IB Mathematics: Analysis and Approaches Quiz: Vietas Formulas
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Vietas FormulasQuestion 1 of 20

Let α,β,γ\alpha, \beta, \gamma be the roots of the equation 2x3−5x2+4x−1=02x^3 - 5x^2 + 4x - 1 = 0. Find the value of α2+β2+γ2\alpha^2 + \beta^2 + \gamma^2.

9/4
25/4
41/4
17
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IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: Vietas Formulas

Practice Vietas Formulas in IB Mathematics: Analysis and Approaches with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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Question 1

Let α,β,γ\alpha, \beta, \gamma be the roots of the equation 2x3−5x2+4x−1=02x^3 - 5x^2 + 4x - 1 = 0. Find the value of α2+β2+γ2\alpha^2 + \beta^2 + \gamma^2.

  1. 9/4 (correct answer)
  2. 25/4
  3. 41/4
  4. 17
Explanation: We use the identity (α+β+γ)2=α2+β2+γ2+2(αβ+βγ+γα)(\alpha+\beta+\gamma)^2 = \alpha^2+\beta^2+\gamma^2 + 2(\alpha\beta+\beta\gamma+\gamma\alpha). Rearranging this gives α2+β2+γ2=(α+β+γ)2−2(αβ+βγ+γα)\alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha). From Vieta's formulas for the equation 2x3−5x2+4x−1=02x^3 - 5x^2 + 4x - 1 = 0: Sum of roots: α+β+γ=−−52=52\alpha+\beta+\gamma = -\frac{-5}{2} = \frac{5}{2} Sum of products of roots taken two at a time: αβ+βγ+γα=42=2\alpha\beta+\beta\gamma+\gamma\alpha = \frac{4}{2} = 2 Now substitute these values into the rearranged identity: α2+β2+γ2=(52)2−2(2)\alpha^2+\beta^2+\gamma^2 = \left(\frac{5}{2}\right)^2 - 2(2) =254−4= \frac{25}{4} - 4 =254−164= \frac{25}{4} - \frac{16}{4} =94= \frac{9}{4} Distractor B, 25/4, is the value of (α+β+γ)2(\alpha+\beta+\gamma)^2. Distractor C, 41/4, results from a sign error in the identity, calculating (α+β+γ)2+2(αβ+βγ+γα)(\alpha+\beta+\gamma)^2 + 2(\alpha\beta+\beta\gamma+\gamma\alpha). Distractor D, 17, results from ignoring the leading coefficient of 2.

Question 2

The cubic equation x3+px2+qx+r=0x^3 + px^2 + qx + r = 0 has three distinct roots. Two of the roots are equal in magnitude but opposite in sign. Which of the following relationships between the coefficients must be true?

  1. r=p+qr = p+q
  2. p=qrp = qr
  3. pr=qpr = q
  4. pq=rpq = r (correct answer)
Explanation: Let the roots of the equation be α,−α,γ\alpha, -\alpha, \gamma. Using Vieta's formulas:
  1. Sum of the roots: α+(−α)+γ=−p  ⟹  γ=−p\alpha + (-\alpha) + \gamma = -p \implies \gamma = -p.
  2. Sum of the products of the roots taken two at a time: α(−α)+αγ+(−α)γ=q  ⟹  −α2+αγ−αγ=q  ⟹  −α2=q\alpha(-\alpha) + \alpha\gamma + (-\alpha)\gamma = q \implies -\alpha^2 + \alpha\gamma - \alpha\gamma = q \implies -\alpha^2 = q.
  3. Product of the roots: α(−α)γ=−r  ⟹  −α2γ=−r  ⟹  α2γ=r\alpha(-\alpha)\gamma = -r \implies -\alpha^2\gamma = -r \implies \alpha^2\gamma = r.
Now we can use these three equations to find a relationship between p,q,rp, q, r. From (2), we have α2=−q\alpha^2 = -q. Substitute γ=−p\gamma = -p and α2=−q\alpha^2 = -q into equation (3): (−q)(−p)=r(-q)(-p) = r pq=rpq = r This relationship must hold true.

Question 3

The roots of the polynomial P(x)=x3−6x2+11x−5P(x) = x^3 - 6x^2 + 11x - 5 are α,β,γ\alpha, \beta, \gamma. The roots of its derivative P′(x)P'(x) are r1,r2r_1, r_2. What is the value of α+β+γ3−r1+r22\frac{\alpha+\beta+\gamma}{3} - \frac{r_1+r_2}{2}?

