 4 \checkmark \\ \hline \end{array}$$

The number is 4

heureka  Feb 24, 2017

25 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details