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August 31, 2019

Problems on quadratic roots: ISI MMA 2018 Question 9

[et_pb_section fb_built="1" _builder_version="3.22.4" fb_built="1" _i="0" _address="0"][et_pb_row _builder_version="3.25" _i="0" _address="0.0"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||" _i="0" _address="0.0.0"][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_padding="20px|20px|20px|20px" _i="0" _address="0.0.0.0"]

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

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.27" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff" hover_enabled="0" _i="2" _address="0.1.0.2"][et_pb_tab title="Hint 0" _builder_version="3.22.4" _i="0" _address="0.1.0.2.0"]Do you really need a hint? Try it first!

[/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"]

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Similar Problems

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2 comments on “Problems on quadratic roots: ISI MMA 2018 Question 9”

    1. Hi,

      Yes, you are right but this method came into my mind first so I typed it down :). You can write your solution so that the readers can see that as well. Thanks a lot for seeing this. Keep reading and don't forget to comment. Your valuable comments/appreciations inspire us to create more contents.

      Regards,
      Arnab

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