Systems and Polynomials with Technology

Real problems often lead to equations that are slow or impossible to solve by hand: three unknown prices, the curve through three data points, or a cubic from a volume. Your GDC has built-in solvers that do the algebra in seconds. The real skill, and what IB questions test, is setting up the equations correctly from the context and interpreting what the solver tells you.

Key ideas

Systems of linear equations

A system is a set of equations that must all be true at the same time. A solution gives a value for every variable that satisfies every equation. With $2$ unknowns you need $2$ equations; with $3$ unknowns you need $3$.

To solve on a GDC, use the simultaneous equations (linear system) solver:

1. Write every equation in the standard form $ax + by + cz = d$, with the variables in the same order and the constant on the right.
2. Enter the number of unknowns, then the coefficients row by row. Put $0$ for any missing variable.
3. Read off the solution, and check it in at least one original equation.

In IB exams, any method is allowed, and a system will always have a unique solution. You can also solve systems by elimination or, at higher level, with inverse matrices.

When there isn’t one solution

Outside exams, a solver may report something other than a single answer:

• No solution (inconsistent): the equations contradict each other, like $x + 2y = 4$ and $x + 2y = 9$. Graphically the lines (or planes) are parallel and never all meet.
• Infinitely many solutions: one equation is really a combination of the others, like $x + 2y = 4$ and $2x + 4y = 8$. They describe the same line, so every point on it works. The GDC may give the answer in terms of a parameter, or just say “infinite solutions”.

In a context, either message usually means the information doesn’t pin down the answer, or there’s an error in setting up the equations. (See intersection of planes for the 3-D picture.)

Polynomial equations

A polynomial equation has the form $a_n x^n + \dots + a_1 x + a_0 = 0$. Its solutions are called roots of the equation, or zeros of the polynomial $p(x)$. Real roots are the $x$-intercepts of $y = p(x)$.

To solve on a GDC, use the polynomial root finder:

1. Rearrange so that one side is $0$, and expand any brackets.
2. Enter the degree, then the coefficients from the highest power down. Put $0$ for any missing power.
3. Read off the roots. Some GDCs also show complex roots (with $i$). Real-life answers use only the real roots.

You can also graph $y = p(x)$ and find its zeros, or graph both sides of the original equation and find the intersections. A graph is a useful check that you’ve found every real root.

Interpreting answers in context

The solver doesn’t know what your variables mean. After solving, ask:

• Is each value possible? Lengths, times and prices can’t be negative; numbers of people must be whole numbers.
• Does it fit the domain of the model? A box cut from a $20$ cm wide sheet can’t have $17$ cm corners.
• Is the answer rounded sensibly (3 s.f. unless the context needs something else, like money to the cent)?

Worked examples

Example 1: Ticket sales

A school concert sold $640$ tickets. Adult tickets cost $18 and student tickets cost $11. Ticket sales were $9805. How many of each ticket were sold?

Solution. Let $a$ be the number of adult tickets and $s$ the number of student tickets.

$$\begin{aligned} a + s &= 640 && \text{number of tickets} \\ 18a + 11s &= 9805 && \text{money from tickets} \end{aligned}$$

Enter the coefficients $1, 1, 640$ and $18, 11, 9805$ in the $2 \times 2$ solver. It gives $a = 395$ and $s = 245$.

So $395$ adult and $245$ student tickets were sold.

Check: $395 + 245 = 640$ ✓ and $18(395) + 11(245) = 7110 + 2695 = 9805$ ✓.

Example 2: A quadratic model through three points

A ball is thrown upward. Its height $h$ metres after $t$ seconds is modelled by $h = at^2 + bt + c$. The ball is $14.6$ m high at $t = 1$, $17.9$ m at $t = 2$ and $11.4$ m at $t = 3$.

• (a) Find $a$, $b$ and $c$.
• (b) Find when the ball hits the ground.

Solution.

(a) Substitute each point into $h = at^2 + bt + c$. Each one gives a linear equation in $a$, $b$, $c$:

$$\begin{aligned} a + b + c &= 14.6 && t = 1 \\ 4a + 2b + c &= 17.9 && t = 2 \\ 9a + 3b + c &= 11.4 && t = 3 \end{aligned}$$

The $3 \times 3$ solver gives $a = -4.9$, $b = 18$, $c = 1.5$, so $h = -4.9t^2 + 18t + 1.5$.

(b) The ball hits the ground when $h = 0$:

$$-4.9t^2 + 18t + 1.5 = 0$$

The polynomial root finder (degree $2$, coefficients $-4.9$, $18$, $1.5$) gives $t = 3.754\,99\ldots$ and $t = -0.081\,5\ldots$. Time after the throw can’t be negative, so the ball lands after about $3.75$ s (3 s.f.).

