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ESAT Maths 2 Algebra questions

12 original ESAT Maths 2 algebra questions in the real exam style, each with a step-by-step worked solution and an explanation of why the tempting wrong answers are wrong. Try each one before opening the solution.

Methods on this page: Use the discriminant to find a range of valuesSplit a modulus into casesCoefficients in a product of expansionsConstant term: choose the power of each part so the x's cancel

Question 1ESAT Maths 2Warm-up

How many real solutions does the equation ∣x2−6∣=9\lvert x^2 - 6 \rvert = 9 have?
  1. A00
  2. B11
  3. C22
  4. D33
  5. E44
  6. F55
  7. G66
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Answer: C · Method: Split a modulus into cases

  1. Split the modulus: x2−6=9x^2 - 6 = 9 or x2−6=−9x^2 - 6 = -9.
  2. The first gives x2=15x^2 = 15: two solutions.
  3. The second gives x2=−3x^2 = -3, which is impossible: no solutions.
  4. That's 22 solutions in total.

Why the tempting wrong answers are wrong

  • E: x2=−3x^2 = -3 has no real solutions, because a square can't be negative.

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Question 2ESAT Maths 2Warm-up

What is the coefficient of x4x^{4} in the expansion of (2+x)7(2 + x)^{7}?
  1. A88
  2. B278278
  3. C279279
  4. D280280
  5. E281281
  6. F282282
  7. G283283
  8. H560560
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Answer: D · Method: Coefficients in a product of expansions

  1. The x4x^{4} term is (74)×23×(x)4\binom{7}{4} \times 2^{3} \times (x)^{4}.
  2. That's 35×8×1=28035 \times 8 \times 1 = 280.

Why the tempting wrong answers are wrong

  • A: Don't forget the binomial coefficient (74)=35\binom{7}{4} = 35.
  • H: The power of xx comes from the second term, so it's 23(x)42^{3}(x)^{4}, not the other way round.

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Question 3ESAT Maths 2Standard

For which values of kk is x2+2kx+(−k+12)x^2 + 2kx + (-k + 12) positive for all real values of xx?
  1. A−4<k<3-4 < k < 3
  2. Bk>−4k > -4
  3. Ck<−4 or k>3k < -4 \text{ or } k > 3
  4. D−3<k<4-3 < k < 4
  5. E−4≤k≤3-4 \le k \le 3
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Answer: A · Method: Use the discriminant to find a range of values

  1. A quadratic with a positive x2x^2 coefficient is positive for every xx exactly when it has no real roots: its discriminant is negative.
  2. The discriminant is (2k)2−4(−k+12)=4(k2+k−12)(2k)^2 - 4(-k + 12) = 4(k^2 + k - 12).
  3. That's 4(k+4)(k−3)4(k + 4)(k - 3), which is negative exactly when −4<k<3-4 < k < 3.

Why the tempting wrong answers are wrong

  • C: That's where the discriminant is positive, so the quadratic crosses the xx-axis.
  • E: At k=−4k = -4 or k=3k = 3 the discriminant is 00: the quadratic touches zero, so it isn't positive for every xx.

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Question 4ESAT Maths 2Standard

What is the coefficient of x3x^{3} in the expansion of (1−2x)3(1−x)3(1 - 2x)^{3}(1 - x)^{3}?
  1. A−64-64
  2. B−63-63
  3. C−62-62
  4. D−9-9
  5. E−8-8
  6. F1111
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Answer: B · Method: Coefficients in a product of expansions

  1. Expand each bracket up to x3x^{3}: the coefficients are 1,−6,12,−81, -6, 12, -8 and 1,−3,3,−11, -3, 3, -1.
  2. Add up every pair whose powers total 3: (1)(−1)+(−6)(3)+(12)(−3)+(−8)(1)=−63(1)(-1) + (-6)(3) + (12)(-3) + (-8)(1) = -63.

Why the tempting wrong answers are wrong

  • D: You also need the cross terms, where the powers from each bracket add to the power you want.
  • F: Check the signs in the second bracket.

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Question 5ESAT Maths 2Standard

What is the constant term in the expansion of (x+2x)8\left(x + \dfrac{2}{x}\right)^{8}?
  1. A1616
  2. B7070
  3. C448448
  4. D11181118
  5. E11191119
  6. F11201120
  7. G11211121
  8. H11221122
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Answer: F · Method: Constant term: choose the power of each part so the x's cancel

  1. The general term uses (x)8−r(x)^{8 - r} and (…/x)r(\ldots/x)^{r}; the powers of xx cancel when r=4r = 4.
  2. So the constant term is (84)×(1)4×(2)4=1120\binom{8}{4} \times (1)^{4} \times (2)^{4} = 1120.

Why the tempting wrong answers are wrong

  • A: Don't forget the binomial coefficient (84)=70\binom{8}{4} = 70.
  • B: Include the numbers attached to each part, raised to their powers.

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Question 6ESAT Maths 2Standard

What is the constant term in the expansion of (x−1x)6\left(x - \dfrac{1}{x}\right)^{6}?
  1. A−22-22
  2. B−21-21
  3. C−20-20
  4. D−19-19
  5. E−18-18
  6. F−1-1
  7. G1515
  8. H2020
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Answer: C · Method: Constant term: choose the power of each part so the x's cancel

  1. The general term uses (x)6−r(x)^{6 - r} and (…/x)r(\ldots/x)^{r}; the powers of xx cancel when r=3r = 3.
  2. So the constant term is (63)×(1)3×(−1)3=−20\binom{6}{3} \times (1)^{3} \times (-1)^{3} = -20.

Why the tempting wrong answers are wrong

  • F: Don't forget the binomial coefficient (63)=20\binom{6}{3} = 20.
  • H: Include the numbers attached to each part, raised to their powers.

