Word Problems Involving Quadratics

These word problems involve situations I've discussed in other word problems: The area of a rectangle, motion (time, speed, and distance), and work. However, these problems lead to quadratic equations. You can solve them by factoring or by using the Quadratic Formula.


Example. One number is the square of another. Their sum is 132. 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 132, so

$$A + B = 132.$$

Plug $A = B^2$ into $A + B = 132$ and solve for B:

$$\eqalign{ B^2 + B & = 132 \cr B^2 + B - 132 & = 0 \cr (B + 12)(B - 11) & = 0 \cr}$$

The possible solutions are $B = -12$ and $B = 11$ .

If $B = -12$ , then $A = (-12)^2 = 144$ .

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

So two pairs work: -12 and 144, and 11 and 121.


Example. The difference of two numbers is 2 and their product is 224. Find the numbers.

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

$$x - y = 2.$$

Their product is 224, so

$$x y = 224.$$

From $x - y = 2$ , I get $x = y + 2$ . Plug this into $x y = 224$ and solve for y:

$$\eqalign{ (y + 2)y & = 224 \cr y^2 + 2 y & = 224 \cr y^2 + 2 y - 224 & = 0 \cr (y + 16)(y - 14) & = 224 \cr}$$

If $y = -16$ , then $x = -16 + 2 = -14$ .

If $y = 14$ , then $x = 14 + 2 = 16$ .

So two pairs work: -14 and -16, and 14 and 16.


Example. The area of a rectangle is 560 square inches. The length is 3 more than twice the width. Find the length and the width.

Let L be the length and let W be the width. The length is 3 more than twice the width, so

$$L = 2W + 3.$$

The area is 560, so

$$LW = 560.$$

Plug in $L = 2W + 3$ and solve for W:

$$\eqalign{ LW & = 560 \cr (2W + 3)W & = 560 \cr 2W^2 + 3W & = 560 \cr 2W^2 + 3W - 560 & = 0 \cr}$$

Use the Quadratic Formula:

$$W = \dfrac{-3 \pm \sqrt{9 - 4(2)(-560)}}{2 \cdot 2} = \dfrac{-3 \pm \sqrt{4489}}{4} = \dfrac{-3 \pm 67}{4} = \dfrac{-70}{4} \quad\hbox{or}\quad \dfrac{64}{4}.$$

Since the width can't be negative, I get $W = \dfrac{64}{4} = 16$ . The length is $L = 2 \cdot 16 + 3 = 35$ .


Example. Calvin and Bonzo can eat 1260 hamburgers in 12 hours. Eating by himself, it would take Calvin 7 hours longer to eat 1260 hamburgers than it would take Bonzo to eat 1260 hamburgers. How long would it take Bonzo to eat 1260 hamburgers by himself?

Let c be Calvin's rate, in hamburgers per hour. Let b be Bonzo's rate, in hamburgers per hour.

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & & & Hours & & $\cdot$ & & Burgers per hour & & $=$ & & Burgers & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & Calvin & & $t + 7$ & & $\cdot$ & & c & & $=$ & & 1260 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & Bonzo & & t & & $\cdot$ & & b & & $=$ & & 1260 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & Together & & 12 & & $\cdot$ & & $c + b$ & & $=$ & & 1260 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

The last equation gives

$$\eqalign{ 12(c + b) & = 1260 \cr c + b & = 105 \cr c & = 105 - b \cr}$$

The second equation gives

$$\eqalign{ t b & = 1260 \cr \noalign{\vskip2pt} b & = \dfrac{1260}{t} \cr}$$

Plug $b =
   \dfrac{1260}{t}$ into $c = 105 - b$ :

$$c = 105 - \dfrac{1260}{t}.$$

Plug $c = 105 -
   \dfrac{1260}{t}$ into the first equation $(t + 7)c =
   1260$ and solve for t:

$$\eqalign{ (t + 7)\left(105 - \dfrac{1260}{t}\right) & = 1260 \cr 105(t + 7)\left(1 - \dfrac{12}{t}\right) & = 1260 \cr (t + 7)\left(1 - \dfrac{12}{t}\right) & = 12 \cr (t + 7)\left(1 - \dfrac{12}{t}\right) \cdot t & = 12 \cdot t\cr (t + 7)(t - 12) & = 12 t \cr t^2 - 5 t - 84 & = 12 t \cr t^2 - 17 t - 84 & = 0 \cr (t - 21)(t + 4) & = 0 \cr}$$

The solution $t = -4$ doesn't make sense, since time can't be negative. The solution is 21. Bonzo takes 21 hours, and Calvin takes $21 + 7 =
   28$ hours.