  1. -2
  2. 0 (correct answer)
  3. 2
  4. 4
Explanation: For the polynomial P(x)=x3−6x2+11x−5P(x) = x^3 - 6x^2 + 11x - 5, the sum of the roots α,β,γ\alpha, \beta, \gamma is given by Vieta's formulas: α+β+γ=−−61=6\alpha+\beta+\gamma = -\frac{-6}{1} = 6 The arithmetic mean of these roots is α+β+γ3=63=2\frac{\alpha+\beta+\gamma}{3} = \frac{6}{3} = 2. Next, find the derivative of P(x)P(x): P′(x)=3x2−12x+11P'(x) = 3x^2 - 12x + 11 Let the roots of this quadratic be r1,r2r_1, r_2. The sum of the roots of P′(x)P'(x) is given by Vieta's formulas for quadratics: r1+r2=−−123=4r_1+r_2 = -\frac{-12}{3} = 4 The arithmetic mean of these roots is r1+r22=42=2\frac{r_1+r_2}{2} = \frac{4}{2} = 2. The question asks for the value of α+β+γ3−r1+r22\frac{\alpha+\beta+\gamma}{3} - \frac{r_1+r_2}{2}. Value=2−2=0\text{Value} = 2 - 2 = 0 Note: For any cubic polynomial, the arithmetic mean of its roots is equal to the arithmetic mean of the roots of its derivative. This value corresponds to the x-coordinate of the point of inflection.

Question 4

The quadratic equation x2−4x+2=0x^2 - 4x + 2 = 0 has roots α\alpha and β\beta. What is the value of α3+β3\alpha^3 + \beta^3?

  1. 16
  2. 40 (correct answer)
  3. 48
  4. 64
Explanation: From the quadratic equation x2−4x+2=0x^2 - 4x + 2 = 0, we can use Vieta's formulas to find the sum and product of the roots α\alpha and β\beta. Sum of roots: α+β=−−41=4\alpha + \beta = -\frac{-4}{1} = 4. Product of roots: αβ=21=2\alpha \beta = \frac{2}{1} = 2. To find α3+β3\alpha^3 + \beta^3, we can use the identity: α3+β3=(α+β)3−3αβ(α+β)\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) Substitute the values we found: α3+β3=(4)3−3(2)(4)\alpha^3 + \beta^3 = (4)^3 - 3(2)(4) =64−24= 64 - 24 =40= 40 Alternatively, use the identity α3+β3=(α+β)(α2−αβ+β2)\alpha^3+\beta^3 = (\alpha+\beta)(\alpha^2 - \alpha\beta + \beta^2). First find α2+β2=(α+β)2−2αβ=42−2(2)=16−4=12\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = 4^2 - 2(2) = 16-4 = 12. Then α3+β3=(4)(12−2)=4(10)=40\alpha^3+\beta^3 = (4)(12-2) = 4(10) = 40.

Question 5

The roots of the polynomial P(x)=x3−6x2+11x−5P(x) = x^3 - 6x^2 + 11x - 5 are α,β,γ\alpha, \beta, \gamma. The roots of its derivative P′(x)P'(x) are r1,r2r_1, r_2. What is the value of α+β+γ3−r1+r22\frac{\alpha+\beta+\gamma}{3} - \frac{r_1+r_2}{2}?

  1. -2
  2. 0 (correct answer)
  3. 2
  4. 4
Explanation: For the polynomial P(x)=x3−6x2+11x−5P(x) = x^3 - 6x^2 + 11x - 5, the sum of the roots α,β,γ\alpha, \beta, \gamma is given by Vieta's formulas: α+β+γ=−−61=6\alpha+\beta+\gamma = -\frac{-6}{1} = 6 The arithmetic mean of these roots is α+β+γ3=63=2\frac{\alpha+\beta+\gamma}{3} = \frac{6}{3} = 2. Next, find the derivative of P(x)P(x): P′(x)=3x2−12x+11P'(x) = 3x^2 - 12x + 11 Let the roots of this quadratic be r1,r2r_1, r_2. The sum of the roots of P′(x)P'(x) is given by Vieta's formulas for quadratics: r1+r2=−−123=4r_1+r_2 = -\frac{-12}{3} = 4 The arithmetic mean of these roots is r1+r22=42=2\frac{r_1+r_2}{2} = \frac{4}{2} = 2. The question asks for the value of α+β+γ3−r1+r22\frac{\alpha+\beta+\gamma}{3} - \frac{r_1+r_2}{2}. Value=2−2=0\text{Value} = 2 - 2 = 0 Note: For any cubic polynomial, the arithmetic mean of its roots is equal to the arithmetic mean of the roots of its derivative. This value corresponds to the x-coordinate of the point of inflection.

Question 6

Let α,β,γ\alpha, \beta, \gamma be the roots of the equation 2x3−5x2+4x−1=02x^3 - 5x^2 + 4x - 1 = 0. Find the value of α2+β2+γ2\alpha^2 + \beta^2 + \gamma^2.

