Lesson 15: Solving Rational and Radical Equations

This lesson turns Lesson 14's expressions into equations — and adds radical equations, whose solutions carry a warning label. Both types share a theme: the standard solving move (clearing denominators, squaring) can manufacture *fake* solutions, so checking in the original equation is part of the method, not an optional extra.

Rational equations: clear the denominators

Multiply every term by the LCD of all the fractions, and the fractions vanish. Worked example 1. Solve x/3 + x/4 = 7. LCD 12: 4x + 3x = 84 → 7x = 84 → x = 12. Check: 4 + 3 = 7 ✓. Worked example 2 (proportions). A map scale gives 4/9 = x/27: cross-multiply (legitimate here — a single fraction equals a single fraction): 9x = 108 → x = 12. Proportions are just the two-fraction case of the LCD method. Worked example 3. Solve 5/x = 3/(x − 2). Cross-multiply: 5(x − 2) = 3x → 5x − 10 = 3x → 2x = 10 → x = 5. Check: 5/5 = 1 and 3/(5 − 2) = 1 ✓. Also note the exclusions x ≠ 0, 2 — our solution is neither, so it stands.

Work-rate problems: rates add

If one pipe fills a tank in 6 hours, its rate is 1/6 of a tank per hour; a second pipe's is 1/3. Working together, rates *add*: in t hours they fill t/6 + t/3 of the tank. Setting the job equal to 1 whole tank: t/6 + t/3 = 1 → multiply by 6: t + 2t = 6 → t = 2 hours. Sanity: the fast pipe alone needs 3 hours, so together must beat 3 — and 2 does. The structure (amount done) = (rate) × (time), summed over workers, handles every variant: two painters, two printers, a pipe filling while another drains (a negative rate).

Radical equations: isolate, square, check

Strategy: isolate the radical on one side, square both sides, solve, then check *every* candidate in the original. Worked example 4. √(x + 7) = 4 → x + 7 = 16 → x = 9. Check: √16 = 4 ✓. Worked example 5. √(3x − 2) − 1 = 3. Isolate first: √(3x − 2) = 4 → 3x − 2 = 16 → x = 6. Check: √16 − 1 = 4 − 1 = 3 ✓. Worked example 6 (the trap). √(2x − 1) = x − 2. Square both sides: 2x − 1 = (x − 2)² = x² − 4x + 4. Rearrange: 0 = x² − 6x + 5 = (x − 1)(x − 5), so candidates x = 1 and x = 5. Check x = 1: √1 = 1, but the right side is 1 − 2 = −1. 1 ≠ −1 — extraneous, reject. Check x = 5: √9 = 3, and 5 − 2 = 3 ✓. Only x = 5 is a solution. Why did a fake appear? Squaring erases sign: a² = b² holds when a = b *or* a = −b. The squared equation can't distinguish √(2x − 1) = x − 2 from √(2x − 1) = −(x − 2), and x = 1 solves the second of those. The check at the end isn't caution — it's the step that separates the two equations again.

Common mistakes

• Multiplying only the fraction terms by the LCD. In x/3 + x/4 = 7, the 7 must also be multiplied (by 12, giving 84). Every term means *every* term — the same discipline as clearing fractions in Lesson 1.
• Cross-multiplying across a sum. 5/x = 3/(x − 2) + 1 cannot be cross-multiplied as written; cross products are valid for one fraction = one fraction only. Combine the right side first, or clear with the LCD x(x − 2).
• Accepting every squared candidate. As Worked example 6 shows, an unchecked radical-equation answer can be exactly half wrong. Budget the check into your working time.
• Squaring before isolating. Squaring √(3x − 2) − 1 = 3 directly produces (√(3x − 2) − 1)² — a mess containing a radical still. Isolate the radical first so squaring eliminates it in one move.

Exam Experience

Question 1. Solve 6/x = 2 (x ≠ 0). A. x = 12 B. x = 3 C. x = 1/3 D. x = −3 Answer: B. Clear the fraction: 6 = 2x, so x = 3. Check: 6/3 = 2 ✓. Option A multiplies instead of dividing — the classic "move x up" slip.

Question 2. Solve √(x − 3) = 5. A. x = 8 B. x = 22 C. x = 28 D. x = 2 Answer: C. Square: x − 3 = 25, so x = 28. Check: √25 = 5 ✓. Option A adds 3 to 5 without squaring; B squares 5 but subtracts.

