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Vieta's Formulas - IB Mathematics Analysis and Approaches

  1. 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?
    Answer: x3−3x2+8x−16=0x^3 - 3x^2 + 8x - 16 = 0 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.
  2. The equation x4−10x3+kx2−50x+24=0x^4 - 10x^3 + kx^2 - 50x + 24 = 0 has four real roots. The product of two of its roots is 2. Find the value of kk.
    Answer: 35 Let the roots be α,β,γ,δ\alpha, \beta, \gamma, \delta. We are given that αβ=2\alpha\beta = 2. From Vieta's formulas: 1. Product of all roots: αβγδ=24\alpha\beta\gamma\delta = 24. Since αβ=2\alpha\beta = 2, we have 2γδ=24  ⟹  γδ=122\gamma\delta = 24 \implies \gamma\delta = 12. 2. Sum of all roots: α+β+γ+δ=10\alpha+\beta+\gamma+\delta = 10. 3. Sum of products of roots (3 at a time): αβγ+αβδ+αγδ+βγδ=−(−50)=50\alpha\beta\gamma + \alpha\beta\delta + \alpha\gamma\delta + \beta\gamma\delta = -(-50) = 50. We can factor this as αβ(γ+δ)+γδ(α+β)=50\alpha\beta(\gamma+\delta) + \gamma\delta(\alpha+\beta) = 50. Substitute αβ=2\alpha\beta=2 and γδ=12\gamma\delta=12: 2(γ+δ)+12(α+β)=502(\gamma+\delta) + 12(\alpha+\beta) = 50. Let S1=α+βS_1 = \alpha+\beta and S2=γ+δS_2 = \gamma+\delta. We have a system of linear equations: (i) S1+S2=10S_1 + S_2 = 10 (ii) 12S1+2S2=50  ⟹  6S1+S2=2512S_1 + 2S_2 = 50 \implies 6S_1 + S_2 = 25 Subtract (i) from (ii): 5S1=15  ⟹  S1=35S_1 = 15 \implies S_1 = 3. So α+β=3\alpha+\beta = 3. Then from (i), 3+S2=10  ⟹  S2=73 + S_2 = 10 \implies S_2 = 7. So γ+δ=7\gamma+\delta = 7. 4. We need to find kk, which is the sum of products of roots (2 at a time): k=αβ+αγ+αδ+βγ+βδ+γδk = \alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \gamma\delta. We can regroup this sum: k=(αβ)+(γδ)+(αγ+αδ+βγ+βδ)k = (\alpha\beta) + (\gamma\delta) + (\alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta) k=(αβ)+(γδ)+α(γ+δ)+β(γ+δ)k = (\alpha\beta) + (\gamma\delta) + \alpha(\gamma+\delta) + \beta(\gamma+\delta) k=(αβ)+(γδ)+(α+β)(γ+δ)k = (\alpha\beta) + (\gamma\delta) + (\alpha+\beta)(\gamma+\delta) Substitute the values we found: k=2+12+(3)(7)=14+21=35k = 2 + 12 + (3)(7) = 14 + 21 = 35.
  3. 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?
    Answer: pq=rpq = r 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.
  4. 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.
    Answer: 20 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.
  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}?
    Answer: 0 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.
  6. 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.
    Answer: -6 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.
  7. Find the sum of the absolute values of the roots of the equation x4−13x2+36=0x^4 - 13x^2 + 36 = 0.
    Answer: 10 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.
  8. 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.
    Answer: 9/4 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.
  9. 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)?
    Answer: -10 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.
  10. 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.
    Answer: -1 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
  11. The cubic equation 2x3−8x2+kx−9=02x^3 - 8x^2 + kx - 9 = 0 has roots α,β,γ\alpha, \beta, \gamma. Given that one root is the sum of the other two, find the value of kk.
    Answer: 12.5 Let the roots be α,β,γ\alpha, \beta, \gamma. We are given a condition such as α=β+γ\alpha = \beta + \gamma. From Vieta's formulas, the sum of the roots is α+β+γ=−−82=4\alpha + \beta + \gamma = -\frac{-8}{2} = 4. Substituting the given condition into the sum of roots equation: (β+γ)+β+γ=4  ⟹  2(β+γ)=4  ⟹  β+γ=2(\beta + \gamma) + \beta + \gamma = 4 \implies 2(\beta + \gamma) = 4 \implies \beta + \gamma = 2. This means α=2\alpha = 2. Since α=2\alpha=2 is a root of the polynomial, it must satisfy the equation. Substitute x=2x=2 into the equation: 2(2)3−8(2)2+k(2)−9=02(2)^3 - 8(2)^2 + k(2) - 9 = 0 2(8)−8(4)+2k−9=02(8) - 8(4) + 2k - 9 = 0 16−32+2k−9=016 - 32 + 2k - 9 = 0 −16+2k−9=0-16 + 2k - 9 = 0 −25+2k=0-25 + 2k = 0 2k=25  ⟹  k=12.52k = 25 \implies k = 12.5
  12. 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?
    Answer: 40 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.
  13. 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?
    Answer: \{1, 2, 4, 8\} 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.