  1. 9/4 (correct answer)
  2. 25/4
  3. 41/4
  4. 17
Explanation: We use the identity (α+β+γ)2=α2+β2+γ2+2(αβ+βγ+γα)(\alpha+\beta+\gamma)^2 = \alpha^2+\beta^2+\gamma^2 + 2(\alpha\beta+\beta\gamma+\gamma\alpha). Rearranging this gives α2+β2+γ2=(α+β+γ)2−2(αβ+βγ+γα)\alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha). From Vieta's formulas for the equation 2x3−5x2+4x−1=02x^3 - 5x^2 + 4x - 1 = 0: Sum of roots: α+β+γ=−−52=52\alpha+\beta+\gamma = -\frac{-5}{2} = \frac{5}{2} Sum of products of roots taken two at a time: αβ+βγ+γα=42=2\alpha\beta+\beta\gamma+\gamma\alpha = \frac{4}{2} = 2 Now substitute these values into the rearranged identity: α2+β2+γ2=(52)2−2(2)\alpha^2+\beta^2+\gamma^2 = \left(\frac{5}{2}\right)^2 - 2(2) =254−4= \frac{25}{4} - 4 =254−164= \frac{25}{4} - \frac{16}{4} =94= \frac{9}{4} Distractor B, 25/4, is the value of (α+β+γ)2(\alpha+\beta+\gamma)^2. Distractor C, 41/4, results from a sign error in the identity, calculating (α+β+γ)2+2(αβ+βγ+γα)(\alpha+\beta+\gamma)^2 + 2(\alpha\beta+\beta\gamma+\gamma\alpha). Distractor D, 17, results from ignoring the leading coefficient of 2.

Question 7

The cubic equation x3+px2+qx+r=0x^3 + px^2 + qx + r = 0 has three distinct roots. Two of the roots are equal in magnitude but opposite in sign. Which of the following relationships between the coefficients must be true?

  1. r=p+qr = p+q
  2. p=qrp = qr
  3. pr=qpr = q
  4. pq=rpq = r (correct answer)
Explanation: Let the roots of the equation be α,−α,γ\alpha, -\alpha, \gamma. Using Vieta's formulas:
  1. Sum of the roots: α+(−α)+γ=−p  ⟹  γ=−p\alpha + (-\alpha) + \gamma = -p \implies \gamma = -p.
  2. Sum of the products of the roots taken two at a time: α(−α)+αγ+(−α)γ=q  ⟹  −α2+αγ−αγ=q  ⟹  −α2=q\alpha(-\alpha) + \alpha\gamma + (-\alpha)\gamma = q \implies -\alpha^2 + \alpha\gamma - \alpha\gamma = q \implies -\alpha^2 = q.
  3. Product of the roots: α(−α)γ=−r  ⟹  −α2γ=−r  ⟹  α2γ=r\alpha(-\alpha)\gamma = -r \implies -\alpha^2\gamma = -r \implies \alpha^2\gamma = r.
Now we can use these three equations to find a relationship between p,q,rp, q, r. From (2), we have α2=−q\alpha^2 = -q. Substitute γ=−p\gamma = -p and α2=−q\alpha^2 = -q into equation (3): (−q)(−p)=r(-q)(-p) = r pq=rpq = r This relationship must hold true.

Question 8

The roots of the equation x4−15x3+70x2−120x+64=0x^4 - 15x^3 + 70x^2 - 120x + 64 = 0 are known to form a geometric sequence. Which of the following sets represents the roots?

  1. {\frac{1}{2}, 1, 2, 4}
  2. {1, 2, 4, 8} (correct answer)
  3. {-8, -4, -2, -1}
  4. {2, 4, 6, 8}
Explanation: Let the roots be α,β,γ,δ\alpha, \beta, \gamma, \delta. Using Vieta's formulas: Sum of roots: α+β+γ+δ=−−151=15\alpha+\beta+\gamma+\delta = -\frac{-15}{1} = 15 Product of roots: αβγδ=641=64\alpha\beta\gamma\delta = \frac{64}{1} = 64 We are looking for a set of four numbers in a geometric sequence that satisfy these conditions. Let's check the options: A. {12,1,2,4}\{\frac{1}{2}, 1, 2, 4\}: This is a geometric sequence. Sum = 0.5+1+2+4=7.5≠150.5+1+2+4 = 7.5 \neq 15. Incorrect. B. {1,2,4,8}\{1, 2, 4, 8\}: This is a geometric sequence with first term 1 and common ratio 2. Sum = 1+2+4+8=151+2+4+8 = 15. This matches the sum of roots. Product = 1×2×4×8=641 \times 2 \times 4 \times 8 = 64. This matches the product of roots. Since the sum and product match, this is the correct set of roots. We can verify the other Vieta's sums as well. C. {−8,−4,−2,−1}\{-8, -4, -2, -1\}: This is a geometric sequence. Sum = −8−4−2−1=−15≠15-8-4-2-1 = -15 \neq 15. Incorrect. D. {2,4,6,8}\{2, 4, 6, 8\}: This is an arithmetic sequence, not a geometric one. Sum = 2+4+6+8=20≠152+4+6+8 = 20 \neq 15. Incorrect.

Question 9

For the cubic equation kx3−7x2+(2k+1)x−3=0kx^3 - 7x^2 + (2k+1)x - 3 = 0, where k≠0k \neq 0, the sum of the roots is equal to the sum of the reciprocals of the roots. Find a possible value of kk.