Example. If Calvin and Bonzo eat together, they can eat 480 hot dogs in 6 hours.

Eating alone, Bonzo takes 16 hours longer than Calvin would to eat 480 hot dogs.

How long does it take Calvin to eat 480 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 480 hot dogs.

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & & & hours & & $\cdot$ & & hot dogs per hour & & $=$ & & hot dogs & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & Calvin and Bonzo & & 6 & & $\cdot$ & & $x + y$ & & $=$ & & 480 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & Calvin & & t & & $\cdot$ & & x & & $=$ & & 480 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & Bonzo & & $t + 16$ & & $\cdot$ & & y & & $=$ & & 480 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

The second equation says $xt = 480$ , so $x = \dfrac{480}{t}$ .

The third equation says $y(t + 16) = 480$ , so $y = \dfrac{480}{t + 16}$ .

Plug these into the first equation $6(x + y) = 480$ and solve for t:

$$\eqalign{ 6(x + y) & = 480 \cr 6\left(\dfrac{480}{t} + \dfrac{480}{t + 16}\right) & = 480 \cr \dfrac{480}{t} + \dfrac{480}{t + 16} & = 80 \cr t(t + 16) \cdot \left(\dfrac{480}{t} + \dfrac{480}{t + 16}\right) & = t(t + 16) \cdot 80 \cr 480(t + 16) + 480t & = 80t(t + 16) \cr 6(t + 16) + 6t & = t(t + 16) \cr 6t + 96 + 6t & = t^2 + 16t \cr 0 & = t^2 + 4t - 96 \cr 0 & = (t + 12)(t - 8) \cr}$$

The solutions are $t =
   -12$ and $t = 8$ . Since t can't be negative, the answer is $t = 8$ hours.


Example. Calvin rides his power boat up and down a drainage ditch. The water in the drainage ditch flows at 6 miles per hour. Calvin takes 5 hours longer to travel 360 miles against the current than he does to travel 360 miles with the current. What is the speed of Calvin's boat in still water?

Let x be the speed of Calvin's boat in miles per hour in still water, and let t be the time in hours it takes him to travel 360 miles with the current. Thus, it takes him $t + 5$ hours to travel 360 miles against the current.

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & & & time & & $\cdot$ & & speed & & $=$ & & distance & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & against current & & $t + 5$ & & $\cdot$ & & $x - 6$ & & $=$ & & 360 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & with current & & t & & $\cdot$ & & $x + 6$ & & $=$ & & 360 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

The equations are

$$(t + 5)(x - 6) = 360 \quad\hbox{and}\quad t(x + 6) = 360.$$

Solve the second equation for t:

$$t = \dfrac{360}{x + 6}.$$

Plug this into the first equation and solve for x:

$$\eqalign{ \left(\dfrac{360}{x + 6} + 5\right)(x - 6) & = 360 \cr (x + 6)\left(\dfrac{360}{x + 6} + 5\right)(x - 6) & = 360(x + 6) \cr \left[360 + 5(x + 6)\right](x - 6) &= 360(x + 6) \cr (5x + 390)(x - 6) &= 360(x + 6) \cr 5x^2 + 360x - 2340 &= 360x + 2160 \cr 5x^2 &= 4500 \cr x^2 &= 900 \cr}$$

The solutions are $x =
   \pm 30$ . Since the speed can't be negative, the answer is 30 miles per hour.


Example. The hypotenuse of a right triangle is 4 times the smallest side. The third side is $\sqrt{735}$ . Find the hypotenuse and the smallest side.

Let s be the smallest side and let h be the hypotenuse. By Pythagoras,

$$h^2 = s^2 + (\sqrt{753})^2, \quad\hbox{so}\quad h^2 = s^2 + 735.$$

The hypotenuse is 4 times the smallest side, so

$$h = 4 s.$$

Plug $h = 4 s$ into $h^2 = s^2 + 735$ and solve for s:

$$\eqalign{ (4 s)^2 &= s^2 + 735 \cr 16 s^2 & = s^2 + 735 \cr 15 s^2 & = 735 \cr s^2 & = 49 \cr s & = \pm 7 \cr}$$

Since $s = -7$ doesn't make sense, the solution is $s = 7$ . Then $h = 4 \cdot 7 = 28$ .


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