International Baccalaureate · DP 1: Applications and Interpretation SL

Equations

⌚ ~146 min ▢ 16 blocks ✓ 70 exercises ➣ Prerequisites : Straight Lines, Surds and Exponents

I Equations of the Form $x^2 \equal k$

I.1 Solutions of $x^2 \equal k$

Recappreviously studied — click to reread

Proposition — Solutions of $x^2 \equal a$

Let $a$ be a number. The solutions of the equation $x^2 = a$ depend on the sign of $a$:

if $a > 0$, the equation has two solutions: $x = \sqrt{a}$ and $x = -\sqrt{a}$;
if $a = 0$, the equation has exactly one solution: $x = 0$;
if $a < 0$, the equation has no solution: the square of a number is never negative.

Example

Solve each equation:

$x^2 = 49$
$x^2 = 7$
$x^2 = -9$

Answer

$49 > 0$: two solutions, $x = \sqrt{49} = 7$ and $x = -\sqrt{49} = -7$.
$7 > 0$: two solutions, $x = \sqrt{7}$ and $x = -\sqrt{7}$. These are the exact values; a calculator gives $\sqrt{7} \approx 2.65$.
$-9 < 0$: no solution, because the square of a number is never negative.

Method — Solving $x^2 \equal a$ Graphically

To solve the equation $x^2 = a$ graphically:

On the given parabola $y = x^2$, draw the horizontal line $y = a$.
Read the $x$-coordinates of their intersection points. When they seem to fall on grid lines, check them by calculation: they are then the solutions; otherwise the reading only gives an approximate value — the exact solutions are then $-\sqrt{a}$ and $\sqrt{a}$.

The line cuts the parabola at two points when $a > 0$, touches it at one point (the origin) when $a = 0$, and does not meet it when $a < 0$.

Skills to practice

Solving $x^2 \equal a$ by Calculation
Solving $x^2 \equal a$ Graphically

II Power Equations

II.1 Solving Power Equations

Definition — Nth Root

The concept of a root can be generalised. For a positive integer $n$, the nth root of a real number $a$, written $\sqrt[n]{a}$, is a number which, when raised to the power of $n$, gives $a$, chosen as follows: for an odd $n$, it is the only real number whose $n$th power is $a$; for an even $n$, it exists only when $a \ge 0$, and it is the non-negative number whose $n$th power is $a$. Whenever it exists, $$\left(\sqrt[n]{a}\right)^n = a.$$ The rules for their domains depend on whether the index $n$ is even or odd.

Even roots (e.g.\ $\sqrt[2]{a}, \sqrt[4]{a}, \dots$): an even root of a negative number is not a real number. For even $n$, $\sqrt[n]{a}$ is only defined for $\boldsymbol{a \ge 0}$, and it denotes the non-negative number whose $n$th power is $a$ (e.g.\ $\sqrt[4]{16} = 2$, not $-2$).
Odd roots (e.g.\ $\sqrt[3]{a}, \sqrt[5]{a}, \dots$): an odd root is defined for all real numbers $a$.

Definition — Power Equation

A power equation is an equation that can be written in the form $x^n = k$, where the exponent $n$ is a nonzero integer and $k$ is a real number.
For example, $x^3 = 5$, $2x^4 = 20$ (that is, $x^4 = 10$) and $x^{-2} = 9$ are power equations.

Counting the Solutions on a Graph

When the exponent $n$ is a positive integer, the solutions of $x^n = k$ are the $x$-coordinates of the points where the curve $y = x^n$ meets the horizontal line $y = k$. What decides their number is the parity of the exponent:

with an odd exponent, the curve rises from left to right through every height: each horizontal line meets it at exactly one point;
with an even exponent, the curve is symmetric about the $y$-axis and never goes below the $x$-axis: a line above the axis meets it at two points, the axis itself at one point (the origin), a line below the axis at no point.

For the exponent $2$, this is the rule already known for $x^2 = a$; the Proposition below extends it to every integer exponent $n \geq 2$, and the Method reduces a negative exponent to a positive one.

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Proposition — Solutions of $x^n \equal k$

Let $n \geq 2$ be an integer and $k$ a real number.

If $n$ is odd, the equation $x^n = k$ has exactly one solution, whatever the sign of $k$: $x = \sqrt[n]{k}$.
If $n$ is even and $k > 0$, the equation has two solutions: $x = -\sqrt[n]{k}$ and $x = \sqrt[n]{k}$.
If $n$ is even and $k = 0$, the equation has exactly one solution: $x = 0$.
If $n$ is even and $k < 0$, the equation has no solution: an even power of a real number is never negative.