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Question 7ESAT Maths 2Standard

What is the constant term in the expansion of (3x+2x)6\left(3x + \dfrac{2}{x}\right)^{6}?
  1. A2020
  2. B216216
  3. C43184318
  4. D43194319
  5. E43204320
  6. F43214321
  7. G43224322
  8. H48604860
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Answer: E · Method: Constant term: choose the power of each part so the x's cancel

  1. The general term uses (3x)6−r(3x)^{6 - r} and (…/x)r(\ldots/x)^{r}; the powers of xx cancel when r=3r = 3.
  2. So the constant term is (63)×(3)3×(2)3=4320\binom{6}{3} \times (3)^{3} \times (2)^{3} = 4320.

Why the tempting wrong answers are wrong

  • A: Include the numbers attached to each part, raised to their powers.
  • B: Don't forget the binomial coefficient (63)=20\binom{6}{3} = 20.

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Question 8ESAT Maths 2Standard

For which values of kk is x2+2kx+(−6k−8)x^2 + 2kx + (-6k - 8) positive for all real values of xx?
  1. Ak>−4k > -4
  2. B−4<k<−2-4 < k < -2
  3. C2<k<42 < k < 4
  4. D−4≤k≤−2-4 \le k \le -2
  5. Ek<−4 or k>−2k < -4 \text{ or } k > -2
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Answer: B · Method: Use the discriminant to find a range of values

  1. A quadratic with a positive x2x^2 coefficient is positive for every xx exactly when it has no real roots: its discriminant is negative.
  2. The discriminant is (2k)2−4(−6k−8)=4(k2+6k+8)(2k)^2 - 4(-6k - 8) = 4(k^2 + 6k + 8).
  3. That's 4(k+4)(k+2)4(k + 4)(k + 2), which is negative exactly when −4<k<−2-4 < k < -2.

Why the tempting wrong answers are wrong

  • D: At k=−4k = -4 or k=−2k = -2 the discriminant is 00: the quadratic touches zero, so it isn't positive for every xx.
  • E: That's where the discriminant is positive, so the quadratic crosses the xx-axis.

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Question 9ESAT Maths 2Standard

How many real solutions does the equation ∣x2−8x∣=5\lvert x^2 - 8x \rvert = 5 have?
  1. A11
  2. B22
  3. C33
  4. D44
  5. E55
  6. F66
  7. G77
  8. H88
Show the answer and worked solution

Answer: D · Method: Split a modulus into cases

  1. Split the modulus: x2−8x−5=0x^2 - 8x - 5 = 0 or x2−8x+5=0x^2 - 8x + 5 = 0.
  2. The first has discriminant 64+20>064 + 20 > 0: two solutions.
  3. The second has discriminant 64−20=44>064 - 20 = 44 > 0: two more.
  4. That's 44 solutions in total.

Why the tempting wrong answers are wrong

  • B: Don't forget the case where the inside of the modulus is negative.

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Question 10ESAT Maths 2Standard

What is the coefficient of x3x^{3} in the expansion of (1−3x)3(1+x)2(1 - 3x)^{3}(1 + x)^{2}?
  1. A−90-90
  2. B−27-27
  3. C1616
  4. D1717
  5. E1818
  6. F1919
  7. G2020
  8. H2121
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Answer: E · Method: Coefficients in a product of expansions

  1. Expand each bracket up to x3x^{3}: the coefficients are 1,−9,27,−271, -9, 27, -27 and 1,2,1,01, 2, 1, 0.
  2. Add up every pair whose powers total 3: (1)(0)+(−9)(1)+(27)(2)+(−27)(1)=18(1)(0) + (-9)(1) + (27)(2) + (-27)(1) = 18.

Why the tempting wrong answers are wrong

  • A: Check the signs in the second bracket.
  • B: You also need the cross terms, where the powers from each bracket add to the power you want.

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Question 11ESAT Maths 2Standard

What is the coefficient of x2x^{2} in the expansion of (1+3x)6(1−x)2(1 + 3x)^{6}(1 - x)^{2}?
  1. A100100
  2. B101101
  3. C135135
  4. D136136
  5. E172172
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Answer: A · Method: Coefficients in a product of expansions

  1. Expand each bracket up to x2x^{2}: the coefficients are 1,18,1351, 18, 135 and 1,−2,11, -2, 1.
  2. Add up every pair whose powers total 2: (1)(1)+(18)(−2)+(135)(1)=100(1)(1) + (18)(-2) + (135)(1) = 100.

Why the tempting wrong answers are wrong

  • D: You also need the cross terms, where the powers from each bracket add to the power you want.
  • E: Check the signs in the second bracket.

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Question 12ESAT Maths 2Standard

What is the constant term in the expansion of (x2−2x)6\left(\dfrac{x}{2} - \dfrac{2}{x}\right)^{6}?
  1. A−21-21
  2. B−20-20
  3. C−19-19
  4. D−18-18
  5. E−1-1
  6. F154\dfrac{15}{4}
  7. G2020
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Answer: B · Method: Constant term: choose the power of each part so the x's cancel

  1. The general term uses (x2)6−r(\dfrac{x}{2})^{6 - r} and (…/x)r(\ldots/x)^{r}; the powers of xx cancel when r=3r = 3.
  2. So the constant term is (63)×(12)3×(−2)3=−20\binom{6}{3} \times (\dfrac{1}{2})^{3} \times (-2)^{3} = -20.

Why the tempting wrong answers are wrong

  • E: Don't forget the binomial coefficient (63)=20\binom{6}{3} = 20.
  • G: Include the numbers attached to each part, raised to their powers.

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