  1. -3.5
  2. -3
  3. 3 (correct answer)
  4. 10
Explanation: Let the roots be α,β,γ\alpha, \beta, \gamma. From Vieta's formulas: Sum of roots: Σ1=α+β+γ=−−7k=7k\Sigma_1 = \alpha+\beta+\gamma = -\frac{-7}{k} = \frac{7}{k} Sum of products of roots (2 at a time): Σ2=αβ+βγ+γα=2k+1k\Sigma_2 = \alpha\beta+\beta\gamma+\gamma\alpha = \frac{2k+1}{k} Product of roots: Σ3=αβγ=−−3k=3k\Sigma_3 = \alpha\beta\gamma = -\frac{-3}{k} = \frac{3}{k} The sum of the reciprocals of the roots is: 1α+1β+1γ=βγ+αγ+αβαβγ=Σ2Σ3\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\beta\gamma+\alpha\gamma+\alpha\beta}{\alpha\beta\gamma} = \frac{\Sigma_2}{\Sigma_3} =(2k+1)/k3/k=2k+13= \frac{(2k+1)/k}{3/k} = \frac{2k+1}{3} We are given that the sum of the roots equals the sum of the reciprocals: 7k=2k+13\frac{7}{k} = \frac{2k+1}{3} Cross-multiply to solve for kk: 21=k(2k+1)21 = k(2k+1) 21=2k2+k21 = 2k^2 + k 2k2+k−21=02k^2 + k - 21 = 0 We can factor this quadratic equation: (2k+7)(k−3)=0(2k+7)(k-3) = 0 This gives two possible values for kk: k=−72=−3.5k = -\frac{7}{2} = -3.5 or k=3k=3. From the answer choices, k=3k=3 is a possible value.

Question 10

A monic cubic polynomial P(x)=x3+ax2+bx+cP(x) = x^3+ax^2+bx+c has roots α,β,γ\alpha, \beta, \gamma. Given ∑α=6\sum \alpha = 6, ∑α2=14\sum \alpha^2 = 14, and ∑α3=36\sum \alpha^3 = 36, find the value of cc.

  1. -6 (correct answer)
  2. 6
  3. 11
  4. 18
Explanation: Let the polynomial be x3+ax2+bx+c=0x^3+ax^2+bx+c=0. From Vieta's formulas, α+β+γ=−a\alpha+\beta+\gamma = -a and αβ+βγ+γα=b\alpha\beta+\beta\gamma+\gamma\alpha=b and αβγ=−c\alpha\beta\gamma=-c. Given ∑α=6\sum \alpha = 6, we have a=−6a = -6. We use the identity ∑α2=(∑α)2−2(∑αβ)\sum \alpha^2 = (\sum \alpha)^2 - 2(\sum \alpha\beta). Substituting the given values: 14=(6)2−2b14 = (6)^2 - 2b. 14=36−2b  ⟹  2b=22  ⟹  b=1114 = 36 - 2b \implies 2b = 22 \implies b = 11. Now the polynomial is x3−6x2+11x+c=0x^3 - 6x^2 + 11x + c = 0. Since α,β,γ\alpha, \beta, \gamma are roots, they satisfy the equation: α3−6α2+11α+c=0\alpha^3 - 6\alpha^2 + 11\alpha + c = 0 β3−6β2+11β+c=0\beta^3 - 6\beta^2 + 11\beta + c = 0 γ3−6γ2+11γ+c=0\gamma^3 - 6\gamma^2 + 11\gamma + c = 0 Summing these three equations gives: (α3+β3+γ3)−6(α2+β2+γ2)+11(α+β+γ)+3c=0(\alpha^3+\beta^3+\gamma^3) - 6(\alpha^2+\beta^2+\gamma^2) + 11(\alpha+\beta+\gamma) + 3c = 0 ∑α3−6∑α2+11∑α+3c=0\sum \alpha^3 - 6\sum \alpha^2 + 11\sum \alpha + 3c = 0 Substitute the given sums: 36−6(14)+11(6)+3c=036 - 6(14) + 11(6) + 3c = 0 36−84+66+3c=036 - 84 + 66 + 3c = 0 18+3c=018 + 3c = 0 3c=−18  ⟹  c=−63c = -18 \implies c = -6. The polynomial is x3−6x2+11x−6=0x^3-6x^2+11x-6=0, and its roots are 1, 2, 3.

Question 11

The roots of the equation 8x3−42x2+63x−k=08x^3 - 42x^2 + 63x - k = 0 are in a geometric progression. Find the value of kk.

  1. 1.5
  2. 3.375
  3. 21
  4. 27 (correct answer)
Explanation: Let the roots in geometric progression be ar,a,ar\frac{a}{r}, a, ar. Using Vieta's formulas:
  1. Sum of roots: ar+a+ar=a(1r+1+r)=−−428=214\frac{a}{r} + a + ar = a(\frac{1}{r} + 1 + r) = -\frac{-42}{8} = \frac{21}{4}.
  2. Sum of products of roots (2 at a time): (ar)(a)+(a)(ar)+(ar)(ar)=a2(1r+r+1)=638(\frac{a}{r})(a) + (a)(ar) + (\frac{a}{r})(ar) = a^2(\frac{1}{r} + r + 1) = \frac{63}{8}.
  3. Product of roots: (ar)(a)(ar)=a3=−−k8=k8(\frac{a}{r})(a)(ar) = a^3 = -\frac{-k}{8} = \frac{k}{8}.
We have a system of equations from (1) and (2): (i) a(1r+1+r)=214a(\frac{1}{r} + 1 + r) = \frac{21}{4} (ii) a2(1r+1+r)=638a^2(\frac{1}{r} + 1 + r) = \frac{63}{8} Divide equation (ii) by equation (i): a2(1r+1+r)a(1r+1+r)=63/821/4\frac{a^2(\frac{1}{r} + 1 + r)}{a(\frac{1}{r} + 1 + r)} = \frac{63/8}{21/4} a=638×421=3×212×4×421=32a = \frac{63}{8} \times \frac{4}{21} = \frac{3 \times 21}{2 \times 4} \times \frac{4}{21} = \frac{3}{2} Now use the product of roots formula (3) to find kk: a3=k8a^3 = \frac{k}{8} k=8a3=8(32)3=8(278)=27k = 8a^3 = 8\left(\frac{3}{2}\right)^3 = 8\left(\frac{27}{8}\right) = 27