Proof

Order. If $0 \le a < b$, then $a^n < b^n$: multiplying inequalities between non-negative numbers keeps the order, and each of the $n$ factors of $a^n = a \times a \times \dots \times a$ is smaller than the matching factor of $b^n$. So two different non-negative numbers never have the same $n$th power: when $k \ge 0$, the only non-negative solution of $x^n = k$ is $\sqrt[n]{k}$, by the definition of the $n$th root.
Even $n$. An even number of minus signs gives $(-x)^n = x^n$, and one of the numbers $x$ and $-x$ is non-negative, so $x^n \ge 0$ for every real $x$: there is no solution when $k < 0$. When $k \ge 0$, a non-negative solution is $\sqrt[n]{k}$, and a negative solution $x$ makes $-x$ a positive solution, so $-x = \sqrt[n]{k}$, that is $x = -\sqrt[n]{k}$. Both numbers are solutions, since $\left(-\sqrt[n]{k}\right)^n = \left(\sqrt[n]{k}\right)^n = k$: they are two different numbers when $k > 0$, and both equal $0$ when $k = 0$.
Odd $n$. An odd number of minus signs gives $(-x)^n = -x^n$, so $x^n$ has the sign of $x$. If $k \ge 0$, a solution is non-negative, so it is $\sqrt[n]{k}$. If $k < 0$, a solution $x$ is negative and $-x$ is a positive solution of $(-x)^n = -k$, so $-x = \sqrt[n]{-k}$, that is $x = -\sqrt[n]{-k}$; this number is $\sqrt[n]{k}$, since $\left(-\sqrt[n]{-k}\right)^n = -(-k) = k$. In both cases the equation has exactly one solution, $\sqrt[n]{k}$.

Method — Solving a Power Equation

To solve an equation that can be written $X^n = k$, where $X$ is $x$ or an expression such as $x - 2$:

Isolate the power: undo the additions, then the multiplications, until the equation reads $X^n = k$.
Remove a negative exponent: if $n = -m$ with $m > 0$, then $X \neq 0$ and $X^{-m} = \dfrac{1}{X^m}$; a quotient $\dfrac{1}{X^m}$ is never $0$, so for $k = 0$ there is no solution, and for $k \neq 0$ the equation becomes $X^m = \dfrac{1}{k}$.
Read the parity of the exponent and the sign of $k$: the exponent $1$ gives $X = k$ directly; an odd exponent $n \geq 3$ gives $X = \sqrt[n]{k}$; an even exponent gives $X = -\sqrt[n]{k}$ or $X = \sqrt[n]{k}$ if $k > 0$, $X = 0$ if $k = 0$, and no solution if $k < 0$.
If $X$ is an expression, solve each equation $X = \ldots$ for $x$.
Give the exact values; if a decimal value is asked, round only at the end.

Example

Solve each equation:

$x^3 = -64$
$x^4 = 81$
$x^6 = -5$
$x^4 = 0$
$x^5 = 7$

Answer

The exponent $3$ is odd: exactly one solution, $x = \sqrt[3]{-64} = -4$, since $(-4)^3 = -64$.
The exponent $4$ is even and $81 > 0$: two solutions, $x = -\sqrt[4]{81} = -3$ and $x = \sqrt[4]{81} = 3$, since $3^4 = 81$.
The exponent $6$ is even and $-5 < 0$: no solution, because a sixth power is never negative.
The exponent $4$ is even and the right side is $0$: exactly one solution, $x = 0$.
The exponent $5$ is odd: exactly one solution, $x = \sqrt[5]{7}$. This is the exact value; a calculator gives $\sqrt[5]{7} \approx 1.48$.

Example

Solve each equation:

$2x^3 + 5 = -11$
$(x - 1)^4 = 16$
$x^{-2} = \dfrac{1}{25}$
$x^{-3} = -\dfrac{1}{8}$
$x^{-4} = 0$

