Lesson Math circle level problems on finding polynomials with prescribed roots
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<H2>Math circle level problems on finding polynomials with prescribed roots</H2> <H3>Problem 1</H3>Find the polynomial with roots {{{alpha}}}, {{{beta}}} and {{{gamma}}}, if {{{alpha*beta*gamma}}} = 6, {{{alpha + beta + gamma}}} = 5, and {{{alpha^2 + beta^2 + gamma^2}}} =21. <B>Solution</B> <pre> We are given {{{alpha}}} + {{{beta}}} + {{{gamma}}} = 5, (1) {{{alpha^2}}} + {{{beta^2}}} + {{{gamma^2}}} = 21. (2) It implies {{{2*alpha*beta + 2*alpha*gamma + 2*beta*gamma}}} = {{{(alpha + beta + gamma)^2}}} - ({{{alpha^2}}} + {{{beta^2}}} + {{{gamma^2}}}) = {{{5^2}}} - {{{21}}} = 4. Hence, {{{alpha*beta + alpha*gamma + beta*gamma}}} = 2. (3) Since {{{alpha}}}, {{{beta}}} and {{{gamma}}} satisfy equations {{{alpha}}} + {{{beta}}} + {{{gamma}}} = 5, {{{alpha*beta + alpha*gamma + beta*gamma}}} = 2, {{{alpha}}}.{{{beta}}}.{{{gamma}}} = 6, then by the Vieta's theorem, {{{alpha}}}, {{{beta}}} and {{{gamma}}} are the roots of the cubic equation {{{x^3 -5x^2 + 2x - 6}}} = 0. <U>Answer</U>. The polynomial with roots {{{alpha}}}, {{{beta}}} and {{{gamma}}} satisfying given conditions is {{{x^3 -5x^2 + 2x - 6}}} = 0. </pre> <H3>Problem 2</H3>Let the roots of the equation x^3 -2x^2 -3x-7=0 be r, s, and t. Find an equation whose roots are r^2, s^2 and t^2. <B>Solution</B> <pre> The given equation is {{{x^3 - 2x^2 - 3x - 7}}} = 0 (1) Equation (1) has the roots r, s and t. Therefore, due to to Vieta's theorem r + s + t = 2, r*s + r*t + s*t = -3, r*s*t = 7. (2) Now, an equation with the roots {{{r^2}}}, {{{s^2}}} and {{{t^2}}} is {{{(x-r^2)*(x-s^2)*(x-t^2)}}} = 0. (3) By the Vieta's theorem (or by applying FOIL directly), the coefficients of the left side polynomial are {{{-(r^2 + s^2 + t^2)}}} at x^2; (4) {{{r^2*s^2 + r^2*t^2 + s^2*t^2}}} at x; and (5) {{{-r^2*s^2*t^2}}} as the constant term. (6) So, my task now is to express the coefficient (4), (5) and (6) via the coefficients (2) of the equation (1). Regarding {{{(r^2 + s^2 + t^2)}}}, it is easy: {{{r^2 + s^2 + t^2}}} = {{{(r + s + t)^2}}} - {{{2*(r*s + r*t + s*t)}}} = {{{2^2}}} - {{{2*(-3)}}} = 4 + 6 = 10. So, the coefficient at x^2 of the polynomial (3) is {{{-(r^2 + s^2 + t^2)}}} = -10. Regarding {{{-r^2*s^2*t^2}}}, it is easy, too : {{{r^2*s^2*t^2}}} = {{{(r*s*t)^2}}} = {{{7^2}}} = 49. So, the constant term of the polynomial (3) is {{{-(r^2*s^2*t^2)}}} = -49. Regarding {{{r^2*s^2 + r^2*s^2 + s^2*t^2}}}, it is slightly more long way : {{{r*s + r*t + s*t}}} = -3 of (2) implies (squaring both sides) 9 = {{{ r^2*s^2 + r^2*s^2 + s^2*t^2 + 2*r^2*s*t + 2*r*s^2*t + 