Question 12

The equation x3−x−2=0x^3 - x - 2 = 0 has roots α,β,γ\alpha, \beta, \gamma. Find the value of 1α+1+1β+1+1γ+1\frac{1}{\alpha+1} + \frac{1}{\beta+1} + \frac{1}{\gamma+1}.

  1. -1
  2. -0.5
  3. 1 (correct answer)
  4. 2
Explanation: Let the required value be SS. A quick method is to find a new polynomial whose roots are α+1,β+1,γ+1\alpha+1, \beta+1, \gamma+1. Let y=x+1y = x+1, so x=y−1x = y-1. Substitute this into the original equation: (y−1)3−(y−1)−2=0(y-1)^3 - (y-1) - 2 = 0 (y3−3y2+3y−1)−y+1−2=0(y^3 - 3y^2 + 3y - 1) - y + 1 - 2 = 0 y3−3y2+2y−2=0y^3 - 3y^2 + 2y - 2 = 0 This new polynomial in yy has roots r1=α+1,r2=β+1,r3=γ+1r_1 = \alpha+1, r_2 = \beta+1, r_3 = \gamma+1. We want to find 1r1+1r2+1r3\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}. For any polynomial, the sum of the reciprocals of the roots is given by sum of products of roots (n-1 at a time)product of roots\frac{\text{sum of products of roots (n-1 at a time)}}{\text{product of roots}}, which for a cubic is r1r2+r2r3+r3r1r1r2r3\frac{r_1r_2+r_2r_3+r_3r_1}{r_1r_2r_3}. Using Vieta's formulas for y3−3y2+2y−2=0y^3 - 3y^2 + 2y - 2 = 0:
  • Sum of products of roots (2 at a time): r1r2+r2r3+r3r1=21=2r_1r_2+r_2r_3+r_3r_1 = \frac{2}{1} = 2.
  • Product of roots: r1r2r3=−−21=2r_1r_2r_3 = -\frac{-2}{1} = 2.
Therefore, the sum of the reciprocals is 22=1\frac{2}{2} = 1.

Question 13

The equation x3+3x2−7x+1=0x^3 + 3x^2 - 7x + 1 = 0 has roots α,β,γ\alpha, \beta, \gamma. What is the value of (α+1)(β+1)(γ+1)(\alpha+1)(\beta+1)(\gamma+1)?

  1. -10 (correct answer)
  2. -2
  3. 4
  4. 10
Explanation: Let P(x)=x3+3x2−7x+1P(x) = x^3 + 3x^2 - 7x + 1. We can expand the expression (α+1)(β+1)(γ+1)(\alpha+1)(\beta+1)(\gamma+1): (α+1)(β+1)(γ+1)=(αβ+α+β+1)(γ+1)(\alpha+1)(\beta+1)(\gamma+1) = (\alpha\beta + \alpha + \beta + 1)(\gamma+1) =αβγ+αβ+αγ+α+βγ+β+γ+1= \alpha\beta\gamma + \alpha\beta + \alpha\gamma + \alpha + \beta\gamma + \beta + \gamma + 1 =(αβγ)+(αβ+βγ+γα)+(α+β+γ)+1= (\alpha\beta\gamma) + (\alpha\beta + \beta\gamma + \gamma\alpha) + (\alpha+\beta+\gamma) + 1 Using Vieta's formulas for P(x)P(x): Sum of roots: α+β+γ=−31=−3\alpha+\beta+\gamma = -\frac{3}{1} = -3 Sum of products of roots taken two at a time: αβ+βγ+γα=−71=−7\alpha\beta + \beta\gamma + \gamma\alpha = \frac{-7}{1} = -7 Product of roots: αβγ=−11=−1\alpha\beta\gamma = -\frac{1}{1} = -1 Substituting these values into the expanded expression: −1+(−7)+(−3)+1=−10-1 + (-7) + (-3) + 1 = -10 Alternatively, consider a new polynomial Q(y)Q(y) whose roots are α+1,β+1,γ+1\alpha+1, \beta+1, \gamma+1. Let y=x+1y = x+1, which means x=y−1x = y-1. Substitute this into P(x)P(x): Q(y)=P(y−1)=(y−1)3+3(y−1)2−7(y−1)+1=0Q(y) = P(y-1) = (y-1)^3 + 3(y-1)^2 - 7(y-1) + 1 = 0 The value we seek is the product of the roots of Q(y)Q(y), which is −constant termleading coefficient-\frac{\text{constant term}}{\text{leading coefficient}}. The constant term of Q(y)Q(y) is found by setting y=0y=0, which is P(−1)P(-1): P(−1)=(−1)3+3(−1)2−7(−1)+1=−1+3+7+1=10P(-1) = (-1)^3 + 3(-1)^2 - 7(-1) + 1 = -1 + 3 + 7 + 1 = 10 So the constant term is 10. The leading coefficient is 1. The product of the roots is −101=−10-\frac{10}{1} = -10.

