#1**+2 **

If the question is asking you to make W the subject, this is my method

W= x

length = 2x + 8

as it is double W + 8

and we know that width times the length is 72

therefore 72/lenth must be W

W= 72/2x+8

YEEEEEET
Apr 26, 2018

#2**+3 **

Let w = width of shoebox

Let l = length of the shoebox

Since the length is 8 centimeteres more than twice the width, we can conclude that l = 2w+8.

The area of a rectangle is given, so we can solve for the now one-variable equation.

\(w(2w+8)=72\) | Expand completely. |

\(2w^2+8w=72\) | |

\(2w^2+8w-72=0\) | Factor out the GCF of every term, which is 2 in this case. |

^{\(2(w^2+4w-36)=0\)} | Unfortunately, this trinomial is not factorable. Let's use the quadratic formula instead. |

\(a=1,b=4,c=-36;\\ x_{1,2} = {-b \pm \sqrt{b^2-4ac} \over 2a}\) | Now, this is a matter of simplification. |

\(x_{1,2} = {-4 \pm \sqrt{4^2-4(1)(-36)} \over 2(1)}\) | |

\(x_{1,2} = {-4 \pm \sqrt{160} \over 2}\) | It is possible to simplify the radical. |

\(x_{1,2} = {\frac{-4\pm4\sqrt{10}}{2}}\) | Simiplify the fraction as every term has a common factor of 2. |

\(x_{1,2} = -2\pm2\sqrt{10}\) | Reject the negative answer since a width can never be negative in the context of geometry. |

\(x=-2+2\sqrt{10}\approx 4.32 \text{cm}\) |

When I read the question more closely, I realized that all this solving was completely superfluous. You can still use the equation from the beginning, though! It represents the situation perfectly.

TheXSquaredFactor
Apr 26, 2018