Solving Problems with Quadratic Models

You now have all the tools: graphs, factoring, the quadratic formula, and finding the vertex. This page puts them to work on real questions. How high does a ball go? When does it land? How wide should a path be? The math is the same as before. The new skill is setting up the equation and interpreting the answers, including throwing out the ones that don’t make sense.

Key ideas

A plan for word problems

1. Define your variable, with units: “Let $t$ be the time in seconds.”
2. Write an equation (or read the graph you’re given).
3. Solve using the best method: factoring, the formula, or the vertex.
4. Interpret: check each answer against the situation. Reject negative times and lengths, and answer the question that was asked, with units.

The projectile model

When an object is thrown or launched (and air resistance is small), its height $h$ in metres after $t$ seconds is modelled by

$$h = -4.9t^2 + vt + h_0$$

• $-4.9$ comes from gravity on Earth (half of $9.8$ m/s²). It’s negative, so the parabola opens down.
• $v$ is the starting upward speed in metres per second.
• $h_0$ is the starting height in metres (the $h$-intercept).

Each kind of question matches a feature of the parabola:

Question	What to find
What is the maximum height? When?	the vertex: its $h$-value is the height, its $t$-value is the time
When does it hit the ground?	the positive root of $h = 0$
When is it at a height of $8$ m?	solve $h = 8$ (there are usually two times: going up and coming down)
When is it above $8$ m?	the time between the two roots of $h = 8$

Finding the vertex by partial factoring

When the equation doesn’t factor, you can still find the axis of symmetry quickly. Factor $t$ out of the first two terms only:

$$h = -4.9t^2 + 19.6t + 2 = t(-4.9t + 19.6) + 2$$

The bracketed part is $0$ when $t = 0$ or when $-4.9t + 19.6 = 0$, that is, $t = 4$. At both of these times, $h = 2$. Two points at the same height are mirror images, so the axis of symmetry is halfway between them, at $t = 2$. Substitute $t = 2$ to get the maximum height. (You can also use the shortcut $t = -\dfrac{b}{2a}$ from maximum and minimum of a quadratic, which gives the same answer.)

Reading the answers from a graph

The ball from Example 1: $h = -4.9t^2 + 19.6t + 2$. It is above $12$ m between the two green points.

Interpreting answers

• A quadratic equation usually gives two roots. In context, often only one makes sense: time can’t be negative before the launch, and a length can’t be negative.
• Some questions are only true between two answers (the time above a certain height), so give an interval.
• Answers from a graph are estimates. Answers from the formula can be as precise as you need; round sensibly (for example, to the nearest hundredth of a second).

Worked examples

Example 1: Reading a graph

A ball is thrown upward from the top of a $2$ m high wall. Its height is shown in the graph above. Use the graph to estimate:

• (a) the maximum height and when it happens
• (b) when the ball hits the ground
• (c) the time interval when the ball is higher than $12$ m

Then check (a) algebraically, using $h = -4.9t^2 + 19.6t + 2$.

Solution.

(a) The highest point is $(2, 21.6)$: a maximum height of about $21.6$ m, after $2$ s.

(b) The curve reaches the time axis at about $4.1$ s.

(c) The curve is above the line $h = 12$ between about $0.6$ s and $3.4$ s, so for about $2.8$ s.

Check (a). By partial factoring (from Key ideas), the axis of symmetry is $t = 2$. Then

$$h = -4.9(2)^2 + 19.6(2) + 2 = -19.6 + 39.2 + 2 = 21.6$$

The maximum height is $21.6$ m at $t = 2$ s. ✓

Example 2: A model rocket

A model rocket is launched straight up from the ground. Its height in metres after $t$ seconds is $h = -4.9t^2 + 39.2t$.

• (a) When does the rocket land?
• (b) What is its maximum height?
• (c) For how long is the rocket higher than $60$ m? Round to two decimal places.

Solution.

(a) Set $h = 0$ and factor out $t$:

$$\begin{aligned} -4.9t^2 + 39.2t &= 0 \\ t(-4.9t + 39.2) &= 0 \end{aligned}$$

So $t = 0$ (the launch) or $-4.9t + 39.2 = 0$, which gives $t = \dfrac{39.2}{4.9} = 8$. The rocket lands after $8$ s.

(b) The maximum is halfway between the roots, at $t = 4$:

$$h = -4.9(4)^2 + 39.2(4) = -78.4 + 156.8 = 78.4$$

The maximum height is $78.4$ m, after $4$ s.