2*r*s*t^2}}} = = {{{ r^2*s^2 + r^2*s^2 + s^2*t^2}}} + {{{2*(r*s*t)*( r + s + t)}}} = substituting the known values from (4) = = {{{ r^2*s^2 + r^2*s^2 + s^2*t^2}}} + 2*7*2, which implies {{{r^2*s^2 + r^2*s^2 + s^2*t^2}}} = 9 - 28 = -19. So, the coefficient at x of the polynomial (3) is {{{r^2*s^2 + r^2*s^2 + s^2*t^2}}} = -11. Thus we know all three coefficients of the polynomial (3) {{{-(r^2+ s^2+ t^2)}}} = -10 at x^2; {{{r^2* s^2+ r^2* t^2+ s^2* t^2}}} = -19 at x; and {{{-r^2* s^2* t^2}}} = -49 as the constant term. <U>Answer</U>. The polynomial equation under the question is {{{x^3 -10x^2 - 19x - 49}}} = 0. </pre> <H3>Problem 3</H3>The roots of the polynomial equation {{{2x^3 - 8x^2 + 3x + 5}}} = 0 are {{{alpha}}}, {{{beta}}} and {{{gamma}}}. Find the polynomial equation with roots {{{alpha^2}}}, {{{beta^2}}}, {{{gamma^2}}}. <B>Solution</B> <pre> The given equation {{{2x^3 - 8x^2 + 3x + 5}}} = 0 (1) is equivalent to {{{x^3 - 4x^2 + 1.5x + 2.5}}} = 0 (2) (all the coefficients of (1) are divided by 2) Equation (2) has the same roots {{{alpha}}}, {{{beta}}} and {{{gamma}}} as equation (1). Therefore, {{{x^3 - 4x^2 + 1.5x + 2.5}}} = {{{(x-alpha)*(x-beta)*(x-gamma)}}}, (3) and, according to Vieta's theorem {{{alpha + beta + gamma}}} = 4, {{{alpha*beta+alpha*gamma+beta*gamma}}} = 1.5, {{{alpha*beta*gamma}}} = -2.5. (4) Now, an equation with the roots {{{alpha^2}}}, {{{beta^2}}} and {{{gamma^2}}} is {{{(x-alpha^2)*(x-beta^2)*(x-gamma^2)}}} = 0. (5) By the Vieta's theorem (or by applying FOIL directly), the coefficients of the left side polynomial are {{{-(alpha^2+beta^2+gamma^2)}}} at x^2; (6) {{{alpha^2*beta^2+alpha^2*gamma^2+beta^2*gamma^2}}} at x; and (7) {{{-alpha^2*beta^2*gamma^2}}} as the constant term. (8) So, my task now is to express the coefficient (6), (7) and (8) via the coefficients (4) of the equation (2). Regarding {{{(alpha^2+beta^2+gamma^2)}}}, it is easy: {{{(alpha^2+beta^2+gamma^2)}}} = {{{(alpha+beta+gamma)^2-2*(alpha*beta+alpha*gamma+beta*gamma)}}} = {{{4^2 - 2*1.5}}} = 16-3 = 13. So, the coefficient at x^2 of the polynomial (5) is {{{-(alpha^2+beta^2+gamma^2)}}} = -13. Regarding {{{-alpha^2*beta^2*gamma^2}}}, it is easy, too : {{{alpha^2*beta^2*gamma^2}}} = {{{(alpha*beta*gamma)^2}}} = {{{(-2.5)^2}}} = 6.25. So, the constant term of the polynomial (5) is {{{-(alpha^2*beta^2*gamma^2)}}} = -6.25. Regarding {{{alpha^2*beta^2+alpha^2*gamma^2+beta^2*gamma^2}}}, it is slightly more long way : {{{alpha*beta+alpha*gamma+beta*gamma}}} = 1.5 of (4) implies (squaring both sides) 2.25 = {{{alpha^2*beta^2 + alpha^2*beta^2 + beta^2*gamma^2 + 2*alpha^2*beta*gamma + 2*alpha*beta^2*gamma + 2*alpha*beta*gamma^2}}} = = {{{alpha^2*beta^2 + alpha^2*beta^2 + beta^2*gamma^2}}} + {{{2*(alpha*beta*gamma)*(alpha+beta+gamma)}}} = substituting the known values from (4) = = {{{alpha^2*beta^2 + alpha^2*beta^2 + beta^2*gamma^2}}} + 2*(-2.5)*4, which implies {{{alpha^2*beta^2 + alpha^2*beta^2 + beta^2*gamma^2}}} = 2.25 + 20 = 22.25. Thus we know all three coefficients of the polynomial (5) {{{-(alpha^2+beta^2+gamma^2)}}} = -13 at x^2; {{{alpha^2*beta^2+alpha^2*gamma^2+beta^2*gamma^2}}} = 22.25 at x; and {{{-alpha^2*beta^2*gamma^2}}} = -6.25 as the constant term. <U>Answer</U>. The polynomial equation under the question is {{{x^3 -13x^2 + 22.25x - 6.25}}} = 0. </pre> <H3>Problem 4</H3>Use this identify tan4Q = {{{(4tanQ-4(tanQ)^3)/(1-6(tanQ)^2+(tanQ)^4)}}} to find the polynomial of least degree that has zeroes {{{(tan(pi/24))^2}}}, {{{(tan(7pi/24))^2}}}, {{{(tan(13pi/24))^2}}}, {{{(tan(19pi/24))^2)}}}. <B>Solution</B> So, according to the problem formulation, I should accept the given identity "as is", and based on it, to construct the polynomial. Since this problem is designed/intendent for advanced students, I will only show the major idea and the major steps without going in details. Let me start noticing that of four numbers {{{(tan(pi/24))^2}}}, {{{(tan(7pi/24))^2}}}, {{{(tan(13pi/24))^2}}}, {{{(tan(19pi/24))^2)}}}, the first is equal to the third, while the second is equal to the fourth. So, I will construct the polynomial which has two zeroes as the first and the second numbers of the four numbers listed: then the third and the fourth numbers will be the roots of this polynomial automatically . . . <pre> Take Q = {{{pi/24}}}. Then 4Q = {{{pi/6}}} and tan(4Q) = {{{tan(pi/6)}}} = {{{sqrt(3)/3}}}. According to the given identity, I have then {{{sqrt(3)/3}}} = {{{(4tan(Q)-4(tan(Q))^3)/(1-6(tan(Q))^2+(tan(Q))^4)}}} or {{{sqrt(3)*(1-6(tan(Q))^2+(tan(Q))^4)}}} = {{{3*(4tan(Q)-4(tan(Q))^3)}}}. (2) Next, in (2), I will replace (tan(Q))^2 by x everywhere (at each appearance). I will get then {{{sqrt(3)*(1 - 6x + x^2)}}} = {{{12*tan(Q)*(1-x)}}}. (3) Now, for {{{tan(pi/24)}}} there is the formula (the known expression) {{{tan(pi/24)}}} = {{{-2 - sqrt(3) + 2*sqrt(2 + sqrt(3))}}} (see the link https://brainly.in/question/9263846 ) and/or {{{tan(pi/24)}}} = {{{-2 + sqrt(2) - sqrt(3) + sqrt(6)}}} (see the link https://mathworld.wolfram.com/TrigonometryAnglesPi24.html) Replacing {{{tan(pi/24)}}} in formula (3) by any of these constant expressions (they are equal (!)), I get the polynomial of the degree 2 {{{sqrt(3)*(1 - 6x + x^2)}}} = {{{12c(1-x)}}}, (4) whose two roots are {{{(tan(pi/24))^2}}} and {{{(tan(7pi/24))^2}}}. 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