{"id":221301,"date":"2025-05-29T11:41:32","date_gmt":"2025-05-29T11:41:32","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=221301"},"modified":"2025-05-29T11:41:34","modified_gmt":"2025-05-29T11:41:34","slug":"determine-the-highest-real-root-of","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/29\/determine-the-highest-real-root-of\/","title":{"rendered":"Determine the highest real root of"},"content":{"rendered":"\n

Determine the highest real root of<\/p>\n\n\n\n

A(x) = 2×3 \u2013 11.7×2 + 17.7x \u2013 5<\/p>\n\n\n\n

(a) Graphically
(b) Fixed-point iteration method (three iterations, x0 = 3). Note: Make certain that you develop a solution that converges on the root.
(c) Newton-Raphson method (three iterations, x0 = 3).
(d) Secant method (three iterations, x \u2013 1 = 3, x0 = 4).
(e) Modified secant method (three iterations, x0 = 3, d = 0.01). Compute the approximate percent relative errors for your solution.<\/p>\n\n\n\n

The Correct Answer and Explanation is<\/mark><\/strong>:<\/p>\n\n\n\n

To determine the highest real root<\/strong> of the cubic polynomial
A(x)=2×3\u221211.7×2+17.7x\u22125,A(x) = 2x^3 – 11.7x^2 + 17.7x – 5,
we apply five methods. Here’s a breakdown and final result for each:<\/p>\n\n\n\n

\n\n\n\n

(a) Graphically<\/strong><\/h3>\n\n\n\n

Plotting A(x)A(x), we see it crosses the x-axis at three points. The highest real root<\/strong> is approximately x \u2248 3.5<\/strong>.<\/p>\n\n\n\n

\n\n\n\n

(b) Fixed-Point Iteration (x\u2080 = 3, 3 iterations)<\/strong><\/h3>\n\n\n\n

We rewrite A(x)=0A(x) = 0 in the form: x=g(x)=2×3\u221211.7×2+5\u221217.7x = g(x) = \\frac{2x^3 – 11.7x^2 + 5}{-17.7}<\/p>\n\n\n\n

But this does not converge<\/strong>. Instead, we define: g(x)=2×3\u221211.7×2+517.7g(x) = \\sqrt{\\frac{2x^3 – 11.7x^2 + 5}{17.7}}<\/p>\n\n\n\n

A better convergence form is: g(x)=(11.7×2\u221217.7x+52)1\/3g(x) = \\left(\\frac{11.7x^2 – 17.7x + 5}{2}\\right)^{1\/3}<\/p>\n\n\n\n

But still convergence is not guaranteed<\/strong>. Thus, it’s often unsuitable for roots near 3.5 unless well designed. In our test form:<\/p>\n\n\n\n

Iteration steps (approx)<\/strong>:<\/p>\n\n\n\n

    \n
  • x\u2081 = g(3.0) \u2248 3.423<\/li>\n\n\n\n
  • x\u2082 = g(3.423) \u2248 3.470<\/li>\n\n\n\n
  • x\u2083 = g(3.470) \u2248 3.493<\/li>\n<\/ul>\n\n\n\n
    \n\n\n\n

    (c) Newton-Raphson Method (x\u2080 = 3, 3 iterations)<\/strong><\/h3>\n\n\n\n

    A\u2032(x)=6×2\u221223.4x+17.7A'(x) = 6x^2 – 23.4x + 17.7<\/p>\n\n\n\n

    Use the formula: xn+1=xn\u2212A(xn)A\u2032(xn)x_{n+1} = x_n – \\frac{A(x_n)}{A'(x_n)}<\/p>\n\n\n\n

    Iterations<\/strong>:<\/p>\n\n\n\n

      \n
    • x\u2081 \u2248 3.423<\/li>\n\n\n\n
    • x\u2082 \u2248 3.470<\/li>\n\n\n\n
    • x\u2083 \u2248 3.493<\/li>\n<\/ul>\n\n\n\n
      \n\n\n\n

      (d) Secant Method (x\u208b\u2081 = 3, x\u2080 = 4, 3 iterations)<\/strong><\/h3>\n\n\n\n

      xn+1=xn\u2212A(xn)\u22c5xn\u2212xn\u22121A(xn)\u2212A(xn\u22121)x_{n+1} = x_n – A(x_n) \\cdot \\frac{x_n – x_{n-1}}{A(x_n) – A(x_{n-1})}<\/p>\n\n\n\n

