Two cars leave an intersection at the same time. One car travels north and the other car travels west. When the car traveling north had gone 24 miles, the distance between the cars was four miles more than three times the distance traveled by the car heading west. Find the distance between the cars at that time.
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Khan Academy Applying Quadratic Equations
Guidance
Quadratic functions can be used to help solve many different real world problems. Here are two hints for solving quadratic word problems:
 It is often helpful to start by drawing a picture in order to visualize what you are asked to solve.
 Once you have solved the problem, it is important to make sure that your answers are realistic given the context of the problem. For example, if you are solving for the age of a person and one of your answers is a negative number, that answer does not make sense in the context of the problem and is not actually a solution.
Example A
The number of softball games that must be scheduled in a league with \begin{align*}n\end{align*} teams is given by \begin{align*}G(n)= \frac{n^2  n}{2}\end{align*}. Each team can only play every other team exactly once. A league schedules 21 games. How many softball teams are in the league?
Solution: You are given the function \begin{align*}G(n)= \frac{n^2n}{2}\end{align*} and you are asked to find \begin{align*}n\end{align*} when \begin{align*}G(n)=21\end{align*}. This means, you have to solve the equation:
\begin{align*}21=\frac{n^2n}{2}\end{align*}
Start by setting the equation equal to zero:
\begin{align*}42&=n^2n\\ n^2n42&=0\end{align*}
Now solve for \begin{align*}n\end{align*} to find the number of teams \begin{align*}(n)\end{align*} in the league. Start by factoring the left side of the equation and rewriting the equation:
\begin{align*}n^2n42=0 \end{align*} becomes \begin{align*}(n7)(n+6)=0\end{align*}
There are 7 teams in the softball league.
Example B
When a homemade rocket is launched from the ground, it goes up and falls in the pattern of a parabola. The height, in feet, of a homemade rocket is given by the equation \begin{align*}h(t) = 160t  16t^2\end{align*} where \begin{align*}t\end{align*} is the time in seconds. How long will it take for the rocket to return to the ground?
Solution: The formula for the path of the rocket is \begin{align*}h(t)=160t16t^2\end{align*}. You are asked to find \begin{align*}t\end{align*} when \begin{align*}h(t)=0\end{align*}, or when the rocket hits the ground and no longer has height. Start by factoring:
\begin{align*}160t16t^2=0\end{align*} becomes \begin{align*}16t(10t)=0\end{align*}
This means \begin{align*}16t=0\end{align*} (so \begin{align*}t=0\end{align*}) or \begin{align*}10t=0\end{align*} (so \begin{align*}t=10\end{align*}). \begin{align*}t=0\end{align*} represents the rocket being on the ground when it starts, so it is not the answer you are looking for. \begin{align*}t=10\end{align*} represents the rocket landing back on the ground.
The rocket will hit the ground after 10 seconds.
Example C
Using the information in Example B, what is the height of the rocket after 2 seconds?
Solution: To solve this problem, you need to replace \begin{align*}t\end{align*} with 2 in the quadratic function.
\begin{align*}h(t)&=160t16t^2\\ h(2)&=160(2)16(2)^2\\ h(2)&=32064\\ h(2)&=256.\end{align*}
Therefore, after 2 seconds, the height of the rocket is 256 feet.
Concept Problem Revisited
Two cars leave an intersection at the same time. One car travels north and the other car travels west. When the car traveling north had gone 24 miles the distance between the cars was four miles more than three times the distance traveled by the car heading west. Find the distance between the cars at that time.
First draw a diagram. Since the cars are traveling north and west from the same starting position, the triangle made to connect the distance between them is a right triangle. Since you have a right triangle, you can use the Pythagorean Theorem to set up an equation relating the lengths of the sides of the triangle.
The Pythagorean Theorem is a geometry theorem that says that for all right triangles, \begin{align*}a^2+b^2=c^2\end{align*} where \begin{align*}a\end{align*} and \begin{align*}b\end{align*} are legs of the triangle and \begin{align*}c\end{align*} is the longest side of the triangle, the hypotenuse. The equation for this problem is:
\begin{align*}x^2+24^2&=(3x+4)^2\\ x^2+576&=(3x+4)(3x+4)\\ x^2+576&=9x^2+24x+16\end{align*}
Now set the equation equal to zero and factor the quadratic expression so that you can use the zero product property.
\begin{align*}x^2+576&=9x^2+24x+16\\ 0&=8x^2+24x560\\ 0&=8(x^2+3x70)\\ 0&=8(x7)(x+10)\end{align*}
So you now know that \begin{align*}x = 7\end{align*}. Since the distance between the cars is represented by the expression \begin{align*}3x + 4\end{align*}, the actual distance between the two cars after the car going north has traveled 24 miles is:
\begin{align*}3x+4&=3(7)+4\\ &=21+4\\ &=25 \ miles\end{align*}
Vocabulary
 Factor
 To factor means to rewrite an expression as a product.
 Pythagorean Theorem
 The Pythagorean Theorem is a right triangle theorem that relates all three sides of a right triangle according to the equation \begin{align*}a^2 + b^2 = c^2\end{align*} where \begin{align*}a\end{align*} and \begin{align*}b\end{align*} are legs of the triangle and \begin{align*}c\end{align*} is the hypotenuse.
 Quadratic Expression
 A quadratic expression is a polynomial of degree 2. The general form of a quadratic expression is \begin{align*}ax^2 + bx + c\end{align*}.
Guided Practice
1. A rectangle is known to have an area of 520 square inches. The lengths of the sides are shown in the diagram below. Solve for both the length and the width.
2. The height of a ball in feet can be found by the quadratic function \begin{align*}h(t)=16t^2+80t+5\end{align*} where \begin{align*}t\end{align*} is the time in seconds that the ball is in the air. Determine the time(s) at which the ball is 69 feet high.
3. A manufacturer measures the number of cell phones sold using the binomial \begin{align*}0.015c + 2.81\end{align*}. She also measures the wholesale price on these phones using the binomial \begin{align*}0.011c +3.52\end{align*}. Calculate her revenue if she sells 100,000 cell phones.
Answers:
1. The rectangle has an area of 520 square inches and you know that the area of a rectangle has the formula: \begin{align*}A = l \times w\end{align*}. Therefore:

\begin{align*}520&=(x+7)(2x)\\
520&=2x^2+14x\\
0&=2x^2+14x520\\
0&=2(x^2+7x260)\\
0&=2(x13)(x+20)\end{align*}
 Therefore the value of \begin{align*}x\end{align*} is 13. This means that the width is \begin{align*}2x\end{align*} or \begin{align*}2(13) = 26 \ inches\end{align*}. The length is \begin{align*}x + 7 = 13 + 7 = 20 \ inches\end{align*}.
2. The equation for the ball being thrown is \begin{align*}h(t)=16t^2+80t+5\end{align*}. If you drew the path of the thrown ball, you would see something like that shown below.
 You are asked to find the time(s) when the ball hits a height of 69 feet. In other words, solve for:

\begin{align*}69=16t^2+80t+5\end{align*}
 To solve for \begin{align*}t\end{align*}, you have to factor the quadratic and then solve for the value(s) of \begin{align*}t\end{align*}.

\begin{align*}&\quad \qquad 0=16t^2+80t64\\
&\quad \qquad 0=16(t^25t+4)\\
&\quad \qquad 0=16(t1)(t4)\\
&\qquad \quad \qquad \swarrow \qquad \qquad \searrow\\
& \quad t  1 = 0 \qquad \qquad t4 = 0\\
&\qquad \ \ t = 1 \quad \qquad \qquad \ \ t = 4\end{align*}
 Since both values are positive, you can conclude that there are two times when the ball hits a height of 69 feet. These times are at 1 second and at 4 seconds.
3. The number of cell phones sold is the binomial \begin{align*}0.015c + 2.81\end{align*}. The wholesale price on these phones is the binomial \begin{align*}0.011c +3.52.\end{align*} The revenue she takes in is the wholesale price times the number that she sells. Therefore:
 \begin{align*}R(c)=(0.015c+2.81)(0.011c+3.52)\end{align*}
 First, let’s expand the expression for \begin{align*}R\end{align*} to get the quadratic expression. Therefore:

\begin{align*}R(c)&=(0.015c+2.81)(0.011c+3.52)\\
R(c)&=0.000165c^2+0.08371c+9.8912\end{align*}
 The question then asks if she sold 100,000 cell phones, what would her revenue be. Therefore what is \begin{align*}R(c)\end{align*} when \begin{align*}c = 100,000\end{align*}.

\begin{align*}R(c)&=0.000165c^2+0.08371c+9.8912\\
R(c)&=0.000165(100,000)^2+0.08371(100,000)+9.8912\\
R(c)&=1,658,380.89\end{align*}
 Therefore she would make $1,658,380.89 in revenue.
Practice
 A rectangle is known to have an area of 234 square feet. The length of the rectangle is given by \begin{align*}x+3\end{align*} and the width of the rectangle is given by \begin{align*}x+8\end{align*}. What is the value of \begin{align*}x\end{align*}?
 Solve for \begin{align*}x\end{align*} in the rectangle below given that the area is 9 units.
 Solve for \begin{align*}x\end{align*} in the triangle below given that the area is 10 units.
A pool is treated with a chemical to reduce the amount of algae. The amount of algae in the pool \begin{align*}t\end{align*} days after the treatment can be approximated by the function \begin{align*}A(t)=40t^2300t+500\end{align*}.
 How many days after treatment will the pool have the no algae?
 How much algae is in the pool before treatments are started?
 How much less algae is in the pool after 1 day?
A football is kicked into the air. The height of the football in meters can be found by the quadratic function \begin{align*}h(t)=5t^2+25t\end{align*} where \begin{align*}t\end{align*} is the time in seconds since the ball has been kicked.
 How high is the ball after 3 seconds? At what other time is the ball the same height?
 When will the ball be 20 meters above the ground?
 After how many seconds will the ball hit the ground?
A ball is thrown into the air. The height of the ball in meters can be found by the quadratic function \begin{align*}h(t)=5t^2+30t\end{align*} where \begin{align*}t\end{align*} is the time in seconds since the ball has been thrown.
 How high is the ball after 3 seconds?
 When will the ball be 25 meters above the ground?
 After how many seconds will the ball hit the ground?
Kim is drafting the windows for a new building. Their shape can be modeled by the function \begin{align*}h(w)=w^2+4\end{align*}, where \begin{align*}h\end{align*} is the height and \begin{align*}w\end{align*} is the width of points on the window frame, measured in meters.
 Find the width of each window at its base.
 Find the width of each window when the height is 3 meters.
 What is the height of the window when the width is 1 meter?