Solutions to Problem Set 28

Math 101-06

10-17-2017

[Quadratic word problems]

1. A number minus eight times its reciprocal equals 2. What is the number? (There may be more than one answer.)

Let x be the number. Since the number minus eight times its reciprocal equals 2,

$$\eqalign{ x - 8 \cdot \dfrac{1}{x} & = 2 \cr \noalign{\vskip2pt} x \cdot \left(x - 8 \cdot \dfrac{1}{x}\right) & = x \cdot 2 \cr \noalign{\vskip2pt} x^2 - 8 & = 2 x \cr x^2 - 2 x - 8 & = 0 \cr (x - 4)(x + 2) & = 0 \cr}$$

The solutions are $x = 4$ and $x = -2$ .


2. One number is the square of another. Their sum is 110. Find the numbers.

Let A and B be the numbers. The first sentence says one is the square of the other, so I can write

$$A = B^2.$$

The sum is 110, so

$$A + B = 110.$$

Plug $A = B^2$ into $A
   + B = 110$ and solve for B:

$$\eqalign{ B^2 + B & = 110 \cr B^2 + B - 110 & = 0 \cr (B + 11)(B - 10) & = 0 \cr}$$

The possible solutions are $B =
   -11$ and $B = 10$ .

If $B = -11$ , then $A =
   (-11)^2 = 121$ .

If $B = 10$ , then $A = 10^2
   = 100$ .

So two pairs work: -11 and 121, and 10 and 100.


3. The difference of two numbers is 2 and their product is 48. Find the numbers.

Let x and y be the numbers. Their difference is 2, so I can write

$$x - y = 2.$$

Their product is 48, so

$$x y = 48.$$

From $x - y = 2$ , I get $x = y + 2$ . Plug this into $x y = 48$ and solve for y:

$$\eqalign{ (y + 2)y & = 48 \cr y^2 + 2 y & = 48 \cr y^2 + 2 y - 48 & = 0 \cr (y + 8)(y - 6) & = 48 \cr}$$

If $y = -8$ , then $x = -8 +
   2 = -6$ .

If $y = 6$ , then $x = 6 +
   2 = 8$ .

So two pairs work: -6 and -8, and 6 and 8.


4. The area of a rectangle is 168 square inches. The length is 5 more than 2 times the width. Find the length and the width.

Let L be the length and let W be the width. The length is 5 more than 2 times the width, so

$$L = 2 W + 5.$$

The area is 168, so

$$L W = 168.$$

Plug in $L = 2 W + 5$ and solve for W:

$$\eqalign{ L W & = 168 \cr (2 W + 5)W & = 168 \cr 2 W^2 + 5 W & = 168 \cr 2 W^2 + 5W - 168 & = 0 \cr}$$

Use the Quadratic Formula:

$$W = \dfrac{-5 \pm \sqrt{25 - 4(2)(-168)}}{2 \cdot 2} = -\dfrac{21}{2} \quad\hbox{or}\quad 8.$$

Since the width can't be negative, I get $W = 8$ . The length is $L = 2 \cdot 8 + 5 = 21$ .


5. If Calvin and Bonzo eat together, they can eat 720 hot dogs in 20 hours.

Eating alone, Bonzo takes 9 hours longer than Calvin would to eat 720 hot dogs.

How long does it take Calvin to eat 720 hot dogs?

Let x be Calvin's rate (in hot dogs per hour), let y be Bonzo's rate, and let t be the time it takes Calvin to eat 720 hot dogs.

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & & & hours & & $\cdot$ & & hot dogs per hour & & $=$ & & hot dogs & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & Calvin and Bonzo & & 20 & & $\cdot$ & & $x + y$ & & $=$ & & 720 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & Calvin & & t & & $\cdot$ & & x & & $=$ & & 720 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & Bonzo & & $t + 9$ & & $\cdot$ & & y & & $=$ & & 720 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

The second equation says $x t =
   720$ , so $x = \dfrac{720}{t}$ .

The third equation says $y(t +
   9) = 720$ , so $y = \dfrac{720}{t + 9}$ .

Plug these into the first equation $20(x + y) = 720$ and solve for t:

$$\eqalign{ 20(x + y) & = 720 \cr 20 \left(\dfrac{720}{t} + \dfrac{720}{t + 9}\right) & = 720 \cr \dfrac{720}{t} + \dfrac{720}{t + 9} & = 36 \cr t(t + 9) \cdot \left(\dfrac{720}{t} + \dfrac{720}{t + 9}\right) & = t(t + 9) \cdot 36 \cr 720(t + 9) + 720 t & = 36 t(t + 9) \cr 20(t + 9) + 20 t & = t(t + 9) \cr 20 t + 180 + 20 t & = t^2 + 9 t \cr 0 & = t^2 - 31 t - 180 \cr 0 & = (t + 5)(t - 36) \cr}$$

The solutions are $t = -5$ and $t = 36$ . Since t can't be negative, the answer is $t = 36$ hours.


6. A plane can fly at a speed of 360 miles per hour in still air. It takes 2 hours longer to fly 3200 miles against the wind as it does to fly the same distance with the wind. Assume that the wind's speed adds to the plane's speed when the plane flies with the wind, and that the wind's speed subtracts from the plane's speed when the plane flies against the wind. Find the speed of the wind.

Let w be the speed of the wind, and let t be the time in hours it takes to fly 3200 miles against the wind. Then it takes $t + 2$ hours to fly the same distance with the wind.

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & & & time & & & & speed & & & & distance & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & against the wind & & $t + 2$ & & $\cdot$ & & $360 - w$ & & $=$ & & 3200 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & with the wind & & t & & $\cdot$ & & $360 + w$ & & $=$ & & 3200 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

The table gives the two equations

$$(t + 2)(360 - w) = 3200 \quad\hbox{and}\quad t (360 + w) = 3200.$$

Solving the second equation for t, I obtain

$$t = \dfrac{3200}{360 + w}.$$

Plug this into the first equation and solve for w:

$$\eqalign{ \left(\dfrac{3200}{360 + w} + 2\right) (360 - w) & = 3200 \cr \noalign{\vskip2pt} (360 + w) \left(\dfrac{3200}{360 + w} + 2\right) (360 - w) & = 3200 (360 + w) \cr \noalign{\vskip2pt} \left(3200 + 2 (360 + w)\right) (360 - w) & = 3200 (360 + w) \cr 3200 (360 - w) + 2 (360 + w) (360 - w) & = 3200 (360 + w) \cr 3200 (360 - w) + 2 (129600 - w^2) & = 3200 (360 + w) \cr 1152000 - 3200 w + 259200 - 2 w^2 & = 1152000 + 3200 w \cr -2 w^2 - 6400 w + 259200 & = 0 \cr 2 w^2 + 6400 w - 259200 & = 0 \cr w^2 + 3200 w - 129600 & = 0 \cr}$$

$$w = \dfrac{-3200 \pm \sqrt{3200^2 - 4(1)(-129600)}}{2} = \dfrac{-3200 \pm \sqrt{10758400}}{2} = \dfrac{-3200 \pm 3280}{2} = -1600 \pm 1640.$$

The roots are 40 and -3240, but the wind speed can't be negative. Hence, the speed of the wind is 40.0 miles per hour.


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