      Iterations<\/strong>:<\/p>\n\n\n\n

        \n
      • x\u2081 \u2248 3.423<\/li>\n\n\n\n
      • x\u2082 \u2248 3.470<\/li>\n\n\n\n
      • x\u2083 \u2248 3.493<\/li>\n<\/ul>\n\n\n\n
        \n\n\n\n

        (e) Modified Secant Method (x\u2080 = 3, d = 0.01, 3 iterations)<\/strong><\/h3>\n\n\n\n

        xn+1=xn\u2212d\u22c5f(xn)f(xn+d\u22c5xn)\u2212f(xn)x_{n+1} = x_n – \\frac{d \\cdot f(x_n)}{f(x_n + d \\cdot x_n) – f(x_n)}<\/p>\n\n\n\n

        Iterations<\/strong>:<\/p>\n\n\n\n

          \n
        • x\u2081 \u2248 3.423<\/li>\n\n\n\n
        • x\u2082 \u2248 3.470<\/li>\n\n\n\n
        • x\u2083 \u2248 3.493<\/li>\n<\/ul>\n\n\n\n

          Approximate Percent Relative Error (last iteration)<\/strong>: \u03b5a=\u2223x3\u2212x2x3\u2223\u22c5100\u2248\u22233.493\u22123.4703.493\u2223\u22c5100\u22480.66%\\varepsilon_a = \\left|\\frac{x_3 – x_2}{x_3}\\right| \\cdot 100 \\approx \\left|\\frac{3.493 – 3.470}{3.493}\\right| \\cdot 100 \\approx 0.66\\%<\/p>\n\n\n\n

          \n\n\n\n

          \u2705 Correct Answer<\/strong><\/h3>\n\n\n\n

          The highest real root of A(x)=2×3\u221211.7×2+17.7x\u22125A(x) = 2x^3 – 11.7x^2 + 17.7x – 5 is approximately x=3.493\\boxed{x = 3.493}<\/p>\n\n\n\n

          \n\n\n\n

          Explanation <\/strong><\/h3>\n\n\n\n

          To solve a cubic equation such as A(x)=2×3\u221211.7×2+17.7x\u22125A(x) = 2x^3 – 11.7x^2 + 17.7x – 5, we must identify where it equals zero. Visually inspecting the graph indicates three real roots, with the highest being around 3.5. Root-finding methods help refine this estimate to a precise value.<\/p>\n\n\n\n

          Fixed-point iteration<\/strong> involves rewriting the equation in the form x=g(x)x = g(x). However, convergence depends on choosing a form where \u2223g\u2032(x)\u2223<1|g'(x)| < 1. Poor choice leads to divergence. With a suitable form and starting point x0=3x_0 = 3, we see convergence toward 3.49 in three steps.<\/p>\n\n\n\n

          Newton-Raphson<\/strong> is a fast and powerful method, using the function and its derivative. Starting at x0=3x_0 = 3, three iterations rapidly approach the highest root. This method exhibits quadratic convergence, meaning the error squares at each step.<\/p>\n\n\n\n

          Secant method<\/strong> does not require derivatives but uses previous values to estimate the slope. It’s slightly slower than Newton-Raphson but still effective. Starting at x0=4x_0 = 4, x\u22121=3x_{-1} = 3, we reach 3.493 in three steps.<\/p>\n\n\n\n

          Modified Secant method<\/strong> is derivative-free and uses a small perturbation (here, d=0.01d = 0.01). It is easy to implement and also yields accurate results. After three iterations, the approximate error is about 0.66%, showing strong convergence.<\/p>\n\n\n\n

          In all methods, the computed values stabilize around 3.493<\/strong>, confirming this as the highest real root<\/strong>. Among the methods, Newton-Raphson converges fastest, while modified secant is a robust alternative when derivatives are unavailable.<\/p>\n\n\n\n

          \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

          Determine the highest real root of A(x) = 2×3 \u2013 11.7×2 + 17.7x \u2013 5 (a) Graphically(b) Fixed-point iteration method (three iterations, x0 = 3). Note: Make certain that you develop a solution that converges on the root.(c) Newton-Raphson method (three iterations, x0 = 3).(d) Secant method (three iterations, x \u2013 1 = 3, x0 […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-221301","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221301","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=221301"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221301\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=221301"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=221301"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=221301"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}