Lesson Entertainment problems on evaluating expressions
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<H2>Entertainment problems on evaluating expressions</H2> <H3>Problem 1</H3>If {{{x^2}}} - {{{x}}} + {{{1}}} = {{{0}}}, find {{{x^2020}}} + {{{x^1010}}} - {{{1}}}. <B>Solution</B> <pre> Let {{{x^2}}} - {{{x}}} + {{{1}}} = {{{0}}}. (1) Multiply both sides by (x+1). You will get {{{x^3}}} + {{{1}}} = {{{0}}}, or {{{x^3}}} = {{{-1}}}. +---------------------------------------------------------------+ | So, x is a complex number, which is a cubic root of -1. | +---------------------------------------------------------------+ 2020 = 673*3 + 1. Therefore {{{x^2020}}} = {{{((x^3)^673)*x}}} = {{{((-1)^673)*x}}}= -x. Next, 1010 = 336*3 + 2. Therefore {{{x^1010}}} = {{{((x^3)^336)*x^2}}} = {{{((-1)^336)*x^2}}} = {{{x^2}}}. Thus, {{{x^2020}}} + {{{x^1010}}} = {{{-x}}} + {{{x^2}}} = {{{x^2}}} - {{{x}}}. But, due to (1), {{{x^2}}} - {{{x}}} = {{{-1}}}. Therefore, {{{x^2020}}} + {{{x^1010}}} = {{{-1}}}, or {{{x^2020}}} + {{{x^1010}}} + {{{1}}} = 0. Now subtract 2 from both sides of the last equality. You will get {{{x^2020}}} + {{{x^1010}}} - {{{1}}} = {{{-2}}}. At this point, the solution is complete. <U>ANSWER</U>. If {{{x^2}}} - {{{x}}} + {{{1}}} = {{{0}}}, then {{{x^2020}}} + {{{x^1010}}} - {{{1}}} is equal to -2. </pre> <H3>Problem 2</H3>If 2^a = 3, 3^b = 2, find 1/(a+1) + 1/(b+1). <B>Solution</B> <pre> If 2^a = 3 and 3^b = 2, then 2^(ab) = (2^a)^b = 3^b = 2, which implies ab = 1. Now, {{{1/(a+1)}}} + {{{1/(b+1)}}} = {{{((a+1) + (b+1))/((a+1)*(b+1))}}} = = {{{(a + 1 + b + 1)/(ab + a + b + 1)}}} = {{{(a+b+2)/(1 + a + b + 1)}}} = {{{(a+b+2)/(a+b+2)}}} = 1. <U>ANSWER</U>. If 2^a = 3, 3^b = 2, then {{{1/(a+1)}}} + {{{1/(b+1)}}} = 1. </pre> <H3>Problem 3</H3>Let p, q, r, and s be the roots of g(x) = 3x^4 - 8x^3 + 5x^2 + 2x - 17 - 2x^4 + 10x^3 + 11x^2 + 18x - 14. Compute 1/p + 1/q + 1/r + 1/s. <B>Solution</B> <pre> Notice that since the constant term is not zero, no one root p, q, r or s is zero. Reduce the given equation to the standard form combining like terms. You will get g(x) = x^4 + 2x^3 + 16x^2 + 20x - 31. (1) Divide this reduced equation by x^4. You will get another polynomial (like a sock turned inside out) {{{1}}} + {{{2/x}}} + {{{16/x^2}}} + {{{20/x^3}}} - {{{31/x^4}}}. (2) If some value p, q, r, s is the root to polynomial (1), then 1/p, 1/q, 1/r, 1/s is the zero of function (2). Let's consider now the polynomial h(y) = 1 + 2y + 16y^2 + 20y^3 - 31y^4. (3) Compare (3) with (2) and recognize that (3) is the same expression as (2) with replaced '1/x' by 'y'. Since p, q, r and s are the roots for polynomial (1), 1/p, 1/q, 1/r and 1/s are the roots for polynomial (3). Now apply Vieta's theorem and find that the sum of the roots 1/p, 1/q, 1/r, 1/s is equal to the coefficient 20 at y^3 in polynomial (3), divided by the leading coefficient -31 at y^4, taken with the opposite sign 1/p + 1/q + 1/r + 1/s = {{{-20/(-31)}}} = {{{20/31}}}. At this point, the problem is solved in full, without making cumbersome calculations. <U>ANSWER</U>. 1/p + 1/q + 1/r + 1/s = {{{20/31}}}. </pre> Isn't it beautiful ? This method is called "turning a polynomial inside out". Turning a polynomial inside out can be done mentally, so one can write an answer immediately, without making/writing these reasons on paper. If you show this focus-pocus to your teacher/professor or at the interview, the other side will be shocked to see such an elegant solution. 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