Notice that $c = 1.5$ is the height at $t = 0$: the ball left the thrower’s hand $1.5$ m above the ground.

Example 3: An open box

An open box is made from a $30$ cm by $20$ cm sheet of card by cutting a square of side $x$ cm from each corner and folding up the sides. Find the values of $x$ for which the volume is $1000$ cm³.

Solution. The base is $(30 - 2x)$ by $(20 - 2x)$ and the height is $x$, so

$$x(30 - 2x)(20 - 2x) = 1000$$

Expand and rearrange so one side is $0$:

$$\begin{aligned} x(600 - 100x + 4x^2) &= 1000 \\ 4x^3 - 100x^2 + 600x - 1000 &= 0 \end{aligned}$$

The root finder (degree $3$, coefficients $4, -100, 600, -1000$) gives

$$x = 2.93 \text{ (3 s.f.)}, \qquad x = 5, \qquad x = 17.1 \text{ (3 s.f.)}$$

The width $20 - 2x$ must be positive, so $0 \lt x \lt 10$. That rules out $x = 17.1$. So $x = 2.93$ cm or $x = 5$ cm. Both boxes have volume $1000$ cm³.

The volume equals $1000$ cm³ where the curve meets the line $V = 1000$: at $x \approx 2.93$ and $x = 5$.

Check: $5(20)(10) = 1000$ ✓.

Example 4: Three unknown prices

At a bakery, three orders cost:

• $2$ coffees, $1$ muffin and $1$ juice: $12.35
• $1$ coffee and $3$ muffins: $11.50
• $3$ coffees, $2$ muffins and $2$ juices: $21.45

Find the price of each item.

Solution. Let $c$, $m$ and $j$ be the prices in dollars of a coffee, a muffin and a juice.

$$\begin{aligned} 2c + m + j &= 12.35 \\ c + 3m + 0j &= 11.50 \\ 3c + 2m + 2j &= 21.45 \end{aligned}$$

Note the $0$ for juice in the second order: you must enter it. The $3 \times 3$ solver gives $c = 3.25$, $m = 2.75$, $j = 3.10$.

A coffee costs $3.25, a muffin $2.75 and a juice $3.10.

Check the third order: $3(3.25) + 2(2.75) + 2(3.10) = 9.75 + 5.50 + 6.20 = 21.45$ ✓.

Common mistakes

Entering equations that aren’t in standard form. The solver needs every equation as $ax + by + cz = d$, with the variables in the same order. An equation like $y = 2z$ must become $0x + 1y - 2z = 0$ before you enter it.

Leaving out a zero. If a variable or a power is missing, its coefficient is $0$ and you must type it. Skipping it shifts every later coefficient into the wrong place.

Not rearranging a polynomial to equal zero. The root finder solves $p(x) = 0$. For $x^3 = 5x + 3$, enter $x^3 - 5x - 3 = 0$: coefficients $1, 0, -5, -3$.

Keeping roots that don’t fit the context. In Example 3, $x = 17.1$ is a correct root of the cubic, but it’s impossible for the box. Always check the domain.

Giving complex roots as real-world answers. If the GDC shows roots like $1.46 + 0.688i$, those are not $x$-intercepts and can’t be a length or a time. Only real roots count in a real context.

Not checking. Typing one coefficient wrong gives a confident but wrong answer. Substitute your solution back into an original equation; it takes seconds.

Practice

1. (Warm-up) Use technology to solve:

$$\begin{aligned} 3x - 2y &= 7 \\ 5x + 4y &= 19 \end{aligned}$$

Solution

Enter $3, -2, 7$ and $5, 4, 19$. The solver gives $x = 3$, $y = 1$.

Check: $3(3) - 2(1) = 7$ ✓ and $5(3) + 4(1) = 19$ ✓.

2. (Warm-up) Use technology to find the roots of $2x^3 - 3x^2 - 11x + 6 = 0$.

Solution

Degree $3$, coefficients $2, -3, -11, 6$. The root finder gives

$$x = -2, \qquad x = \frac{1}{2}, \qquad x = 3$$

Check $x = 3$: $2(27) - 3(9) - 11(3) + 6 = 54 - 27 - 33 + 6 = 0$ ✓.

3. (Warm-up) Use technology to solve:

$$\begin{aligned} x + y + z &= 6 \\ 2x - y + 3z &= 9 \\ x + 4y - z &= 6 \end{aligned}$$

Solution

The $3 \times 3$ solver gives $x = 1$, $y = 2$, $z = 3$.

Check the second equation: $2(1) - 2 + 3(3) = 9$ ✓.

4. (Core) Solve $x^3 = 5x + 3$, giving your answers to 3 s.f.

