Pre-Calc Diagnostic: Solve the Equation — 7 Techniques (Stewart Problem 8)

Direct answer. Stewart's Problem 8 is a diagnostic on technique selection, not just algebra. Seven equations, seven different shapes, seven different first moves: (a) x + 5 = 14 − 12x is linear (x = 6); (b) 2xx+1 = 2x−1x is rational, cross-multiply and check the domain (x = 1); (c) x² − x − 12 = 0 factors (x = 4, −3); (d) 2x² + 4x + 1 = 0 needs the quadratic formula (x = −1 ± 22); (e) x⁴ − 3x² + 2 = 0 becomes a quadratic after the substitution y = x² (x = ±1, ±2); (f) 3|x − 4| = 10 splits into two cases (x = 223, 23); and (g) 2x(4 − x)^−1/2 − 34 − x = 0 takes the substitution t = 4 − x (x = 125). The lesson isn't any one technique — it's reading the equation's shape and choosing the right tool in under thirty seconds.

(a) Linear: x + 5 = 14 − 12x

Move the x-terms to one side, constants to the other:

x + 12x = 14 − 5

32x = 9 ⟹ x = 9 · 23 = 6

Linear equations are diagnostic floor. A pre-calc student who hesitates on part (a) is missing fluency that should have been automatic since algebra 1. We treat fluency on (a)-style problems as a prerequisite — if it isn't there, we step back and build it before moving on.

(b) Rational with a domain check: 2xx+1 = 2x − 1x

Before doing any algebra, list the excluded values. The denominators are x + 1 and x, so x ≠ −1 and x ≠ 0. Now cross-multiply:

2x · x = (2x − 1)(x + 1)

2x² = 2x² + x − 1

The 2x² cancels off both sides — which is a clue this was set up to test discipline, not algebra:

0 = x − 1 ⟹ x = 1

Check against the excluded values: x = 1 is not 0 and not −1, so it's valid. If the cross-multiply had produced x = 0 or x = −1, those would be extraneous — algebraically valid but invalid in the original equation. The domain check is mandatory, not optional.

(c) Factorable quadratic: x² − x − 12 = 0

Always try factoring first. Two integers that multiply to −12 and add to −1: −4 and +3.

(x − 4)(x + 3) = 0 ⟹ x = 4, −3

The whole skill is the 30-second factoring scan. If you can't spot the pair quickly, move on to the quadratic formula. Don't waste time hunting.

(d) Quadratic formula: 2x² + 4x + 1 = 0

Try factoring: pairs of 2 · 1 = 2 that add to 4? There aren't any clean integer pairs. Switch to the formula. a = 2, b = 4, c = 1:

x = −4 ± 16 − 84 = −4 ± 84 = −4 ± 224

Simplify by dividing every term by 2:

x = −1 ± 22

Discriminant was 16 − 8 = 8 > 0, so two real roots. 8 = 22 — students who leave it as 8 aren't wrong, but they aren't simplified.

(e) Quartic by substitution: x⁴ − 3x² + 2 = 0

Only even powers of x, so let y = x². The equation becomes a clean quadratic:

y² − 3y + 2 = 0 ⟹ (y − 1)(y − 2) = 0

So y = 1 or y = 2. Back-substitute y = x²:

x² = 1 ⟹ x = ±1 and x² = 2 ⟹ x = ±2

x = ±1, ±2

Four real solutions, exactly as expected from a quartic. The substitution turns a "scary" 4th-degree equation into something a student already knows how to solve. This technique recurs in calculus integration: anytime you see the same expression squared and to the first power, substitute.

(f) Absolute value: 3|x − 4| = 10

Isolate the absolute value first:

|x − 4| = 103

Now split into two cases — the expression inside is either positive or negative its target:

x − 4 = 103 or x − 4 = −103

x = 103 + 4 = 223 or x = −103 + 4 = 23

x = 223, 23

The student error to watch for: forgetting to isolate the absolute value first. Splitting 3|x − 4| = 10 into 3(x − 4) = ±10 works algebraically, but isolating the absolute value first keeps the logic crisper. Always isolate before splitting.

