Lesson Miscellaneous problems on Vieta's theorem
Algebra
->
Customizable Word Problem Solvers
->
Evaluation
-> Lesson Miscellaneous problems on Vieta's theorem
Log On
Ad:
Over 600 Algebra Word Problems at edhelper.com
Word Problems: Evaluation, Substitution
Word
Solvers
Solvers
Lessons
Lessons
Answers archive
Answers
Source code of 'Miscellaneous problems on Vieta's theorem'
This Lesson (Miscellaneous problems on Vieta's theorem)
was created by by
ikleyn(52776)
:
View Source
,
Show
About ikleyn
:
<H2>Miscellaneous problems on Vieta's theorem</H2> <H3>Problem 1</H3>Let 'a' and 'b' be the roots of the quadratic equation 5x^2 - 23x - 4 = 0. Compute 1/a^2 + 1/b^2. <B>Solution</B> <pre> The given equation is the standard form quadratic equation 5x^2 - 23x - 4 = 0. (1) Therefore, according to Vieta's theorem, a + b = {{{23/5}}}, (2) ab = {{{-4/5}}}. (3) Next, {{{1/a^2}}} + {{{1/b^2}}} = {{{(a^2 + b^2)/(a^2*b^2)}}}. (4) The numerator in (4) is a^2 + b^2 = (a^2 + 2ab + b^2) - 2ab = (a+b)^2 - 2ab = replace here a+b by {{{23/5}}} and replace ab by {{{-4/5}}} based on (2),(3) and continue = {{{(23/5)^2}}} - {{{2*(-4/5)}}} = {{{529/25}}} + {{{8/5}}} = {{{(529+5*8)/25}}} = {{{569/25}}}. Therefore {{{1/a^2}}} + {{{1/b^2}}} = {{{(a^2 + b^2)/(a^2*b^2)}}} = {{{((569/25))/((16/25))}}} = {{{569/16}}} = 35{{{9/16}}} = 35.5625. <U>ANSWER</U> </pre> <H3>Problem 2</H3>Let 'a' and ' b' be the solutions to 6x^2 - 16x + 7 = 0. Find a^2/b + b^2/a. <B>Solution</B> <pre> The given equation is the standard form quadratic equation 6x^2 - 16x + 7 = 0. (1) According to Vieta's theorem, if "a" and "b" are the roots of this quadratic equation, then a + b = {{{16/6}}} = {{{8/3}}}, (2) ab = {{{7/6}}}. (3) Now, {{{a^2/b}}} + {{{b^2/a}}} = {{{(a^3+b^3)/ab}}}. The numerator is a^3 + b^3 = (a+b)*(a^2 - ab + b^2) = (a+b)*((a^2+2ab+b^2)-3ab) = (a+b)*((a+b)^2-3ab) = (a+b)^3 - 3(a+b)*(ab). Now replace here (a+b) by 8/3 based on (2) and replace ab by 7/6 based on (3). You will get a^3 + b^3 = {{{(8/3)^3}}} - {{{3*(8/3)*(7/6)}}} = {{{512/27}}} - {{{(8/3)*(7/2)}}} = {{{512/27}}} - {{{28/3}}} = {{{(512-9*28)/27}}} = {{{260/27}}}. Therefore, {{{a^2/b + b^2/a}}} = {{{((260/27))/((7/6))}}} = {{{(260*6)/(27*7)}}} = {{{1560/189}}} = {{{520/63}}}. <U>ANSWER</U>. {{{a^2/b + b^2/a}}} = {{{520/63}}}. </pre> <H3>Problem 3</H3>Let 'a' and 'b' be the roots of the quadratic x^2 - 5x + 3 = 0. Find the quadratic whose roots are a^2/b and b^2/a. <B>Solution</B> <pre> According to Vieta's theorem, we have a + b = 5, ab = 3. Next, {{{a^2/b+b^2/a}}} = {{{(a^3+b^3)/ab}}}. In the numerator, a^3 + b^3 = (a+b)*(a^2-ab+b^2) = (a+b)*((a^2 +2ab+b^2)-3ab) = (a+b)*((a+b^2-3ab) = (a+b)^3 - 3ab*(a+b) = substitute here a+b = 5, ab = 3 and continue = 5^3 - 3*3*5 = 125-45 = 80. Therefore, {{{a^2/b}}} + {{{b^2/a}}} = {{{(a^3+b^3)/ab}}} = {{{80/3}}}. Also, {{{a^2/b}}}.{{{b^2/a}}} = {{{(a^2*b^2)/(ab)}}} = ab = 3. So, due to Vieta's theorem (again), the desired quadratic equation has the coefficient {{{-80/3}}} at x and the constant term 3. <U>ANSWER</U>. The desired quadratic equation is x^2 - {{{(80/3)x}}} + 3 = 0. </pre> <H3>Problem 4</H3>There are integers 'b', 'c' for which both roots of the polynomial x^2 - x - 3 are also roots of the polynomial x^3 - bx^2 - c. Determine the ordered pair (b,c). <B>Solution</B> <pre> Let p and q be the roots of the polynomial x^2 - x - 3. Due to Vieta's theorem, p + q = 1, (1) pq = -3. (2) According to the problem, p and q are also the roots of the polynomial x^3 - bx^2 - c. Let r be the third root of this polynomial. Then, due to Vieta's theorem for polynomial x^3 - bx^2 - c p + q + r = b, (3) p*q + p*r + q*r = 0, (4) (coefficient at x in polynomial x^3 - bx^2 - c) p*q*r = c. (5) In (3), replace p+q by 1, based on (3). In (5), replace p*q by -3, based on (2). Then from (3) an (4) you will have 1 + r = b, (6) -3r = c, (7) In equation (4), replace p*q by -3, based on (2). Then equation (4) takes the form -3 + pr + qr = 0, or p*r + q*r = 3, (p + q)*r = 3. In the last equation, replace (p + q) by 1, based on (1). You will get 1*r = 3, i.e. r = 3. Now from (6) b = 1 + r = 1 + 3 = 4; from (7) c = -3r = -3*3 = -9. <U>ANSWER</U>. In polynomial x^3 - bx^2 - c, coefficients are b = 4, c = -9. </pre> <H3>Problem 5</H3>Let 'p', 'q', 'r', and 's' be the roots of g(x) = x^4 + 2x^3 + 16x^2 + 20x - 31. Compute p^2*qrs + pq^2*rs + pqr^2*s + pqrs^2. <B>Solution</B> <pre> Notice that p^2*qrs + pq^2*rs + pqr^2*s + pqrs^2 = pqrs*(p + q + r + s). Also notice that the leading coefficient of the polynomial at x^4 is 1. Due to Vieta's theorem pqrs = -31 (the product of the roots is equal to the constant term) p + q + r + s = -2 (the sum of the roots is the coefficient at x^3 with the opposite sign). Therefore, p^2*qrs + pq^2*rs + pqr^2*s + pqrs^2 = (-31)*(-2) = 62. <U>ANSWER</U>. p^2*qrs + pq^2*rs + pqr^2*s + pqrs^2 = 62. </pre> <H3>Problem 6</H3>Let 'p', 'q', 'r', and 's' be the roots of x^4 + 2x^3 + 16x^2 + 20x - 31. Compute p^2 + q^2 + r^2 + s^2. <B>Solution</B> <pre> Notice that p^2 + q^2 + r^2 + s^2 = (p + q + r + s)^2 - 2*(pq + pr +ps + qr + qs + rs) (standard decomposition for the square of a sum) Also notice that the leading coefficient of the polynomial at x^4 is 1. Due to Vieta's theorem pq + pr + ps + qr + qs + rs = 16 (the sum of in-pairs products of the roots is the coefficient at x^2) p + q + r + s = -2 (the sum of the roots is the coefficient at x^3 with the opposite sign). Therefore, p^2 + q^2 + r^2 + s^2 = (-2)^2 - 2*16 = 4 - 32 = -28. <U>ANSWER</U>. p^2 + q^2 + r^2 + s^2 = -28. (by the way, since the sum of squares is negative, it means that some roots are complex numbers). </pre> My other lessons on Evaluating expressions in this site are - <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/HOW-TO-evaluate-expressions-involving-x%2Binv%28x%29-x2%2Binv%28x2%29-and-x%5E3%2Binv%28x%5E3%29.lesson>HOW TO evaluate expressions involving {{{(x + 1/x)}}}, {{{(x^2+1/x^2)}}} and {{{(x^3+1/x^3)}}}</A> - <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/Advanced-lesson-on-evaluating-expressions.lesson>Advanced lesson on evaluating expressions</A> - <A HREF=http://www.algebra.com/algebra/homework/word/evaluation/HOW-TO-evaluate-functions-of-roots-of-a-square-equation.lesson>HOW TO evaluate functions of roots of a square equation</A> - <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/HOW-TO-evaluate-functions-of-roots-of-a-cubic-and-quartic-equation.lesson>HOW TO evaluate functions of roots of a cubic and quartic equation</A> - <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/Problems-on-Vieta%27s-formulas.lesson>Problems on Vieta's formulas</A> - <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/Advanced-problems-on-Vieta%27s-formulas.lesson>Advanced problems on Vieta's