Answer

The power is isolated first: $$\begin{aligned}[t] 2x^3 + 5 &= -11\\ 2x^3 &= -16 && \text{(subtract $5$ from both sides)}\\ x^3 &= -8 && \text{(divide both sides by $2$)} \end{aligned}$$ The exponent $3$ is odd: $x = \sqrt[3]{-8} = -2$. Check: $2 \times (-2)^3 + 5 = -16 + 5 = -11$.
Here $X = x - 1$ and $X^4 = 16$, with $4$ even and $16 > 0$. Since $\sqrt[4]{16} = 2$: $$x - 1 = -2 \quad \text{or} \quad x - 1 = 2,$$ so $x = -1$ or $x = 3$. Check: $(-2)^4 = 2^4 = 16$.
The negative exponent is removed first: $$\begin{aligned}[t] x^{-2} &= \frac{1}{25}\\ \frac{1}{x^2} &= \frac{1}{25} && \text{(definition of a negative exponent)}\\ x^2 &= 25 && \text{(reciprocals of both sides)} \end{aligned}$$ The exponent $2$ is even and $25 > 0$: $x = -5$ or $x = 5$.
The negative exponent is removed first: $$\begin{aligned}[t] x^{-3} &= -\frac{1}{8}\\ \frac{1}{x^3} &= -\frac{1}{8} && \text{(definition of a negative exponent)}\\ x^3 &= -8 && \text{(reciprocals of both sides)} \end{aligned}$$ The exponent $3$ is odd: exactly one solution, $x = \sqrt[3]{-8} = -2$. Check: $(-2)^{-3} = \dfrac{1}{(-2)^3} = -\dfrac{1}{8}$.
$x^{-4} = \dfrac{1}{x^4}$ with $x \neq 0$, and a quotient $\dfrac{1}{x^4}$ is never $0$: the equation has no solution. With a negative exponent, the right side $0$ gives no solution, whereas $x^4 = 0$ gives $x = 0$.

Skills to practice

Counting the Solutions of a Power Equation
Solving $x^n \equal k$: Level 1
Solving Power Equations: Level 2
Solving Power Equations with a Calculator
Justifying the Solutions of a Power Equation

III Equations in Factored Form

III.1 The Null Factor Law

Recappreviously studied — click to reread

Definition — Quadratic Equation

A quadratic equation is an equation that can be written in the standard form: $$ax^2 + bx + c = 0$$ where $a, b,$ and $c$ are known coefficients, and $a \neq 0$. The condition $a \neq 0$ is what makes the equation quadratic.
A solution or root of the equation is a value of $x$ that makes the statement true.

Proposition — Null Factor Law

If the product of two or more factors is equal to zero, then at least one of the factors must be equal to zero. If $\textcolor{colordef}{A}\,\textcolor{colorprop}{B} = 0$ then $\textcolor{colordef}{A=0}$ or $\textcolor{colorprop}{B=0}$. Note: one or both of the factors can be zero. Conversely, if a factor is zero, the product is zero — so each candidate found this way is indeed a solution.

Method — Solving by Factorisation

The primary strategy for solving quadratic equations is to use the Null Factor Law.

Write the equation in standard form, $ax^2+bx+c=0$.
Factorise the quadratic expression completely.
Apply the Null Factor Law: set each factor equal to zero.
Solve each resulting linear equation.

Example

Solve $(x - 1)(x + 2) = 0$.

Answer

$$\begin{aligned}(x-1)(x+2) &= 0\\ x-1 = 0 &\quad \text{or} \quad x+2=0 &&\color{gray}\text{(Null factor law)} \\ x = 1 &\quad \text{or} \quad x = -2 &&\color{gray}\text{(Solve each equation)}\\ \end{aligned}$$

Skills to practice

Finding Solution Sets of Factored Equations
Solving Factored Equations

IV Solving Equations Graphically

IV.1 Reading Solutions on Two Curves

Method — Solving Graphically

To solve equations and inequalities using the graphs of $f$ and $g$:

For $f(x) = k$: Draw the horizontal line $y = k$. The solutions are the x-coordinates of its intersection points with the curve of $f$ — that is, the antecedents of $k$ under $f$.
For $f(x) = g(x)$: Locate the intersection points of the two curves. The solutions are the x-coordinates of these points.
For $f(x) < g(x)$: Identify the intervals of $x$ where the curve of $f$ is below the curve of $g$.
For $f(x) > g(x)$: Identify the intervals of $x$ where the curve of $f$ is above the curve of $g$.

For a non-strict inequality ($\leq$ or $\geq$), the $x$-coordinates of the intersection points are included in the solution set. Write the solutions as an interval or a union of intervals.

Example

The functions $f$ and $g$ are represented below. Solve $f(x) = g(x)$ and $f(x) \leq g(x)$.