(g) Negative fractional exponent — the hardest one: 2x(4 − x)^−1/2 − 34 − x = 0

This is where students freeze. The expressions (4 − x)^−1/2 and 4 − x are both functions of the same thing, so substitute. Let t = 4 − x (so t > 0). Then:

x = 4 − t² and (4 − x)^−1/2 = 1t

Plug into the original:

2(4 − t²) · 1t − 3t = 0

8 − 2t²t − 3t = 0 ⟹ 8t − 2t − 3t = 0 ⟹ 8t − 5t = 0

Multiply through by t (we know t > 0):

8 − 5t² = 0 ⟹ t² = 85

Now back-substitute. t² = 4 − x, so:

4 − x = 85 ⟹ x = 4 − 85 = 20 − 85 = 125

Domain check: at x = 125 = 2.4, 4 − x = 1.6 > 0, so the square root is defined. Valid. The whole equation collapses to a single solution because of how the substitution clears the negative exponent.

The one big-picture lesson: pattern recognition is the entire skill

Every problem on Stewart's diagnostic Problem 8 has a different first move, and that's the point. AP Calc AB students who got their pre-calc grades by memorizing one or two techniques and brute-forcing every problem hit a wall the first time they see an integral that requires substitution. AP Calc BC students with the same gap hit a wall on improper integrals. Pre-calc is the course where students should be cataloging shapes of equations and the techniques that match each shape — so that by Calc I the matching is automatic.

The shapes worth memorizing from Problem 8:

• Linear (one x at most once on each side): isolate.
• Rational (variable in a denominator): list excluded values, cross-multiply, check at the end.
• Quadratic, factorable: 30-second scan for the factor pair, then factor.
• Quadratic, not factorable: quadratic formula.
• Only even powers of x: substitute y = x².
• Absolute value: isolate, then split into two cases.
• Same expression appearing in multiple forms (square root and negative half-power, or squared and linear): substitute a single letter for that expression.

These seven patterns cover most of what's tested in AP Calc AB and BC equation-solving. A pre-calc student who has them automatic walks into Calc with a real advantage.

How SOMATH teaches the equation-solving toolkit on the Upper West Side

We start every new pre-calc student with a diagnostic identical in structure to Stewart's. Five to seven equations of different shapes, no calculator, no help. Then we sit with the student and ask — before they pick up a pencil — what shape each one is and what the first move would be. The algebra is downstream of the recognition; the recognition is the whole game. From our classroom on 226 W 79th St we run small groups of 3–5 students, every lesson on whiteboards, every step shown out loud. Every new student gets a free 30-minute evaluation and a written diagnostic within 48 hours, even if they don't enroll.

FAQ

How do I know which solving technique to pick?

Read the equation's shape first. Linear, rational, factorable quadratic, quadratic formula, even-power substitution, absolute value, or matched-expression substitution. The shape tells you the first move.

Why check the domain on rational equations?

Cross-multiplying can produce solutions that make the original equation undefined. Listing excluded values up front lets you discard them at the end.

Factor or quadratic formula?

Try factoring for 30 seconds. If no clean factor pair appears, switch to the formula. Don't lose 5 minutes hunting for a factorization that doesn't exist.

What's the quartic substitution trick?

If only even powers of x appear, let y = x², solve the resulting quadratic in y, then back-substitute and take square roots. Twice as many x answers as y answers.

Why is part (g) the hardest equation?

It mixes a negative half-power and a square root of the same expression. The substitution t = 4 − x collapses both into the same letter and turns the whole thing into a clean equation in t.

Related posts

• Pre-Calc Diagnostic: Rewrite by Completing the Square (Stewart Problem 7)
• Pre-Calc Diagnostic: Rationalize the Expression (Stewart Problem 6)
• Pre-Calc Diagnostic: Simplify Rational Expressions (Stewart Problem 5)