Chapter 1.5 — Exercise 1.4 — Operations on Real Numbers

Four Expansion Identities for Surds

Exercise 1.4 leans on algebraic identities adapted for square roots, useful for expanding and later rationalising expressions:

• √a · √b = √(ab) — the product rule for non-negative numbers.
• (√a + √b)(√a − √b) = a − b — the difference-of-squares identity, and the entire basis of rationalisation.
• (√a + √b)² = a + 2√(ab) + b — squaring a binomial surd.
• (√a + √b)(√c + √d) = √(ac) + √(ad) + √(bc) + √(bd) — term-by-term expansion for two surd binomials.

Two Worked Expansions From Question 1

Expanding (√5 + √7)(2 + √5) term by term: √5·2 + √5·√5 + √7·2 + √7·√5 = 2√5 + 5 + 2√7 + √35. None of these four terms combine further, so the simplified form is 5 + 2√5 + 2√7 + √35.

A second part multiplies (5 + √5)(5 − √5). This fits the difference-of-squares pattern directly with a = 5 and b = 5 (note: the rational 5, not √5): a² − b = 25 − 5 = 20.

Two further parts of the same question apply the other identities directly: (√3 + √7)² = 3 + 2√21 + 7 = 10 + 2√21, and (√11 + √7)(√11 − √7) = 11 − 7 = 4.

Across all four parts, the pattern worth carrying forward is that expanding a surd expression never removes the square roots by itself — only the specific difference-of-squares pairing (√a+√b)(√a−√b) does that, because the cross terms it produces cancel exactly. Any other pairing, like (√5+√7)(2+√5), leaves the square roots sitting in the answer, just combined differently.

Rationalising Factors and Why Conjugates Work

A rationalising factor (R.F.) of an irrational number is a second irrational number whose product with the first is rational. √3 and √27 are R.F.s of each other, since √3 × √27 = √81 = 9. An irrational number's R.F. is never unique — √3 × √12 = √36 = 6 works too — so the simplest available option is the practical choice.

For a binomial surd like √a + √b, the R.F. is always its conjugate, √a − √b, because their product is the rational value a − b. This single fact drives every denominator-rationalising problem in this exercise.

Four Denominators, Rationalised

Question 5 applies the conjugate method directly, multiplying top and bottom by the denominator's R.F.:

• 1/(√3 + √2) → multiply by (√3 − √2): denominator becomes 3 − 2 = 1, giving √3 − √2.
• 1/(√7 − √6) → multiply by (√7 + √6): denominator becomes 7 − 6 = 1, giving √7 + √6.
• 1/√7 → multiply by √7: denominator becomes 7, giving √7/7.
• 6/(√3 − √2) → multiply by (√3 + √2): denominator becomes 3 − 2 = 1, giving 6(√3 + √2).

Four More, With Heavier Surds

Question 6 repeats the technique on denominators carrying coefficients and larger radicands, where the arithmetic after multiplying out takes more care:

• (6 − 4√2)/(6 + 4√2) → conjugate (6 − 4√2): denominator 36 − 32 = 4, giving 17 − 12√2 after simplifying.
• (√7 − √5)/(√7 + √5) → conjugate (√7 − √5): denominator 7 − 5 = 2, numerator (√7−√5)² = 12 − 2√35, giving 6 − √35.
• 1/(3√2 − 2√3) → conjugate (3√2 + 2√3): denominator 18 − 12 = 6, giving (3√2 + 2√3)/6.
• (3√5 − √7)/(3√3 + √2) → conjugate (3√3 − √2): denominator 27 − 2 = 25, giving (9√15 − 3√10 − 3√21 + √14)/25.

A Numeric Evaluation to Three Decimals

Question 7 asks for the value of (√10 − √5)/(2√2), using the approximations √2 ≈ 1.414 and √5 ≈ 2.236. Rationalising first — multiply by √2 over √2 — turns the expression into √5(2 − √2)/4, since √10 = √2·√5. Substituting the given approximations: 2.236 × (2 − 1.414)/4 = 2.236 × 0.586/4 ≈ 0.328.

Rationalising before substituting the decimal approximations, rather than after, is what keeps this calculation accurate. Plugging 1.414 and 2.236 directly into the original unrationalised expression and only then simplifying would compound the small rounding error from each approximation across more arithmetic steps than necessary.

Surds Written as Powers

A surd is the positive nth root of a positive rational number that isn't itself a perfect nth power — written ⁿ√a or a^1/n, with 'a' the radicand, '√' the radical sign, and n the degree. √5, ∛4, ⁴√7 are surds; √9 = 3 and ∛27 = 3 are not, since they reduce to whole numbers.

Question 8 evaluates six such expressions using the law ⁿ√(aⁿ) = a: 64^1/6 = (2⁶)^1/6 = 2, 32^1/5 = (2⁵)^1/5 = 2, 625^1/4 = (5⁴)^1/4 = 5, 16^3/2 = (2⁴)^3/2 = 2⁶ = 64, 243^2/5 = (3⁵)^2/5 = 3² = 9, and 46656^−1/6 = (6⁶)^−1/6 = 1/6.

Question 9 combines several such roots in one expression: 4√81 − 8·∛343 + 15·⁵√32 + √225 evaluates term by term as 3 − 8(7) + 15(2) + 15 = 3 − 56 + 30 + 15 = −8.

Solving for Unknown Rationals a and b

Question 10 gives an equation with surds on one side and unknowns a, b (rational) on the other, then asks for their values by rationalising and matching parts. For (√3 + √2)/(√3 − √2) = a + b√6: rationalising the left side gives 5 + 2√6, so comparing rational and irrational parts directly yields a = 5, b = 2.

The second part, (√5 + √3)/(2√5 − 3√3) = a − b√15, works the same way but with messier arithmetic after rationalising: the left side simplifies to −19/7 − (5/7)√15, giving a = −19/7 and b = 5/7.

Question 11 asks for the square root of 11 + 2√30 directly, without a calculator. Matching it to the identity a + b + 2√(ab) = (√a + √b)² means finding a and b with a + b = 11 and ab = 30 — that's 6 and 5. So √(11 + 2√30) = √6 + √5.

Into Polynomials Next

The rationalising and identity-matching techniques built across this exercise reappear constantly once Chapter 2 — Polynomials introduces algebraic identities in a purely symbolic form, without the square roots. The number-classification and irrationality checks from earlier in this chapter also get a formal, contradiction-based proof treatment in Class 10's irrationality proofs, revisiting exactly the numbers rationalised here.