Question 14

The roots of the equation x3−12x2+cx+48=0x^3 - 12x^2 + cx + 48 = 0 form an arithmetic sequence. Find the value of cc.

  1. -44
  2. -20
  3. 20 (correct answer)
  4. 44
Explanation: Let the roots be a−d,a,a+da-d, a, a+d, as they form an arithmetic sequence. From Vieta's formulas, the sum of the roots is: (a−d)+a+(a+d)=−−121=12(a-d) + a + (a+d) = -\frac{-12}{1} = 12 3a=12  ⟹  a=43a = 12 \implies a = 4 So, one of the roots is 4. Since 4 is a root, it must satisfy the polynomial equation: (4)3−12(4)2+c(4)+48=0(4)^3 - 12(4)^2 + c(4) + 48 = 0 64−12(16)+4c+48=064 - 12(16) + 4c + 48 = 0 64−192+4c+48=064 - 192 + 4c + 48 = 0 112−192+4c=0112 - 192 + 4c = 0 −80+4c=0-80 + 4c = 0 4c=80  ⟹  c=204c = 80 \implies c = 20 Alternatively, using another of Vieta's formulas: The sum of the products of the roots taken two at a time is c/1=cc/1 = c. c=(a−d)a+a(a+d)+(a−d)(a+d)c = (a-d)a + a(a+d) + (a-d)(a+d) c=a2−ad+a2+ad+a2−d2=3a2−d2c = a^2 - ad + a^2 + ad + a^2 - d^2 = 3a^2 - d^2 The product of the roots is (a−d)a(a+d)=a(a2−d2)=−481=−48(a-d)a(a+d) = a(a^2-d^2) = -\frac{48}{1} = -48. Substitute a=4a=4: 4(16−d2)=−48  ⟹  16−d2=−12  ⟹  d2=284(16-d^2) = -48 \implies 16-d^2 = -12 \implies d^2 = 28. Now substitute a=4a=4 and d2=28d^2=28 into the expression for cc: c=3(4)2−28=3(16)−28=48−28=20c = 3(4)^2 - 28 = 3(16) - 28 = 48 - 28 = 20.

Question 15

The polynomial P(x)=x3−2x2+3x−4=0P(x) = x^3 - 2x^2 + 3x - 4 = 0 has roots α,β,γ\alpha, \beta, \gamma. Which of the following polynomials has roots αβ,βγ,γα\alpha\beta, \beta\gamma, \gamma\alpha?

  1. x3+3x2+8x+16=0x^3 + 3x^2 + 8x + 16 = 0
  2. x3−3x2+2x−16=0x^3 - 3x^2 + 2x - 16 = 0
  3. x3−3x2+8x−16=0x^3 - 3x^2 + 8x - 16 = 0 (correct answer)
  4. x3−8x2+3x−16=0x^3 - 8x^2 + 3x - 16 = 0
Explanation: Let the new polynomial be Q(y)=y3−S1y2+S2y−S3=0Q(y) = y^3 - S_1 y^2 + S_2 y - S_3 = 0, where the roots are r1=αβ,r2=βγ,r3=γαr_1 = \alpha\beta, r_2 = \beta\gamma, r_3 = \gamma\alpha. From the original polynomial x3−2x2+3x−4=0x^3 - 2x^2 + 3x - 4 = 0, Vieta's formulas give: σ1=α+β+γ=−(−2)/1=2\sigma_1 = \alpha+\beta+\gamma = -(-2)/1 = 2 σ2=αβ+βγ+γα=3/1=3\sigma_2 = \alpha\beta+\beta\gamma+\gamma\alpha = 3/1 = 3 σ3=αβγ=−(−4)/1=4\sigma_3 = \alpha\beta\gamma = -(-4)/1 = 4 Now we find the coefficients for Q(y)Q(y): Sum of new roots: S1=r1+r2+r3=αβ+βγ+γα=σ2=3S_1 = r_1+r_2+r_3 = \alpha\beta+\beta\gamma+\gamma\alpha = \sigma_2 = 3. Product of new roots: S3=r1r2r3=(αβ)(βγ)(γα)=(αβγ)2=(σ3)2=42=16S_3 = r_1 r_2 r_3 = (\alpha\beta)(\beta\gamma)(\gamma\alpha) = (\alpha\beta\gamma)^2 = (\sigma_3)^2 = 4^2 = 16. Sum of products of new roots (2 at a time): S2=r1r2+r2r3+r3r1=(αβ)(βγ)+(βγ)(γα)+(γα)(αβ)S_2 = r_1 r_2 + r_2 r_3 + r_3 r_1 = (\alpha\beta)(\beta\gamma) + (\beta\gamma)(\gamma\alpha) + (\gamma\alpha)(\alpha\beta) S2=αβ2γ+αβγ2+α2βγ=αβγ(β+γ+α)=σ3σ1=(4)(2)=8S_2 = \alpha\beta^2\gamma + \alpha\beta\gamma^2 + \alpha^2\beta\gamma = \alpha\beta\gamma(\beta+\gamma+\alpha) = \sigma_3 \sigma_1 = (4)(2) = 8. Constructing the polynomial Q(y)Q(y): y3−(3)y2+(8)y−(16)=0y^3 - (3)y^2 + (8)y - (16) = 0 Replacing yy with xx gives x3−3x2+8x−16=0x^3 - 3x^2 + 8x - 16 = 0.

