Algebra — Identities, Surds & Equations — Practice

Question 1

If x + 1/x = 4, then x² + 1/x² equals:
(a) 12  (b) 14  (c) 16  (d) 18

Accepted Answer

x² + 1/x² = k² − 2 = 16 − 2 = 14.

Answer: (b) 14.

Question 2

If x + 1/x = 6, then x³ + 1/x³ equals:
(a) 198  (b) 216  (c) 210  (d) 234

Accepted Answer

k³ − 3k = 216 − 18 = 198.

Answer: (a) 198. Trap: (b) 216 = k³ alone, forgetting −3k.

Question 3

If x + 1/x = 1, the value of x⁹⁹ + 1/x⁹⁹ is:
(a) 1  (b) −1  (c) 2  (d) −2

Accepted Answer

x + 1/x = 1 ⇒ x² − x + 1 = 0 ⇒ x³ = −1.

x⁹⁹ = (x³)³³ = (−1)³³ = −1, and 1/x⁹⁹ = −1.

Sum = −2.

Answer: (d) −2.

Question 4

If a + b + c = 12 and ab + bc + ca = 47, then a² + b² + c² equals:
(a) 50  (b) 52  (c) 47  (d) 144

Accepted Answer

Σa² = (a+b+c)² − 2Σab = 144 − 94 = 50.

Answer: (a) 50.

Question 5

The value of (2.3)³ + (3.7)³ − (6)³ + 3 × 2.3 × 3.7 × 6 is:
(a) 0  (b) 1  (c) 25.53  (d) 76.59

Accepted Answer

Write as a³ + b³ + c³ − 3abc with a = 2.3, b = 3.7, c = −6.

a + b + c = 2.3 + 3.7 − 6 = 0 ⇒ a³ + b³ + c³ = 3abc, so the expression is 0.

Answer: (a) 0. (Note +3abc in the question equals −3abc with c = −6 — signs absorb.)

Question 6

If a² + b² + c² = ab + bc + ca and a = 5, then the value of a + b + c is:
(a) 5  (b) 10  (c) 15  (d) 25

Accepted Answer

The condition gives ½[(a−b)² + (b−c)² + (c−a)²] = 0 ⇒ a = b = c = 5.

a + b + c = 15.

Answer: (c) 15.

Question 7

The value of 1/(1+√2) + 1/(√2+√3) + 1/(√3+2) is:
(a) 1  (b) √3  (c) 2  (d) √2 − 1

Accepted Answer

Each term rationalises to a difference: (√2−1) + (√3−√2) + (2−√3).

Everything cancels except 2 − 1 = 1.

Answer: (a) 1.

Question 8

If x = 7 − 4√3, then √x + 1/√x equals:
(a) 2√3  (b) 4  (c) √14  (d) 2 + √3

Accepted Answer

7 − 4√3 = 7 − 2√12 = (2 − √3)², so √x = 2 − √3.

1/√x = 1/(2−√3) = 2 + √3 (conjugate; 4 − 3 = 1).

Sum = 4.

Answer: (b) 4.

Question 9

If x/y = 3/4, the value of (7x − 4y)/(3x + y) is:
(a) 5/13  (b) 4/13  (c) 7/13  (d) 6/13

Accepted Answer

Substitute x = 3t, y = 4t.

(21t − 16t)/(9t + 4t) = 5t/13t = 5/13.

Answer: (a) 5/13. Substitution beats algebraic manipulation every time the ratio is given.

Question 10

The system 2x + 3y = 7 and (k−1)x + (k+2)y = 3k has infinitely many solutions when k equals:
(a) 5  (b) 7  (c) 3  (d) no such k

Accepted Answer

Need 2/(k−1) = 3/(k+2) = 7/3k.

From the first pair: 2(k+2) = 3(k−1) → 2k + 4 = 3k − 3 → k = 7.

Check the third ratio: 7/21 = 1/3, and 2/(k−1) = 2/6 = 1/3 ✓.

Answer: (b) 7.

Question 11

If x − 1/x = 3, then x³ − 1/x³ equals:
(a) 27  (b) 30  (c) 36  (d) 39

Accepted Answer

x³ − 1/x³ = m³ + 3m = 27 + 9 = 36.

Answer: (c) 36. (Plus chain subtracts 3k; minus chain ADDS 3m.)

Question 12

If a + b = 5 and ab = 6, then a³ + b³ equals:
(a) 35  (b) 30  (c) 45  (d) 65

Accepted Answer

a³ + b³ = (a+b)³ − 3ab(a+b) = 125 − 90 = 35.

Answer: (a) 35. (Check: a, b = 2, 3 → 8 + 27 = 35 ✓ — substitution as verification.)

Practice — Algebra — Identities, Surds & Equations

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