Questions on indices and surds

Question 1

Evaluate 16^3/4, 5⁰ and 4⁻² without a calculator.

Accepted Answer

16^3/4 = (⁴√16)³ = 2³ = 8. Then 5⁰ = 1 and 4⁻² = 1/16. B1 B1 B1, one for each value. Root before power keeps the numbers small; going the other way means cubing 16 first. A negative index means reciprocal, never a negative answer, so 4⁻² is 1/16 and not −16.

Question 2

Write √72 in the form k√2.

Accepted Answer

√72 = √(36 × 2) = √36 × √2 = 6√2. M1 for extracting a square factor, A1 for 6√2. Pull out the largest square factor in one step. Pulling out 4 first reaches the same place in two rounds.

Question 3

Simplify (3x²)³ × 2x⁻⁴, giving your answer as a single term.

Accepted Answer

The cube applies to both parts of the bracket: (3x²)³ = 27x⁶. Then 27x⁶ × 2x⁻⁴ = 54x⁶⁻⁴ = 54x². M1 for 27x⁶, A1 for 54x². The power outside a bracket acts on the coefficient as well as on the letter, so 3x⁶ is short of a factor of 9, and multiplying like bases adds the indices, 6 + (−4) = 2, rather than multiplying them.

Question 4

Expand and simplify (3 + √2)(3 − √2) and (2 + √5)².

Accepted Answer

The first pair are conjugates, so the cross terms cancel: (3 + √2)(3 − √2) = 9 − 2 = 7. B1. The second is a square, so the cross term survives: (2 + √5)² = 4 + 2 × 2√5 + 5 = 9 + 4√5. M1 for all three terms, A1 for 9 + 4√5. The two results show the difference between (√x + √y)(√x − √y) = x − y, which is an integer, and a square, which keeps a surd term: the squared terms make up all of the first product but only part of the second.

Question 5

Evaluate 8^−2/3 and 25^3/2 without a calculator.

Accepted Answer

Work outside in. The minus means reciprocal and the 3 in the denominator means cube root, so 8^−2/3 = 1/(∛8)² = 1/4. Likewise 25^3/2 = (√25)³ = 5³ = 125. M1 for treating the negative index as a reciprocal, A1 for 1/4, A1 for 125.

Question 6

Simplify √50 + √32 − √18.

Accepted Answer

Each surd reduces to a multiple of √2. √50 = 5√2, √32 = 4√2 and √18 = 3√2, so the sum is 5√2 + 4√2 − 3√2 = 6√2. M1 for reducing at least one surd, A1 for all three, A1 for 6√2. Like terms in √2 collect exactly as terms in x would. Surds do not add under the root: √28 + √36 is not √64.

Question 7

Express 5/√5 and 4/√6 with rational denominators, in simplest form.

Accepted Answer

5/√5 = 5√5/5 = √5, since multiplying top and bottom by the surd clears it. And 4/√6 = 4√6/6 = 2√6/3. M1 for multiplying top and bottom by the surd, A1 for √5, A1 for 2√6/3. The final cancellation is part of the required form, so simplify 4√6/6 to 2√6/3.

Question 8

Express 10/(√6 − 2) in the form a√6 + b, where a and b are integers.

Accepted Answer

Multiply top and bottom by the conjugate √6 + 2. The denominator becomes 6 − 4 = 2 and the numerator becomes 10(√6 + 2), so the fraction is 10(√6 + 2)/2 = 5√6 + 10, giving a = 5 and b = 10. M1 for multiplying by the conjugate, A1 for the rational denominator, A1 for 5√6 + 10. Multiplying by √6 − 2 instead leaves 10 − 4√6 downstairs, which is no better than where you started.

Question 9

Write (x + 3)/√x as a sum of two terms, each of the form axⁿ.

Accepted Answer

Split the fraction and write the root as a power: (x + 3)/√x = x/x^1/2 + 3/x^1/2. M1 for splitting over the common denominator. Then x ÷ x^1/2 = x^1 − 1/2 = x^1/2, and 3/x^1/2 = 3x^−1/2. So the expression is x^1/2 + 3x^−1/2. A1 A1, one for each term. This rewriting is what makes the expression ready to differentiate or integrate term by term, since neither rule applies to a quotient as it stands.

Question 10

Solve the equation 2x + 3√x − 2 = 0.

Accepted Answer

The equation is a quadratic in √x, since x = (√x)². Put u = √x: 2u² + 3u − 2 = 0. M1 for the substitution. It factorises as (2u − 1)(u + 2) = 0, so u = ½ or u = −2. A1 for both values of u. But u = √x is a non-negative square root, so u = −2 gives no value of x and is discarded. B1 for the rejection with the reason. From u = ½, x = ¼. A1 for x = ¼. Checking, 2(¼) + 3(½) − 2 = ½ + 3/2 − 2 = 0. Squaring u = −2 to produce x = 4 is the slip to guard against: at x = 4 the left side is 8 + 6 − 2 = 12, not 0, because √4 is 2 and not −2.

