4. Equations, inequalities and graphs

Learning objectives

4.1Solve modulus equations• Solve equations of the type |ax + b| = c, |ax + b| = cx + d, |ax + b| = |cx + d| and |ax^2 + bx + c| = d using algebraic or graphical methods.• For graphical solutions, draw an accurate graph.• For algebraic methods, any valid method is acceptable.4.2Solve modulus inequalities• Solve graphically or algebraically inequalities involving k|ax + b| and |ax^2 + bx + c|, including comparisons with constants, linear expressions and other modulus expressions.• Use k > 0 and c >= 0 or c > 0 where specified.• For graphical solutions, draw an accurate graph; for algebraic methods, any valid method is acceptable.4.3Use substitution for related equations• Use substitution to form and solve a quadratic equation in order to solve a related equation.• Identify the appropriate substitution in equations involving powers, logarithms or exponentials.4.4Sketch cubic polynomials and moduli• Sketch graphs of cubic polynomials and their moduli when given as a product of three linear factors.• Clearly label points of intersection with the coordinate axes.4.5Solve cubic inequalities graphically• Solve graphically cubic inequalities of the form f(x) >= d, f(x) > d, f(x) <= d and f(x) < d where f(x) is a product of three linear factors and d is a constant.

An equation $|A(x)|=c$ with $c\geq0$ means $A(x)=c$ or $A(x)=-c$ . Solve both branches and check every result in the original equation.

Form	Algebraic branches	Extra condition
$\|A\|=c$	$A=c$ or $A=-c$	no solution if $c<0$
$\|A\|=B$	$A=B$ or $A=-B$	retained roots must have $B\geq0$
$\|A\|=\|B\|$	$A=B$ or $A=-B$	verify both branches
$\|Q(x)\|=d$	$Q(x)=d$ or $Q(x)=-d$	solve both resulting quadratics

$5|2x-1|+8=23\Rightarrow|2x-1|=3\Rightarrow2x-1=\pm3\Rightarrow x=2\text{ or }x=-1$

Graphically, solutions are the $x$ -coordinates where $y=|A(x)|$ meets the other side. An accurate graph must show all intersections, including tangencies that produce one distinct solution.

Do not write $|A|=c\Rightarrow A=c$ only. When the other side contains $x$ , solving both algebraic branches is not enough: reject any value for which that side is negative.

A modulus measures distance from zero. For $c>0$ , $|A|<c$ places $A$ between $-c$ and $c$ , while $|A|>c$ places it outside those limits; include endpoints for $\leq$ or $\geq$ .

$|A|<c\Longleftrightarrow-c<A<c,\qquad |A|>c\Longleftrightarrow A<-c\text{ or }A>c$

For $|A(x)|$ compared with a linear expression, use piecewise branches or graph $y=|A(x)|$ and the line, then choose where one graph is above or below the other. For $|A|$ compared with $|B|$ , both sides are non-negative, so squaring preserves the comparison.

$4|x-1|\leq|3x+2|\Rightarrow16(x-1)^2\leq(3x+2)^2\Rightarrow7x^2-44x+12\leq0$

The critical values are $2/7$ and $6$ . The upward-opening quadratic is non-positive between them, so $2/7\leq x\leq6$ .

Do not square an inequality against an expression whose sign is unknown: squaring can create false values. On a graph, ‘greater than’ means above, and strict inequalities exclude intersection points.

When an equation repeats one expression and its square, replace that expression by a single variable. Solve the quadratic in the substitute, then return to the original variable and apply its domain conditions.

Repeated expressions	Useful substitute	Condition
$a^{2x}$ and $a^x$	$u=a^x$	$u>0$ for $a>0$
$e^{2x}$ and $e^{-2x}$	multiply to clear the negative power, then $u=e^{2x}$	$u>0$
reciprocal logarithms	$u=\log_a x$	valid log base and argument
$x^{1/3}$ and $x^{1/6}$	$u=x^{1/6}$	respect the real-domain restriction

$3^{2x+1}+8(3^x)-3=0,\quad u=3^x\Rightarrow3u^2+8u-3=(3u-1)(u+3)=0$

Because $u=3^x>0$ , reject $u=-3$ and keep $u=1/3$ . Therefore $3^x=3^{-1}$ and $x=-1$ .

Solving the quadratic is only the middle step. Always translate each admissible $u$ back, and reject roots that violate positivity, logarithm or fractional-power conditions.

For $f(x)=a(x-r_1)(x-r_2)(x-r_3)$ , the factors give the $x$ -intercepts and $f(0)$ gives the $y$ -intercept. These points, root multiplicities and the sign of $a$ determine the sketch.

Feature	Effect on the graph
simple root	crosses the $x$ -axis
repeated root	touches and turns at the $x$ -axis
$a>0$	falls to the left and rises to the right
$a<0$	rises to the left and falls to the right

$f(x)=3(x-3)(x-1)(x+2)\Rightarrow x=-2,1,3,\quad f(0)=18$

To sketch $y=|f(x)|$ , keep every section on or above the $x$ -axis and reflect each negative section upward. The intercepts stay fixed; a simple crossing becomes a sharp turn, while a section already above the axis is unchanged.

Do not reflect the whole cubic or change the $x$ -coordinates. Absolute value acts on the output only, and all axis intersections must be labelled.

To solve $f(x)\square d$ graphically, compare the cubic $y=f(x)$ with the horizontal line $y=d$ . Their intersection $x$ -coordinates divide the graph into intervals with a consistent above/below relationship.

Draw or use the accurate graph, mark every intersection with $y=d$ , then read where the cubic is above the line for $>$ or $\geq$ and below it for $<$ or $\leq$ . Include intersection values only for non-strict inequalities.

$f(x)=-\frac15(x+2)(2x-1)(x+5)\geq0$

The zero-level intersections are $x=-5,-2,1/2$ . Reading the sign of the cubic gives $x\leq-5$ or $-2\leq x\leq1/2$ .

Do not assume the sign switches at every contact: at a repeated root the graph touches the level and turns back. For $d\ne0$ , use the actual intersections with $y=d$ , not the polynomial's roots.

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4. Equations, inequalities and graphs

Learning objectives

Solve modulus equations by controlling both branches

Solve modulus inequalities with valid sign logic

Turn a related equation into a quadratic

Sketch a factored cubic and its modulus

Read a cubic inequality from the graph

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