Question: Solve $|x^2 - 16x + 39| = 24$.

Step 1. Split into two cases:

• Case 1: $x^2 - 16x + 39 = 24 \Rightarrow x^2 - 16x + 15 = 0$
• Case 2: $x^2 - 16x + 39 = -24 \Rightarrow x^2 - 16x + 63 = 0$

Step 2. Solve Case 1: $x^2 - 16x + 15 = (x-1)(x-15) = 0 \Rightarrow x = 1, 15$.

Step 3. Solve Case 2: discriminant $= 256 - 252 = 4$. $x = \dfrac{16 \pm 2}{2} = 7, 9$.

Answer: $x = 1, 7, 9, 15$.

Tip: $|f(x)| = k$ (with $k > 0$) always splits into TWO cases:

• $f(x) = k$
• $f(x) = -k$

Solve each and combine all valid solutions. If $k < 0$: no solutions (modulus is non-negative). If $k = 0$: $f(x) = 0$.

Question: Solve $|2x - 1| = |x + 3|$.

Step 1. Two approaches: case-split OR squaring. Use squaring (cleaner here):

$$(2x - 1)^2 = (x + 3)^2$$ $$4x^2 - 4x + 1 = x^2 + 6x + 9$$ $$3x^2 - 10x - 8 = 0$$

Step 2. Factorise: discriminant $= 100 + 96 = 196 = 14^2$.

$$x = \dfrac{10 \pm 14}{6} = 4 \text{ or } -\dfrac{2}{3}$$

Step 3. Verify (squaring can introduce extraneous roots — though for $|a| = |b|$ squaring is reversible):

• $x = 4$: $|7| = |7|$ $\checkmark$
• $x = -\tfrac{2}{3}$: $|-7/3| = |7/3|$ $\checkmark$

Answer: $x = 4$ or $x = -\dfrac{2}{3}$.

Tip: $|a| = |b| \Leftrightarrow a^2 = b^2 \Leftrightarrow a = \pm b$. Squaring is always reversible here (unlike $|a| =$ non-modulus, where it's not).

Question: Solve $|3x - 5| \leq 7$.

Step 1. Recognise: $|f(x)| \leq k$ is equivalent to $-k \leq f(x) \leq k$ (provided $k \geq 0$).

$$-7 \leq 3x - 5 \leq 7$$

Step 2. Add $5$ throughout:

$$-2 \leq 3x \leq 12$$

Step 3. Divide by $3$:

$$-\dfrac{2}{3} \leq x \leq 4$$

Answer: $x \in \left[-\dfrac{2}{3}, 4\right]$.

Tip: Modulus inequalities:

• $|f(x)| < k$: $-k < f(x) < k$ (inside)
• $|f(x)| > k$: $f(x) < -k$ or $f(x) > k$ (outside, two pieces)
• $|f(x)| \leq |g(x)|$: square both sides to get $f^2 \leq g^2$ (since both sides non-negative).

Question: Sketch $y = |x^2 - 4x + 3|$.

Step 1. First sketch $y = x^2 - 4x + 3 = (x-1)(x-3)$:

• Roots at $x = 1, 3$.
• Vertex at $x = 2$: $y = 4 - 8 + 3 = -1$. So min at $(2, -1)$.
• $y$-intercept: $y = 3$.

The parabola dips BELOW the $x$-axis for $1 < x < 3$.

Step 2. Apply modulus: reflect ANY portion below the $x$-axis to ABOVE. The min $(2, -1)$ becomes max $(2, 1)$.

Resulting shape:

• For $x \leq 1$ or $x \geq 3$: same as the parabola (above axis).
• For $1 \leq x \leq 3$: reflected — opens DOWNWARDS, peak at $(2, 1)$.
• Corners (V-shapes) appear at the roots $(1, 0)$ and $(3, 0)$.

Key features: $y$-intercept $(0, 3)$; corners at $(1, 0)$ and $(3, 0)$; max at $(2, 1)$.

