Pre-Calculus · Radicals & Rationalization
Pre-Calc Diagnostic: Rationalize the Expression (Stewart Problem 6)
Two rationalization problems from Stewart's Pre-Calculus Diagnostic Test — multiply by the conjugate to clear a radical denominator on √10/(√5 − 2), and rationalize the NUMERATOR on (√(4 + h) − 2)/h, the classic AP Calculus derivative-from-the-definition setup that every pre-calc student should master before AP Calc AB. Taught the way we teach it at SOMATH on the Upper West Side.
Direct answer: Problem 6 of Stewart's Pre-Calculus Diagnostic asks you to rationalize and simplify two expressions. The answers are (a) 50 + 210 = 52 + 210 and (b) 14 + h + 2. Both use the same core move — multiply by the conjugate — but they're rationalizing different parts of the expression. Part (a) rationalizes the denominator (the textbook habit). Part (b) rationalizes the numerator, which is the move that makes AP Calculus derivative limits solvable. Below is the full walkthrough we use at SOMATH on the Upper West Side.
The one move behind both parts: multiply by the conjugate
When you see a two-term expression that contains a radical — anything like 5 − 2 or 4 + h + 2 — its conjugate is the same expression with the middle sign flipped. The conjugate of 5 − 2 is 5 + 2. The conjugate of a − b is a + b.
When you multiply a two-term expression by its conjugate, the cross terms cancel via the difference-of-squares pattern:
(a − b)(a + b) = a2 − b2
And since squaring a radical removes it ((n)2 = n), this is exactly how you clear radicals from a denominator (or numerator). To keep the expression's value unchanged, you multiply top and bottom by the same conjugate — i.e., you multiply by 1 in disguise.
(a) Rationalize the denominator: 105 − 2
The denominator is the two-term radical 5 − 2. Multiply top and bottom by its conjugate 5 + 2:
105 − 2 · 5 + 25 + 2 = 10 (5 + 2)(5)2 − 22
The denominator uses the difference-of-squares pattern, with a = 5 and b = 2. The radical disappears because (5)2 = 5:
= 10 · 5 + 2105 − 4 = 50 + 2101 = 50 + 210
Now simplify 50. Pull out the largest perfect square: 50 = 25 · 2, so 50 = 25 · 2 = 52:
= 52 + 210
Two side-notes that catch students. First: when you multiplied 10 · 5 you got 10 · 5 = 50, not 15 — radicals multiply under one radical sign, they don't add. Second: 2 and 10 are not like terms, so they can't be combined further. The final answer has two unrelated radicals — that's correct.
(b) Rationalize the numerator: 4 + h − 2h
This is the problem that pays you back in AP Calculus. The denominator is already rational — it's just h. The radical is in the numerator. We want to clear that radical so we can simplify the expression further (specifically, so we can divide out the h). Multiply top and bottom by the conjugate of the numerator, 4 + h + 2:
4 + h − 2h · 4 + h + 24 + h + 2 = (4 + h)2 − 22h (4 + h + 2)
The numerator is now difference of squares with a = 4 + h and b = 2. Squaring the radical gives back 4 + h:
= (4 + h) − 4h (4 + h + 2) = hh (4 + h + 2)
The h in the numerator and the h in the denominator cancel:
= 14 + h + 2
Done. The radical is now in the denominator (we didn't bother to rationalize it again — once is enough for this problem), the h is gone from the denominator, and the expression is simplified.
Why part (b) is secretly a calculus problem
The expression [4 + h − 2] / h is the difference quotient for f(x) = x at x = 4:
f(4 + h) − f(4)h = 4 + h − 4h = 4 + h − 2h
In AP Calculus AB you take the limit as h → 0 to get the derivative f'(4). If you try to plug in h = 0 right away you get 0/0, which is undefined. But after rationalizing — exactly the move we just did — the expression becomes 1 / (4 + h + 2), and now h = 0 plugs in cleanly:
f'(4) = 14 + 2 = 12 + 2 = 14
This matches the power-rule answer f'(x) = 1/(2x) evaluated at x = 4: 1/(2 · 2) = 1/4. Stewart puts the rationalize-the-numerator problem on the pre-calc diagnostic specifically because students who can't do this algebra cleanly will struggle with every limit-definition derivative problem in the first three weeks of AP Calc.
The decision rule: rationalize numerator or denominator?
Pre-calc tradition: rationalize the denominator. Standardized-test problems usually expect the answer in that form, so part (a) follows the textbook convention. Calculus reality: rationalize whichever side has the radical that's getting in your way. In part (b), the numerator's 4 + h − 2 was the obstacle — once it became (4 + h) − 4 = h, the whole expression simplified.
The mental check: which side, if I cleared the radical, would let me cancel or simplify the most? Rationalize that side. Both moves use the same conjugate trick — the only choice is which one.
How SOMATH teaches rationalization on the Upper West Side
At SOMATH (226 W 79th St, UWS), we teach rationalization with a two-line checklist: (1) identify the two-term radical, (2) multiply top and bottom by its conjugate. Then we drill the difference-of-squares pattern (a − b)(a + b) = a2 − b2 until students can do it in their head. Once those two skills are automatic, both parts of Problem 6 take under a minute each.
We pair this lesson with a 10-problem mixed practice — half pre-calc style (rationalize the denominator), half AP Calc style (rationalize the numerator inside a limit). Students who can do all ten in 15 minutes are ready for the first derivative chapter of Calculus AB.
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FAQ
What does it mean to rationalize an expression?
Rationalizing means rewriting an expression so that a radical (square root, cube root, etc.) is no longer in the denominator (or, sometimes, no longer in the numerator). It doesn't change the value of the expression — it changes the form. The two most common moves are multiplying by the conjugate of a two-term denominator like 5 − 2, and multiplying by the radical itself when the denominator is a single n.
What is a conjugate?
A conjugate is the same two-term expression with the middle sign flipped. The conjugate of 5 − 2 is 5 + 2. The conjugate of a + b is a − b. When you multiply a binomial by its conjugate you get the difference of squares (a − b)(a + b) = a2 − b2, which is why the move works to clear radicals: any radical squared loses its radical sign.
Why do textbooks tell you to rationalize the denominator? It looks the same either way.
Historically, before calculators, rationalized denominators made hand computation easier — dividing by an integer is simpler than dividing by an irrational number. Today the bigger reason is form: standardized tests and textbooks expect simplified answers in a specific form, and most still require a rational denominator. AP Calculus actually flips this — there you often need to rationalize the NUMERATOR (as in part b of this problem) to make a limit solvable.
Why does (4 + h − 2)/h look familiar from calculus?
Because it IS calculus — it's the derivative of f(x) = x at x = 4 using the limit definition. f'(4) = limh→0 [f(4+h) − f(4)] / h = limh→0 [4+h − 2] / h. If you try to plug in h = 0 directly you get 0/0. Rationalizing the numerator turns it into limh→0 1 / (4+h + 2) = 1 / (4 + 2) = 1/4, which matches f'(4) = 1/(24) = 1/4.
When do you multiply by the conjugate vs just by the radical?
If the denominator (or numerator) is a single-term radical like n, multiply top and bottom by n — that turns n into n. If it's a two-term expression that contains a radical, like 5 − 2 or 4+h + 2, multiply by the conjugate so the cross terms cancel by the difference-of-squares pattern. The conjugate move only works on two-term binomials.