Solution

Rearrange: $x^3 - 5x - 3 = 0$. Degree $3$, coefficients $1, 0, -5, -3$ (don’t forget the $0$ for $x^2$). The root finder gives

$$x = -1.83, \qquad x = -0.657, \qquad x = 2.49 \quad \text{(3 s.f.)}$$

5. (Core) Ana invests $20 000 in three funds paying simple interest of $2\%$, $4\%$ and $7\%$ per year. After one year she earns $940 in interest. She put twice as much in the $4\%$ fund as in the $7\%$ fund. How much did she invest in each fund?

Solution

Let $x$, $y$, $z$ be the amounts in dollars in the $2\%$, $4\%$ and $7\%$ funds.

$$\begin{aligned} x + y + z &= 20\,000 \\ 0.02x + 0.04y + 0.07z &= 940 \\ 0x + y - 2z &= 0 \end{aligned}$$

(The last equation is $y = 2z$ rewritten in standard form.) The solver gives $x = 2000$, $y = 12\,000$, $z = 6000$.

She invested $2000 at $2\%$, $12 000 at $4\%$ and $6000 at $7\%$.

Check: interest $= 40 + 480 + 420 = 940$ ✓.

6. (Core) A company’s weekly profit $P$ dollars from selling $x$ items is modelled by $P = ax^2 + bx + c$. When $x = 20$, $P = 250$; when $x = 30$, $P = 400$; and when $x = 50$, $P = 400$.

• (a) Find $a$, $b$ and $c$.
• (b) Find the break-even points (where $P = 0$).

Solution

(a) Substitute the three points:

$$\begin{aligned} 400a + 20b + c &= 250 \\ 900a + 30b + c &= 400 \\ 2500a + 50b + c &= 400 \end{aligned}$$

The solver gives $a = -0.5$, $b = 40$, $c = -350$, so $P = -0.5x^2 + 40x - 350$.

(b) Solve $-0.5x^2 + 40x - 350 = 0$: the root finder gives $x = 10$ and $x = 70$.

The company breaks even when it sells $10$ items or $70$ items, and makes a profit between those values.

7. (Core) A GDC gives the message “no solution” for system A and “infinitely many solutions” for system B. Explain why in each case.

$$\text{A: } \begin{aligned} x + 2y - z &= 3 \\ 2x + 4y - 2z &= 5 \\ x - y + z &= 1 \end{aligned} \qquad\qquad \text{B: } \begin{aligned} x + y + z &= 3 \\ x - y + 2z &= 4 \\ 2x + 0y + 3z &= 7 \end{aligned}$$

Solution

A: Doubling the first equation gives $2x + 4y - 2z = 6$, but the second equation says the same expression equals $5$. Both can’t be true, so there is no solution. (The first two planes are parallel.)

B: Adding the first two equations gives $2x + 0y + 3z = 7$, which is exactly the third equation. So the third equation adds no new information: there are really only two equations for three unknowns, and infinitely many solutions (all the points on a line).

8. (Challenge) Find the coordinates of the points where the curve $y = x^3 - 2x$ meets the line $y = x + 1$, to 3 s.f.

Solution

At an intersection both $y$-values are equal:

$$x^3 - 2x = x + 1 \quad\Rightarrow\quad x^3 - 3x - 1 = 0$$

Root finder (coefficients $1, 0, -3, -1$): $x = -1.532\ldots$, $x = -0.347\ldots$, $x = 1.879\ldots$

Use $y = x + 1$ (with the unrounded $x$-values) for the $y$-coordinates:

$$(-1.53,\ -0.532), \qquad (-0.347,\ 0.653), \qquad (1.88,\ 2.88) \quad \text{(3 s.f.)}$$

9. (Challenge) The curve $y = x^3 + ax^2 + bx + c$ passes through $(1, 2)$, $(-1, 6)$ and $(2, 3)$.

• (a) Find $a$, $b$ and $c$.
• (b) Solve $x^3 + ax^2 + bx + c = 0$. How many real roots are there?

Solution

(a) Substitute each point, and move the known $x^3$ term to the right:

$$\begin{aligned} a + b + c &= 2 - 1 = 1 && (1, 2) \\ a - b + c &= 6 + 1 = 7 && (-1, 6) \\ 4a + 2b + c &= 3 - 8 = -5 && (2, 3) \end{aligned}$$

The solver gives $a = -1$, $b = -3$, $c = 5$, so $y = x^3 - x^2 - 3x + 5$.

(b) Root finder (coefficients $1, -1, -3, 5$): $x = -1.92$ (3 s.f.), and two complex roots $x \approx 1.46 \pm 0.688i$.

There is only one real root, $x \approx -1.92$, so the curve crosses the $x$-axis once.