Tip: Modulus graph procedure:

1. Sketch $y = f(x)$ ignoring modulus.
2. Flip any part below $x$-axis to above (mirror in $x$-axis).
3. Mark "corners" where $f(x) = 0$ — these are V-shaped kinks.

Question: State the values of $k$ for which $|x^2 - 16x + 39| = k$ has exactly $2$ solutions.

Step 1. Sketch $y = |x^2 - 16x + 39|$. The parabola $y = x^2 - 16x + 39$ has roots at $x = 3, 13$ (use quadratic formula: $x = \dfrac{16 \pm \sqrt{256-156}}{2} = \dfrac{16 \pm 10}{2}$). Vertex at $x = 8$: $y = 64 - 128 + 39 = -25$.

After reflection: $|x^2 - 16x + 39|$ has corners at $(3, 0)$ and $(13, 0)$, and a peak at $(8, 25)$.

Step 2. Horizontal line $y = k$ intersects the modulus graph at:

• $0$ points if $k < 0$ (modulus is non-negative).
• $2$ points if $k = 0$ (only at the two corners).
• $4$ points if $0 < k < 25$ (line crosses both branches twice).
• $3$ points if $k = 25$ (line tangent to peak; meets at $1$ point on top, $2$ on outer branches).
• $2$ points if $k > 25$ (only outer branches).

Step 3. Exactly $2$ solutions: $k = 0$ or $k > 25$.

Answer: $k = 0$ or $k > 25$.

Tip: Count intersections of $y = k$ (horizontal) with the modulus graph. Critical $k$-values are the peak of the reflected portion (max of $|f|$ over the reflected range) and $0$.

Question: The graph $y = |f(x)|$ where $f(x) = x^2 - 4x + 3$ meets the line $y = 2$ at $P$, $Q$, $R$ (in increasing $x$-order). Find the coordinates.

Step 1. Solve $|x^2 - 4x + 3| = 2$. Two cases:

Case A: $x^2 - 4x + 3 = 2 \Rightarrow x^2 - 4x + 1 = 0$. Discriminant $= 16 - 4 = 12$. $x = \dfrac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3}$. So $x \approx 0.27, 3.73$.

Case B: $x^2 - 4x + 3 = -2 \Rightarrow x^2 - 4x + 5 = 0$. Discriminant $= 16 - 20 = -4 < 0$. No real solutions.

Step 2. But we also need $x$-values where the reflected branch ($1 \leq x \leq 3$, where original is negative) gives $y = 2$. The reflected branch satisfies $-(x^2 - 4x + 3) = 2 \Rightarrow x^2 - 4x + 5 = 0$ — same as Case B above. No real solutions.

So only TWO intersection points, not three. Re-state — perhaps the question wants $y = 0.5$ or similar.

Try $y = 0.5$: Case A: $x^2 - 4x + 2.5 = 0$, $x = 2 \pm \sqrt{1.5}$. Case B: $-(x^2 - 4x + 3) = 0.5 \Rightarrow x^2 - 4x + 2.5 = 0$ — same. Two solutions only.

Try $y = 0.9$: Same issue.

For $y = k$ to give $3$ solutions, $k$ must equal the peak height of the reflected portion, which is the absolute value of the vertex of $f$: $|-1| = 1$. At $k = 1$: Case A gives $x^2 - 4x + 2 = 0 \Rightarrow x = 2 \pm \sqrt{2}$; Case B gives $x^2 - 4x + 4 = 0 \Rightarrow (x-2)^2 = 0 \Rightarrow x = 2$. THREE solutions.

Answer: $P = (2 - \sqrt{2}, 1)$, $Q = (2, 1)$, $R = (2 + \sqrt{2}, 1)$.

Tip: Multistep modulus problems: identify the critical $k$-value (peak of reflected portion). Three-solution scenarios always have $k$ = max of $|f|$ on the reflected interval.