Question 3. Solve (x + 1)/2 = (x − 1)/3. A. x = 5 B. x = −5 C. x = 1 D. x = −1 Answer: B. Cross-multiply: 3(x + 1) = 2(x − 1) → 3x + 3 = 2x − 2 → x = −5. Check: (−4)/2 = −2 and (−6)/3 = −2 ✓. A negative answer is not a sign of error — the check confirms it.

Question 4 (constructed response). Ana can paint a room in 4 hours; Ben can paint the same room in 6 hours. (a) Write each painter's rate as a fraction of the room per hour. (b) Working together, they paint for t hours. Write and solve an equation for the time needed to paint the whole room. (c) State the time in hours and minutes, and explain how you know it must be less than 4 hours. (d) Ana paints alone for the first hour, then Ben joins her. How much longer until the room is finished?

Scoring guide (4 points).

• 1 point: rates 1/4 and 1/6 of a room per hour.
• 1 point: the equation t/4 + t/6 = 1, cleared (×12) to 3t + 2t = 12.
• 1 point: 5t = 12 → t = 12/5 = 2.4 hours = 2 hours 24 minutes, with the reasoning that together they must beat Ana's solo 4 hours (Ben only helps) — 2.4 < 4 ✓.
• 1 point: after 1 hour alone Ana has done 1/4, leaving 3/4; combined rate 5/12 per hour, so extra time = (3/4) ÷ (5/12) = 9/5 = 1.8 hours (1 hour 48 minutes) more.

Deeper still: drains, and radicals on both sides

A rate that subtracts. A tank's inlet pipe fills it in 4 hours; an open drain empties a full tank in 12 hours. With both open, the net rate is 1/4 − 1/12 = 3/12 − 1/12 = 2/12 = 1/6 of a tank per hour, so the tank fills in 6 hours — slower than the inlet alone (4 hours), as it must be, but it does fill because the inlet's rate is bigger. If the drain were faster than the inlet, the equation's "solution" would come out negative — the algebra reporting, correctly, that the job never completes. Negative or impossible answers in rate problems are information, not arithmetic errors: they describe a race being lost. Radicals on both sides. Solve √(x + 3) = √(2x − 1). Squaring removes both radicals at once: x + 3 = 2x − 1 → x = 4. Check: √7 = √7 ✓ — and this one needs no sign vigilance beyond the usual, since both sides are principal roots (automatically nonnegative). Contrast √(5x + 6) = x: squaring gives 5x + 6 = x², so x² − 5x − 6 = 0 and (x − 6)(x + 1) = 0 with candidates 6 and −1. Only x = 6 survives the check (√36 = 6 ✓); x = −1 gives √1 = 1 ≠ −1. Same course, two adjacent problems: one where squaring is safe, one where it smuggles in an impostor. The distinguishing question is always "can the non-radical side be negative for my candidate?" — if yes, check with extra care.

One more idea: when the cleared solution is the excluded one

Clearing denominators has a failure mode all its own. Solve x/(x − 3) = 3/(x − 3). Cross-multiplying gives x = 3 — but x = 3 is precisely the excluded value, the one that makes both denominators zero. The equation has no solution: the only candidate the algebra offers is the one value the expressions forbid. Compare Worked example 6's extraneous radical root: in both families, mechanical solving can produce a candidate that the original equation rejects, which is why the check-and-exclude step is structural, not cautious. An equation whose every candidate is excluded or extraneous is not "unsolved" — "no solution" *is* the solution. A final rate variant with clean numbers: taps that fill a pool in 5 hours and 10 hours run together at 1/5 + 1/10 = 3/10 of a pool per hour, finishing in 10/3 hours = 3 hours 20 minutes — between the two solo times but closer to the faster tap's, exactly where a combined rate must land. That "between, and nearer the stronger" instinct is the rate-problem version of the mixture lean check: build the expectation before the arithmetic, then let the arithmetic confirm it.

One discipline unites both families of this lesson: estimate before you solve. In x/3 + x/4 = 7, the left side is about 0.58x, so x should land near 12 — and it does, exactly. In √(x + 7) = 4, the answer must make x + 7 a perfect square whose root is 4, i.e. 16, so any candidate far from 9 is dead on arrival. Estimation doesn't replace the method; it patrols it, catching the dropped term, the flipped divisor and the mis-squared binomial at the moment they happen instead of after a failed check. Professional solvers estimate first and are therefore rarely surprised — cultivate the same reflex and the check at the end becomes a formality you pass, not a trap you fear.