Answer

Equation $f(x) = g(x)$: The curves intersect at $x = -1$ and $x = 2$. The solution set is $\boldsymbol{S = \{-1 ; 2\}}$.
Inequality $f(x) \leq g(x)$: The curve of $f$ is below or touching the curve of $g$ between the two intersection points. The solution set is the interval $\boldsymbol{S = [-1 ; 2]}$.

Skills to practice

Solving Equations on Two Curves
Solving Inequalities on Two Curves

V Solving Equations with Technology

V.1 Polynomial and Other Equations with Technology

Factorisation, completing the square and the quadratic formula give the roots of every quadratic equation exactly. Many equations are out of their reach: the equation $x^3 - 3x + 1 = 0$ has three real solutions, and none of these methods finds them. A graphing calculator, or graphing software, finds such solutions numerically. It draws graphs and locates the points where a graph crosses the $x$-axis or another graph, and its solvers return the roots of a polynomial equation and the solution of a system of linear equations. Its answers are in general decimal approximations, displayed with many digits.
Menu names differ from one model to another: each Method gives its steps in words, with a mock-up of what the screen displays. Unless a question says otherwise, a value that is not exact is given to 3 significant figures, counted from the first non-zero digit: $2.2573340 \approx 2.26$, $0.0283416 \approx 0.0283$ and $31.962 \approx 32.0$.

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Method — Solving a Polynomial Equation with the Polynomial Solver

The polynomial solver, in the equation menu of a graphing calculator, returns the real roots of an equation $ax^2 + bx + c = 0$ or $ax^3 + bx^2 + cx + d = 0$, and of higher degrees on most models.

Rearrange the equation so that one side is $0$, and expand the other side, its terms in decreasing powers of $x$.
Open the polynomial solver and choose the degree: the highest power of $x$.
Enter the coefficients in order, each with its sign, and $0$ for a missing power.
Solve and read the roots. An equation of degree $n$ has at most $n$ real roots; a model set to complex mode may also display roots containing $i$, which are not real solutions.
Report a root exactly when the screen shows an exact value, such as an integer that substitution confirms; otherwise give it to 3 significant figures.

The mock-up solves $x^3 - 3x + 1 = 0$: degree $3$, with $b = 0$ for the missing term in $x^2$. Its solutions are $x \approx -1.88$, $x \approx 0.347$ and $x \approx 1.53$.

Example

Use technology to solve:

$3x^2 = 5x + 4$;
$x^2(x + 2) = 5x + 3$.

Answer

One side is brought to $0$: $$\begin{aligned}[t] 3x^2 &= 5x + 4 \\ 3x^2 - 5x - 4 &= 0. &&\text{(subtract } 5x + 4 \text{ from both sides)} \end{aligned}$$ The polynomial solver, degree $2$, with $a = 3$, $b = -5$ and $c = -4$, displays $-0.5906673$ and $2.2573340$. The solutions are $x \approx -0.591$ and $x \approx 2.26$.
The equation is quadratic, so the quadratic formula also gives its exact roots, $x = \dfrac{5 \pm \sqrt{73}}{6}$: the screen shows their decimal approximations.
The left side is expanded, then one side is brought to $0$: $$\begin{aligned}[t] x^2(x + 2) &= 5x + 3 \\ x^3 + 2x^2 &= 5x + 3 &&\text{(expand)} \\ x^3 + 2x^2 - 5x - 3 &= 0. &&\text{(subtract } 5x + 3 \text{ from both sides)} \end{aligned}$$ The polynomial solver, degree $3$, with $a = 1$, $b = 2$, $c = -5$ and $d = -3$, displays $-3.2534180$, $-0.5199694$ and $1.7733874$. The solutions are $x \approx -3.25$, $x \approx -0.520$ and $x \approx 1.77$. In $-0.520$ the final zero is kept: it is the third significant figure.

Method — Solving an Equation Graphically with Technology

The solutions of an equation $f(x) = g(x)$ are the $x$-coordinates of the points where the graphs of $f$ and $g$ meet; the solutions of an equation $f(x) = 0$ are the $x$-intercepts of the graph of $f$.

Enter each side as a function: $Y_1 = f(x)$ and $Y_2 = g(x)$. For an equation $f(x) = 0$, enter $f(x)$ alone.
Choose a window that shows every point of intersection, or every $x$-intercept. A table of values of $f(x) - g(x)$ helps: a change of sign between two consecutive values of $x$ signals a solution between them, provided the curves have no break between these two values: this holds for polynomials, and for rational functions away from the values that make a denominator zero.
In the analysis tools of the graph menu, choose intersection (for $f(x) = 0$: zero), select the curves and give a first guess, or bounds, close to one point. Read the $x$-coordinate $X$ of the point displayed: the $y$-coordinate $Y$ is the common value of the two sides, not a solution.
Repeat for each point, then report the solutions: exactly when the screen shows an exact value that substitution confirms, otherwise to 3 significant figures.

