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Problems on quadratic roots: ISI MMA 2018 Question 9

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Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" _i="1" _address="0.0.0.1"]If  \(\alpha\) is a root of  \(x^2\) - x +1 = 0 , then \(\alpha^{2018}\)  + \(\alpha^{-2018}\)  is [/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25" _i="1" _address="0.1"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||" _i="0" _address="0.1.0"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.27" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" _i="0" _address="0.1.0.0"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="3.27" hover_enabled="0" _i="0" _address="0.1.0.0.0"]Sample Questions (MMA) : 2019[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.27" hover_enabled="0" _i="1" _address="0.1.0.0.1" open="off"]Quadratic Roots[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.27" hover_enabled="0" _i="2" _address="0.1.0.0.2" open="off"]Medium [/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.0.3" open="off"]Abstract Algebra - Dummit and Foote [/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.23.3" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" _i="1" _address="0.1.0.1"]

Start with hints

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[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.27" hover_enabled="0" _i="1" _address="0.1.0.2.1"]We have to use algebraic expression and expansion e.g-  \(( x + 1) ^{2}\) = \(x^2\) + 2x + 1 ; \(( x +1)^{3}\) = \(x^{3}\) + 3\(x^2\) +3x + 1 Can you think about this kind of case a bit more ?[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.27" hover_enabled="0" _i="2" _address="0.1.0.2.2"]Note that \(x^3\)+ 1 = (\(x+1\))(\(x^{2} - x + 1 \)) So, \(\alpha\) is a root of \(x^{2} - x + 1\) we have   (\(\alpha^3\) + 1) = (\(\alpha+1\))(\(\alpha^{2}\) - \(\alpha\) + 1 ) = ( \(\alpha\) + 1 ) = 0 Can we think along this line ?[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.2.3"]\(\alpha^{3}\) = -1   ||ly \(\beta^{3}\) = -1 [ when \(\beta\) is another root (\(x^{2} - x + 1 \)) ] So, (\(x^{2} - x + 1 \)) = ( x - \(\alpha\) )( x - \(\beta\)) = \(x^{2}\) - (\(\alpha + \beta\))x + \(\alpha\)\(\beta\). Can you get the expression of \(\beta\) in terms of \(\alpha\) ? Is the cloud getting clear now ?[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.27" hover_enabled="0" _i="4" _address="0.1.0.2.4"]\(\alpha\) + \(\beta\) = 1 \(\alpha\)\(\beta\) =  1  implying   \(\beta\)  =  1/ \(\alpha\)  So, \(\alpha^{2018}\)  + \(\alpha^{-2018}\)  = \(\alpha^{2018}\)  + \(\beta^{2018}\)  Note that : \(\alpha^{2}\)  +  \(\beta^{2}\) =  \(\alpha + \beta )^{2}\) -2\(\alpha\)\(\beta\)  = 1 - 2\(\alpha\)\(\beta\)  = 1 -2 = -1   \(\alpha^{2018}\) +   \(\beta^{2018}\)  =  \(\alpha^{3x672}\) . \(\alpha^{2}\) + \(\beta^{3x672}\) . \(\beta^{2}\) =  (\(\alpha^{3})^{672}\).\(\alpha^{2}\) + (\(\beta^{3})^{672}\).\(\beta^{2}\) = (\(-1^{672}\)).\(\alpha^{2}\) + (\(-1^{672}\)).\(\beta^{2}\) =  \(\alpha^{2}\)  + \(\beta^{2}\)   = -1 [/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.23.3" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" _i="3" _address="0.1.0.3"]

Watch the video

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The higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuing who wish to rediscover the world of mathematics.[/et_pb_blurb][et_pb_button button_url="https://www.cheenta.com/collegeprogram/" button_text="Learn More" button_alignment="center" _builder_version="3.23.3" custom_button="on" button_bg_color="#0c71c3" button_border_color="#0c71c3" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3" background_layout="dark" _i="7" _address="0.1.0.7"][/et_pb_button][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" _i="8" _address="0.1.0.8"]

Similar Problems

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