theorem</A> - <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/Evaluating-expressions-that-contain-infinitely-many-square-roots.lesson>Evaluating expressions that contain infinitely many square roots</A> - <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/Solving-equations-that-contain-infinitely-many-radicals.lesson>Solving equations that contain infinitely many radicals</A> - <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/Problems-on-evaluating-in-Geometry.lesson>Problems on evaluating in Geometry</A> - <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/Evaluating-trigonometric-expressions-.lesson>Evaluating trigonometric expressions</A> - <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/Evaluate-the-sum-of-the-coefficients-of-a-polynomial.lesson>Evaluate the sum of the coefficients of a polynomial</A> - <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/Miscellaneous-evaluating-problems.lesson>Miscellaneous evaluating problems</A> - <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/Advanced-evaluating-problems.lesson>Advanced evaluating problems</A> - <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/Lowering-a-degree-method.lesson>Lowering a degree method</A> - <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/Find-the-number-of-factorable-quadratic-polynomials-of-special-form.lesson>Find the number of factorable quadratic polynomials of special form</A> - <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/Evaluating-a-function-defined-by-functional-equation.lesson>Evaluating a function defined by functional equation</A> - <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/Math-circle-level-problems-on-evaluating-expressions.lesson>Math circle level problems on evaluating expressions</A> - <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/Math-circle-level-problem-on-finding-polynomial-with-assigned-roots.lesson>Math circle level problems on finding polynomials with prescribed roots</A> - <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/Math-Olympiad-level-problem-on-evaluating-of-a-9-degree-polynomial.lesson>Math Olympiad level problem on evaluating a 9-degree polynomial</A> - <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/Upper-league-problem-on-evaluating-the-sum.lesson>Upper league problem on evaluating the sum</A> - <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/Finding-coefficients-of-decomposition-of-a-rational-function.lesson>Finding coefficients of decomposition of a rational function</A> - <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/Upper-level-problem-on-evaluating-an-expression-of-polynomial-roots.lesson>Upper level problem on evaluating an expression of polynomial roots</A> - <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/A-truly-miraculous-evaluating-problem.lesson>A truly miraculous evaluating problem with a truly miraculous solution</A> - <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/Entertainment-problems-on-evaluating-expressions.lesson>Entertainment problems on evaluating expressions</A> - <A HREF=https://www.algebra.com/algebra/homework/word/evaluation/OVERVIEW-of-lessons-on-Evaluation-Substitution.lesson>OVERVIEW of lessons on Evaluating expressions</A> Use this file/link <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-I - YOUR ONLINE TEXTBOOK</A> to navigate over all topics and lessons of the online textbook ALGEBRA-I.