(c) Set $h = 60$ and rearrange:

$$-4.9t^2 + 39.2t - 60 = 0$$

Use the quadratic formula with $a = -4.9$, $b = 39.2$, $c = -60$:

$$\begin{aligned} b^2 - 4ac &= 39.2^2 - 4(-4.9)(-60) = 1536.64 - 1176 = 360.64 \\ t &= \frac{-39.2 \pm \sqrt{360.64}}{2(-4.9)} = \frac{-39.2 \pm 18.991}{-9.8} \end{aligned}$$

So $t \approx \dfrac{-39.2 + 18.991}{-9.8} \approx 2.06$ or $t \approx \dfrac{-39.2 - 18.991}{-9.8} \approx 5.94$.

The rocket passes $60$ m on the way up at about $2.06$ s and on the way down at about $5.94$ s. It’s higher than $60$ m for about $5.94 - 2.06 = 3.88$ s.

Example 3: A path around a garden

A rectangular garden measures $8$ m by $6$ m. A path of the same width goes all the way around it. The garden and path together cover $120$ m². How wide is the path?

Solution. Let $x$ be the width of the path in metres. The path adds $x$ to each side, so the outer rectangle is $8 + 2x$ by $6 + 2x$:

$$\begin{aligned} (8 + 2x)(6 + 2x) &= 120 \\ 48 + 16x + 12x + 4x^2 &= 120 \\ 4x^2 + 28x - 72 &= 0 \\ x^2 + 7x - 18 &= 0 && \text{divide by } 4 \\ (x + 9)(x - 2) &= 0 \end{aligned}$$

So $x = -9$ or $x = 2$. A width can’t be negative, so reject $x = -9$. The path is $2$ m wide.

Check: the outer rectangle is $12$ m by $10$ m, and $12 \times 10 = 120$. ✓

Example 4: A number problem

The sum of the squares of two consecutive positive integers is $145$. Find the integers.

Solution. Let the smaller integer be $n$. The next one is $n + 1$.

$$\begin{aligned} n^2 + (n + 1)^2 &= 145 \\ n^2 + n^2 + 2n + 1 &= 145 \\ 2n^2 + 2n - 144 &= 0 \\ n^2 + n - 72 &= 0 && \text{divide by } 2 \\ (n + 9)(n - 8) &= 0 \end{aligned}$$

So $n = -9$ or $n = 8$. The integers must be positive, so reject $n = -9$. The integers are $8$ and $9$.

Check: $8^2 + 9^2 = 64 + 81 = 145$. ✓ (Without the word “positive”, $-9$ and $-8$ would also work.)

Common mistakes

Keeping a negative time or length. A root like $t = -0.10$ s or $x = -9$ m is a correct solution of the equation, but it doesn’t fit the situation. Reject it, and say why.

Answering a different question. “When is the maximum height?” asks for a time; “What is the maximum height?” asks for a height. Read the question again before you write the final answer.

Forgetting to rearrange before solving. To find when $h = 60$, solve $-4.9t^2 + 39.2t - 60 = 0$, not $-4.9t^2 + 39.2t = 0$.

Adding the path width only once. A path of width $x$ around a garden adds $x$ on both ends, so each dimension grows by $2x$.

Giving one time for “above a height”. The ball is above $60$ m for a whole interval, from the first root to the second. Give both times, or the length of the interval if that’s what’s asked.

Rounding in the middle of the formula. Keep the full value of the square root until the end, then round once.

Practice

1. (Warm-up) A ball is kicked from the ground. Its height in metres after $t$ seconds is $h = -4.9t^2 + 14.7t$. When does it land?

Solution

$$t(-4.9t + 14.7) = 0$$

So $t = 0$ (the kick) or $t = \dfrac{14.7}{4.9} = 3$. It lands after $3$ s.

2. (Warm-up) A stone is dropped from a cliff $44.1$ m above the water. Its height in metres after $t$ seconds is $h = -4.9t^2 + 44.1$. When does it hit the water?

Solution

$$\begin{aligned} -4.9t^2 + 44.1 &= 0 \\ 4.9t^2 &= 44.1 \\ t^2 &= 9 \\ t &= \pm 3 \end{aligned}$$

Time can’t be negative, so reject $t = -3$. The stone hits the water after $3$ s.

3. (Warm-up) When a number is squared and then $3$ times the number is added, the result is $40$. Find the number.

Solution

Let the number be $x$:

$$\begin{aligned} x^2 + 3x &= 40 \\ x^2 + 3x - 40 &= 0 \\ (x + 8)(x - 5) &= 0 \end{aligned}$$

So $x = 5$ or $x = -8$. Both work: $25 + 15 = 40$ and $64 - 24 = 40$. Nothing in the question rules out a negative number, so both are answers.

4. (Core) A volleyball is hit upward from a height of $1.5$ m. Its height in metres after $t$ seconds is $h = -4.9t^2 + 9.8t + 1.5$.

• (a) Find the maximum height and when it happens.
• (b) When does the ball hit the floor, if no one touches it? Round to two decimal places.