  14. 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.
    Answer: 3 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.
  15. 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}.
    Answer: 1 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.
  16. 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.
    Answer: 27 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
  17. The polynomial P(x)=x4−8x3+18x2+ax+b=0P(x) = x^4 - 8x^3 + 18x^2 + ax + b = 0 has two distinct double roots. Find the value of aa.
    Answer: -8 Let the two distinct double roots be α\alpha and β\beta. The roots of the polynomial are α,α,β,β\alpha, \alpha, \beta, \beta. From Vieta's formulas: 1. Sum of roots: α+α+β+β=2α+2β=−(−8)/1=8\alpha+\alpha+\beta+\beta = 2\alpha + 2\beta = -(-8)/1 = 8. This simplifies to α+β=4\alpha+\beta = 4. 2. Sum of products of roots (2 at a time): The pairs are α2,αβ,αβ,αβ,αβ,β2\alpha^2, \alpha\beta, \alpha\beta, \alpha\beta, \alpha\beta, \beta^2. The sum is α2+4αβ+β2=18\alpha^2 + 4\alpha\beta + \beta^2 = 18. We can write α2+β2=(α+β)2−2αβ\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta. Substituting this gives: (α+β)2−2αβ+4αβ=18  ⟹  (α+β)2+2αβ=18(\alpha+\beta)^2 - 2\alpha\beta + 4\alpha\beta = 18 \implies (\alpha+\beta)^2 + 2\alpha\beta = 18. Substitute α+β=4\alpha+\beta=4: 42+2αβ=18  ⟹  16+2αβ=18  ⟹  2αβ=2  ⟹  αβ=14^2 + 2\alpha\beta = 18 \implies 16 + 2\alpha\beta = 18 \implies 2\alpha\beta = 2 \implies \alpha\beta = 1. 3. Sum of products of roots (3 at a time): The triplets are α2β,α2β,αβ2,αβ2\alpha^2\beta, \alpha^2\beta, \alpha\beta^2, \alpha\beta^2. The sum is 2α2β+2αβ22\alpha^2\beta + 2\alpha\beta^2. This sum is equal to −a/1=−a-a/1 = -a. −a=2α2β+2αβ2=2αβ(α+β)-a = 2\alpha^2\beta + 2\alpha\beta^2 = 2\alpha\beta(\alpha+\beta) Substitute the values we found for α+β\alpha+\beta and αβ\alpha\beta: −a=2(1)(4)=8-a = 2(1)(4) = 8 Therefore, a=−8a = -8.
  18. Consider the two polynomial equations P(x)=x3−7x+6=0P(x) = x^3 - 7x + 6 = 0 and Q(x)=x3−2x2−5x+6=0Q(x) = x^3 - 2x^2 - 5x + 6 = 0. Find the sum of the roots that are not common to both equations.
    Answer: 0 Let the roots of P(x)P(x) be r1,r2,r3r_1, r_2, r_3 and the roots of Q(x)Q(x) be s1,s2,s3s_1, s_2, s_3. From Vieta's formulas for P(x)P(x): r1+r2+r3=−01=0r_1+r_2+r_3 = -\frac{0}{1} = 0. From Vieta's formulas for Q(x)Q(x): s1+s2+s3=−−21=2s_1+s_2+s_3 = -\frac{-2}{1} = 2. To find the common roots, we can set P(x)=Q(x)P(x) = Q(x), since for a common root xcx_c, P(xc)=0P(x_c)=0 and Q(xc)=0Q(x_c)=0. x3−7x+6=x3−2x2−5x+6x^3 - 7x + 6 = x^3 - 2x^2 - 5x + 6 −7x=−2x2−5x-7x = -2x^2 - 5x 2x2−2x=02x^2 - 2x = 0 2x(x−1)=02x(x-1) = 0 This implies that any common roots must be either x=0x=0 or x=1x=1. Test x=0x=0: P(0)=6≠0P(0) = 6 \neq 0. So x=0x=0 is not a common root. Test x=1x=1: P(1)=1−7+6=0P(1) = 1-7+6=0 and Q(1)=1−2−5+6=0Q(1) = 1-2-5+6=0. So x=1x=1 is a common root. Let the common root be α=1\alpha=1. The roots of P(x)P(x) are α,β,γ\alpha, \beta, \gamma. The sum is α+β+γ=0\alpha+\beta+\gamma = 0. So 1+β+γ=0  ⟹  β+γ=−11+\beta+\gamma=0 \implies \beta+\gamma = -1. The roots of Q(x)Q(x) are α,δ,ϵ\alpha, \delta, \epsilon. The sum is α+δ+ϵ=2\alpha+\delta+\epsilon=2. So 1+δ+ϵ=2  ⟹  δ+ϵ=11+\delta+\epsilon=2 \implies \delta+\epsilon = 1. The non-common roots are β,γ,δ,ϵ\beta, \gamma, \delta, \epsilon. We want to find their sum: Sum=(β+γ)+(δ+ϵ)=−1+1=0\text{Sum} = (\beta+\gamma) + (\delta+\epsilon) = -1 + 1 = 0
  19. 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).
    Answer: -2 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.

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