The mock-up shows the equation $x^2 - 2 = \dfrac{1}{x}$: the two curves meet three times, and the point read is the one on the right.

Example

Use technology to solve $x^2 - 2 = \dfrac{1}{x}$, for $x \neq 0$.

Answer

The two sides are entered as $Y_1 = x^2 - 2$ and $Y_2 = \dfrac{1}{x}$, and graphed in the window $-3 \le x \le 3$, $-3 \le y \le 4$. The parabola and the hyperbola meet three times: twice, close together, for negative $x$, and once for positive $x$.

With bounds around each point, the intersection tool displays:

$X = -1$ and $Y = -1$: the value looks exact, and substitution confirms it, since $(-1)^2 - 2 = -1$ and $\dfrac{1}{-1} = -1$;
$X = -0.6180340$, so $x \approx -0.618$;
$X = 1.6180340$, so $x \approx 1.62$.

The solutions are $x = -1$, $x \approx -0.618$ and $x \approx 1.62$. Multiplying both sides by $x \neq 0$ gives $x^3 - 2x - 1 = 0$, and the polynomial solver returns the same three values. An equation of degree $3$ has at most three roots, so none is missing.

Method — Choosing Between Algebra and Technology

Read what the question asks. The words “exact” or “show that” call for an algebraic solution; “use technology”, a required accuracy such as “to 3 significant figures”, or measured data call for technology.
Recognise the type of equation. A linear or quadratic equation, or a product of factors equal to zero, is solved exactly by algebra: the quadratic formula always gives the exact roots. A polynomial equation of degree $3$ or more with no visible factor, or an equation mixing powers of $x$ with fractions such as $\dfrac{1}{x}$, is in general solved with technology.
Use technology to check an algebraic answer: the decimal values of the exact roots must match the values on the screen.
Report exact values exactly; otherwise give 3 significant figures unless told otherwise, and round only the final answer, never an intermediate value.
Answer the question asked: keep only the solutions that the domain or the context allows (a length is positive, a number of objects is a whole number), and give the units.

The mock-up shows the polynomial solver for $x^2 - 6x + 7 = 0$: it displays decimals, whereas the quadratic formula gives the exact roots $3 + \sqrt{2}$ and $3 - \sqrt{2}$.

Example

For each equation, decide whether technology is needed, then solve the equation.

$x^2 - 6x + 7 = 0$, giving exact solutions.
$x^3 + x = 5$.
$(x - 3)(2x + 1) = 0$.

Answer

The equation is quadratic and exact solutions are required: algebra, with the quadratic formula ($a = 1$, $b = -6$, $c = 7$). $$\begin{aligned}[t] \Delta &= (-6)^2 - 4 \times 1 \times 7 &&\text{(discriminant)} \\ &= 36 - 28 = 8. \end{aligned}$$ Since $\Delta > 0$, there are two roots: $$\begin{aligned}[t] x &= \frac{6 \pm \sqrt{8}}{2} &&\text{(quadratic formula)} \\ &= \frac{6 \pm 2\sqrt{2}}{2} &&(\sqrt{8} = 2\sqrt{2}) \\ &= 3 \pm \sqrt{2}. \end{aligned}$$ The solutions are $3 - \sqrt{2}$ and $3 + \sqrt{2}$. Technology only checks them: the polynomial solver displays $1.5857864$ and $4.4142136$, the decimal values of $3 - \sqrt{2}$ and $3 + \sqrt{2}$.
Rearranged, the equation reads $x^3 + x - 5 = 0$: degree $3$, with no visible factor, so neither factorisation nor the quadratic formula applies; technology gives its root numerically. The polynomial solver, degree $3$, with $a = 1$, $b = 0$, $c = 1$ and $d = -5$, displays a single real root, $1.5159802$. The solution is $x \approx 1.52$.
A product of factors equals zero: the null factor law gives $x - 3 = 0$ or $2x + 1 = 0$, that is $x = 3$ or $x = -\dfrac{1}{2}$, both exact. Technology is not needed.

Skills to practice

Interpreting a Calculator Screen
Solving Polynomial Equations with the Solver
Solving Equations Graphically with Technology

VI Going further

Skills to practice

Reasoning about Equations
Solving Problems
Quiz