Question 11

Solve 8^2x − 1 = 4^x + 2.

Accepted Answer

Both sides are powers of 2: 8 = 2³ and 4 = 2², so the equation is 2^3(2x − 1) = 2^2(x + 2), that is 2^6x − 3 = 2^2x + 4. M1 for rewriting both sides over base 2. Equal bases force equal indices: 6x − 3 = 2x + 4. M1 for equating the indices. So 4x = 7 and x = 7/4. A1. The power of a power multiplies the indices, so the 3 multiplies all of 2x − 1 and not just the 2x. Logarithms reach the same answer, but the exact fraction comes out directly here, with no rounding.

Question 12

Given that 2^x = 8^y+1 and 9^y = 3^x−9, find the value of x and the value of y.

Accepted Answer

Rewrite everything over a single base in each equation. The first gives 8^y+1 = 2^3(y+1), so x = 3y + 3. The second gives 9^y = 3^2y, so 2y = x − 9. Substituting the first into the second, 2y = 3y + 3 − 9, so y = 6 and then x = 21. Checking, 2²¹ = 8⁷ and 9⁶ = 3¹², so both hold. M1 for rewriting the first equation over base 2, A1 for x = 3y + 3, M1 for rewriting the second over base 3, A1 for 2y = x − 9, A1 for y = 6 and x = 21. Logs would also work here, but they are slower and the numbers stop being exact.

Question 13

Show that 1/(√(n + 1) + √n) = √(n + 1) − √n for every positive integer n. Hence find the exact value of 1/(√2 + √1) + 1/(√3 + √2) + 1/(√4 + √3) + … + 1/(√100 + √99).

Accepted Answer

Multiply top and bottom by the conjugate √(n + 1) − √n. The denominator becomes (√(n + 1))² − (√n)² = (n + 1) − n = 1, so the fraction is √(n + 1) − √n exactly. M1 for multiplying by the conjugate, A1 for the denominator collapsing to 1 and the identity shown. Now each term of the sum has this shape with n = 1, 2, …, 99, so the sum is (√2 − √1) + (√3 − √2) + (√4 − √3) + … + (√100 − √99). M1 for rewriting every term. Each surd from √2 to √99 appears once with a plus sign and once with a minus sign, and everything cancels except the two ends: −√1 from the first term and +√100 from the last. A1 for the cancellation leaving √100 − √1. The sum is 10 − 1 = 9. A1.

Question 14

Given that (a + b√2)(3 − √2) = 4 + √2, where a and b are rational numbers, find the value of a and the value of b.

Accepted Answer

Expand the left side: 3a − a√2 + 3b√2 − 2b = (3a − 2b) + (3b − a)√2. M1 for the expansion collected into a rational part and a √2 part. Since √2 is irrational, an equation between two such expressions with rational coefficients can hold only if the rational parts agree and the √2 parts agree: 3a − 2b = 4 and 3b − a = 1. A1 for the pair of equations, with that reason. From the second, a = 3b − 1; substituting into the first, 9b − 3 − 2b = 4, so 7b = 7 and b = 1, then a = 2. M1 for solving the pair, A1 for b = 1, A1 for a = 2. A check by the conjugate agrees: dividing, (4 + √2)/(3 − √2) = (4 + √2)(3 + √2)/(9 − 2) = (12 + 4√2 + 3√2 + 2)/7 = (14 + 7√2)/7 = 2 + √2. Comparing parts without stating why is an assertion rather than an argument; if √2 were rational the two equations would not follow.

Question 15

Without using a calculator, determine which of 5 + √3 and 2√11 is the larger, justifying every step.

Accepted Answer

Both numbers are positive, and for positive numbers the order of the squares is the order of the numbers, so compare squares. M1 for that reduction with the positivity stated. (5 + √3)² = 25 + 10√3 + 3 = 28 + 10√3, and (2√11)² = 4 × 11 = 44. A1 for both squares. The comparison is now 28 + 10√3 against 44, that is 10√3 against 16, and both of those are positive, so square again: 100 × 3 = 300 against 256. M1 for reducing and squaring a second time, A1 for 300 against 256. Since 300 > 256, 10√3 > 16, so 28 + 10√3 > 44 and 5 + √3 is the larger. B1 for the conclusion traced back through the chain. The decimal values 6.73 and 6.63 confirm the answer but do not justify it.