Solution

(a) Partial factoring: $h = t(-4.9t + 9.8) + 1.5$. The height is $1.5$ m when $t = 0$ and when $-4.9t + 9.8 = 0$, that is, $t = 2$. The axis is halfway, at $t = 1$:

$$h = -4.9(1)^2 + 9.8(1) + 1.5 = 6.4$$

The maximum height is $6.4$ m, after $1$ s.

(b) Solve $-4.9t^2 + 9.8t + 1.5 = 0$ with $a = -4.9$, $b = 9.8$, $c = 1.5$:

$$b^2 - 4ac = 96.04 + 29.4 = 125.44, \qquad \sqrt{125.44} = 11.2$$$$t = \frac{-9.8 \pm 11.2}{-9.8}$$

So $t = \dfrac{1.4}{-9.8} \approx -0.14$ or $t = \dfrac{-21}{-9.8} \approx 2.14$. Reject the negative time. The ball hits the floor after about $2.14$ s.

5. (Core) For the volleyball in question 4, find the time interval when the ball is higher than $5$ m. Round to two decimal places.

Solution

Set $h = 5$:

$$\begin{aligned} -4.9t^2 + 9.8t + 1.5 &= 5 \\ -4.9t^2 + 9.8t - 3.5 &= 0 \end{aligned}$$$$b^2 - 4ac = 96.04 - 4(-4.9)(-3.5) = 96.04 - 68.6 = 27.44$$$$t = \frac{-9.8 \pm \sqrt{27.44}}{-9.8} \approx \frac{-9.8 \pm 5.238}{-9.8}$$

So $t \approx 0.47$ or $t \approx 1.53$. The ball is higher than $5$ m from about $0.47$ s to about $1.53$ s. (Notice these are the same distance from the vertex time, $t = 1$, as symmetry says they should be.)

6. (Core) A rectangular rink is $5$ m longer than it is wide, and its area is $84$ m². Find its dimensions.

Solution

Let the width be $w$ metres, so the length is $w + 5$:

$$\begin{aligned} w(w + 5) &= 84 \\ w^2 + 5w - 84 &= 0 \\ (w + 12)(w - 7) &= 0 \end{aligned}$$

So $w = -12$ or $w = 7$. Reject the negative width. The rink is $7$ m wide and $12$ m long.

Check: $7 \times 12 = 84$. ✓

7. (Core) One leg of a right triangle is $7$ cm longer than the other leg. The hypotenuse is $13$ cm. Find the lengths of the legs.

Solution

Let the shorter leg be $x$ cm. By the Pythagorean theorem:

$$\begin{aligned} x^2 + (x + 7)^2 &= 13^2 \\ x^2 + x^2 + 14x + 49 &= 169 \\ 2x^2 + 14x - 120 &= 0 \\ x^2 + 7x - 60 &= 0 \\ (x + 12)(x - 5) &= 0 \end{aligned}$$

So $x = -12$ or $x = 5$. Reject the negative length. The legs are $5$ cm and $12$ cm.

Check: $5^2 + 12^2 = 25 + 144 = 169 = 13^2$. ✓

8. (Challenge) A photo is $20$ cm by $30$ cm. It is placed in a frame of the same width all the way around. The area of the frame alone (not counting the photo) is equal to the area of the photo. How wide is the frame?

Solution

Let the frame width be $x$ cm. The photo’s area is $20 \times 30 = 600$ cm², so the photo and frame together cover $600 + 600 = 1200$ cm²:

$$\begin{aligned} (20 + 2x)(30 + 2x) &= 1200 \\ 600 + 40x + 60x + 4x^2 &= 1200 \\ 4x^2 + 100x - 600 &= 0 \\ x^2 + 25x - 150 &= 0 \\ (x + 30)(x - 5) &= 0 \end{aligned}$$

So $x = -30$ or $x = 5$. Reject the negative width. The frame is $5$ cm wide.

Check: the outer size is $30$ cm by $40$ cm, which is $1200$ cm², and $1200 - 600 = 600$. ✓

9. (Challenge) A soccer ball is kicked so that its height in metres after $t$ seconds is $h = -4.9t^2 + 12t + 0.5$. Does the ball ever reach a height of $8$ m? Use the quadratic formula to decide, and explain what the result means.

Solution

Set $h = 8$ and rearrange:

$$-4.9t^2 + 12t - 7.5 = 0$$

With $a = -4.9$, $b = 12$, $c = -7.5$:

$$b^2 - 4ac = 144 - 4(-4.9)(-7.5) = 144 - 147 = -3$$

The value under the square root is negative, so the equation has no real roots. There is no time when the ball is at $8$ m: it never gets that high.

Graphically, the line $h = 8$ is above the vertex of the parabola. (The maximum height is only about $7.85$ m, at $t \approx 1.22$ s.)