Question 16

The polynomial P(x)=x4−2x3+6x2+22x+13=0P(x) = x^4 - 2x^3 + 6x^2 + 22x + 13 = 0 has real coefficients. Given that 2+3i2+3i is a root, find the sum of the real roots of P(x)P(x).

  1. -4
  2. -2 (correct answer)
  3. 2
  4. 4
Explanation: Since the polynomial has real coefficients, complex roots must come in conjugate pairs. If r1=2+3ir_1 = 2+3i is a root, then r2=2−3ir_2 = 2-3i must also be a root. Let the four roots be r1,r2,r3,r4r_1, r_2, r_3, r_4. Let r3r_3 and r4r_4 be the other two roots. From Vieta's formulas, the sum of the roots is: r1+r2+r3+r4=−−21=2r_1+r_2+r_3+r_4 = -\frac{-2}{1} = 2 (2+3i)+(2−3i)+r3+r4=2(2+3i) + (2-3i) + r_3 + r_4 = 2 4+r3+r4=24 + r_3 + r_4 = 2 r3+r4=−2r_3 + r_4 = -2 This is the sum of the remaining two roots. To check if these roots are real, we can use the product of the roots: r1r2r3r4=131=13r_1 r_2 r_3 r_4 = \frac{13}{1} = 13 r1r2=(2+3i)(2−3i)=22−(3i)2=4−(−9)=13r_1 r_2 = (2+3i)(2-3i) = 2^2 - (3i)^2 = 4 - (-9) = 13 13(r3r4)=13  ⟹  r3r4=113(r_3 r_4) = 13 \implies r_3 r_4 = 1 We need to find two numbers r3,r4r_3, r_4 such that their sum is -2 and their product is 1. These are the roots of the quadratic equation t2−(sum)t+(product)=0t^2 - (\text{sum})t + (\text{product}) = 0, which is t2−(−2)t+1=0  ⟹  t2+2t+1=0t^2 - (-2)t + 1 = 0 \implies t^2+2t+1=0. (t+1)2=0(t+1)^2 = 0 The roots are t=−1t = -1 (a repeated root). So, r3=−1r_3 = -1 and r4=−1r_4 = -1. These roots are real. The sum of the real roots is r3+r4=−1+(−1)=−2r_3 + r_4 = -1 + (-1) = -2.

Question 17

The roots of the equation x3+px2+qx−1=0x^3 + px^2 + qx - 1 = 0 are α,β,γ\alpha, \beta, \gamma. Given that α+β+γ=2\alpha+\beta+\gamma = 2 and α2+β2+γ2=6\alpha^2+\beta^2+\gamma^2 = 6, find the value of qq.

  1. -2
  2. -1 (correct answer)
  3. 1
  4. 5
Explanation: From Vieta's formulas for x3+px2+qx−1=0x^3 + px^2 + qx - 1 = 0: Sum of roots: α+β+γ=−p\alpha+\beta+\gamma = -p Sum of products of roots (2 at a time): αβ+βγ+γα=q\alpha\beta+\beta\gamma+\gamma\alpha = q We are given that α+β+γ=2\alpha+\beta+\gamma = 2. Therefore, −p=2  ⟹  p=−2-p = 2 \implies p = -2. We need to find qq. We use the algebraic identity: (α+β+γ)2=α2+β2+γ2+2(αβ+βγ+γα)(\alpha+\beta+\gamma)^2 = \alpha^2+\beta^2+\gamma^2 + 2(\alpha\beta+\beta\gamma+\gamma\alpha) We can substitute the known values into this identity:
  • α+β+γ=2\alpha+\beta+\gamma = 2
  • α2+β2+γ2=6\alpha^2+\beta^2+\gamma^2 = 6
  • αβ+βγ+γα=q\alpha\beta+\beta\gamma+\gamma\alpha = q
So, (2)2=6+2(q)(2)^2 = 6 + 2(q). 4=6+2q4 = 6 + 2q −2=2q-2 = 2q q=−1q = -1

Question 18

A monic cubic polynomial P(x)=x3+ax2+bx+cP(x) = x^3+ax^2+bx+c has roots α,β,γ\alpha, \beta, \gamma. Given ∑α=6\sum \alpha = 6, ∑α2=14\sum \alpha^2 = 14, and ∑α3=36\sum \alpha^3 = 36, find the value of cc.

  1. -6 (correct answer)
  2. 6
  3. 11
  4. 18
Explanation: Let the polynomial be x3+ax2+bx+c=0x^3+ax^2+bx+c=0. From Vieta's formulas, α+β+γ=−a\alpha+\beta+\gamma = -a and αβ+βγ+γα=b\alpha\beta+\beta\gamma+\gamma\alpha=b and αβγ=−c\alpha\beta\gamma=-c. Given ∑α=6\sum \alpha = 6, we have a=−6a = -6. We use the identity ∑α2=(∑α)2−2(∑αβ)\sum \alpha^2 = (\sum \alpha)^2 - 2(\sum \alpha\beta). Substituting the given values: 14=(6)2−2b14 = (6)^2 - 2b. 14=36−2b  ⟹  2b=22  ⟹  b=1114 = 36 - 2b \implies 2b = 22 \implies b = 11. Now the polynomial is x3−6x2+11x+c=0x^3 - 6x^2 + 11x + c = 0. Since α,β,γ\alpha, \beta, \gamma are roots, they satisfy the equation: α3−6α2+11α+c=0\alpha^3 - 6\alpha^2 + 11\alpha + c = 0 β3−6β2+11β+c=0\beta^3 - 6\beta^2 + 11\beta + c = 0 γ3−6γ2+11γ+c=0\gamma^3 - 6\gamma^2 + 11\gamma + c = 0 Summing these three equations gives: (α3+β3+γ3)−6(α2+β2+γ2)+11(α+β+γ)+3c=0(\alpha^3+\beta^3+\gamma^3) - 6(\alpha^2+\beta^2+\gamma^2) + 11(\alpha+\beta+\gamma) + 3c = 0 ∑α3−6∑α2+11∑α+3c=0\sum \alpha^3 - 6\sum \alpha^2 + 11\sum \alpha + 3c = 0 Substitute the given sums: 36−6(14)+11(6)+3c=036 - 6(14) + 11(6) + 3c = 0 36−84+66+3c=036 - 84 + 66 + 3c = 0 18+3c=018 + 3c = 0 3c=−18  ⟹  c=−63c = -18 \implies c = -6. The polynomial is x3−6x2+11x−6=0x^3-6x^2+11x-6=0, and its roots are 1, 2, 3.

Question 19

Find the sum of the absolute values of the roots of the equation x4−13x2+36=0x^4 - 13x^2 + 36 = 0.

  1. 0
  2. 5
  3. 10 (correct answer)
  4. 13
Explanation: The equation x4−13x2+36=0x^4 - 13x^2 + 36 = 0 is a quadratic in x2x^2. Let y=x2y = x^2. The equation becomes y2−13y+36=0y^2 - 13y + 36 = 0. We can factor this quadratic: (y−4)(y−9)=0(y-4)(y-9) = 0 This gives two possible values for yy: y=4y=4 or y=9y=9. Now, substitute back x2=yx^2 = y: Case 1: x2=4  ⟹  x=±4=±2x^2 = 4 \implies x = \pm \sqrt{4} = \pm 2. Case 2: x2=9  ⟹  x=±9=±3x^2 = 9 \implies x = \pm \sqrt{9} = \pm 3. The four roots of the equation are −3,−2,2,3-3, -2, 2, 3. The sum of the absolute values of the roots is: ∣−3∣+∣−2∣+∣2∣+∣3∣=3+2+2+3=10| -3 | + | -2 | + | 2 | + | 3 | = 3 + 2 + 2 + 3 = 10 Distractor A (0) is the sum of the roots. Distractor B (5) is the sum of the positive roots. Distractor D (13) is the absolute value of the coefficient of x2x^2.

Question 20

The quadratic equation x2−4x+2=0x^2 - 4x + 2 = 0 has roots α\alpha and β\beta. What is the value of α3+β3\alpha^3 + \beta^3?

  1. 16
  2. 40 (correct answer)
  3. 48
  4. 64
Explanation: From the quadratic equation x2−4x+2=0x^2 - 4x + 2 = 0, we can use Vieta's formulas to find the sum and product of the roots α\alpha and β\beta. Sum of roots: α+β=−−41=4\alpha + \beta = -\frac{-4}{1} = 4. Product of roots: αβ=21=2\alpha \beta = \frac{2}{1} = 2. To find α3+β3\alpha^3 + \beta^3, we can use the identity: α3+β3=(α+β)3−3αβ(α+β)\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) Substitute the values we found: α3+β3=(4)3−3(2)(4)\alpha^3 + \beta^3 = (4)^3 - 3(2)(4) =64−24= 64 - 24 =40= 40 Alternatively, use the identity α3+β3=(α+β)(α2−αβ+β2)\alpha^3+\beta^3 = (\alpha+\beta)(\alpha^2 - \alpha\beta + \beta^2). First find α2+β2=(α+β)2−2αβ=42−2(2)=16−4=12\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = 4^2 - 2(2) = 16-4 = 12. Then α3+β3=(4)(12−2)=4(10)=40\alpha^3+\beta^3 = (4)(12-2) = 4(10) = 40.