Problem on Equation | AMC-10A, 2007 | Problem 20

Try this beautiful problem from Algebra based on quadratic equation

Problem on Equation - AMC-10A, 2007- Problem 20

Suppose that the number \(a\) satisfies the equation \(4 = a + a^{ - 1}\). What is the value of \(a^{4} + a^{ - 4}\)?

  • \(174\)
  • \(194\)
  • \(156\)

Key Concepts

Algebra

Linear equation

multiplication

Check the Answer

Answer: \(194\)

AMC-10A (2007) Problem 20

Pre College Mathematics

Try with Hints

Given that \(4 = a + a^{ - 1}\). we have to find out the value \(a^{4} + a^{ - 4}\)

Squarring both sides of \(a^{4} + a^{ - 4}\) ...then opbtain...

can you finish the problem........

\((a + a^{ - 1})^2=4^2\) \(\Rightarrow (a^2 + a^{-2} +2)=16\) \(\Rightarrow a^2 + a^{-2}=14\) and now squarring both side again.............

can you finish the problem........

Squarring both sides of \(a^2 + a^{-2}=14\) \(\Rightarrow (a^2 + a^{-2})^2=(14)^2\) \(\Rightarrow a^4 + a^{-4} +2=196\) \(\Rightarrow a^4 + a^{-4}=194\)

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Powers of Numbers AMC 8 ,2013 problem 15

Powers of Numbers AMC 8 ,2013 problem 15

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What are we learning ?

[/et_pb_text][et_pb_text _builder_version="4.1" text_font="Raleway||||||||" text_font_size="20px" text_letter_spacing="1px" text_line_height="1.5em" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]Competency in Focus: Powers of Numbers This problem from American Mathematics contest (AMC 8, 2013) is based on basic  algebra and Powers of Numbers.[/et_pb_text][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

First look at the knowledge graph.

[/et_pb_text][et_pb_image src="https://cheenta.com/wp-content/uploads/2020/01/AMC-8-2013-problem-15-1.png" align="center" force_fullwidth="on" _builder_version="4.1" min_height="298px" height="189px" max_height="207px"][/et_pb_image][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Next understand the problem

[/et_pb_text][et_pb_text _builder_version="4.1" text_font="Raleway||||||||" text_font_size="20px" text_letter_spacing="1px" text_line_height="1.5em" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]If $3^p + 3^4 = 90$, $2^r + 44 = 76$, and $5^3 + 6^s = 1421$, what is the product of $p$, $r$, and $s$?[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="4.0"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="4.1" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="4.1"]American Mathematical Contest 2013, AMC 8 Problem 15[/et_pb_accordion_item][et_pb_accordion_item title="Key Competency" _builder_version="4.1" open="off"]Basic algebra and Powers of Numbers[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.1" open="off"]4/10[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.0.9" open="off"]Challenges and Thrills in Pre College Mathematics Excursion Of Mathematics 

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Start with hints 

[/et_pb_text][et_pb_tabs _builder_version="4.1"][et_pb_tab title="HINT 0" _builder_version="4.0.9"]Do you really need a hint? Try it first![/et_pb_tab][et_pb_tab title="HINT 1" _builder_version="4.1"]First, we're going to solve for $p$. Start with $3^p+3^4=90$. Then, change $3^4$ to $81$. Subtract $81$ from both sides to get $3^p=9$ .Now we can write 9 as \(3^2\) .So, from here we can say that p=2.[/et_pb_tab][et_pb_tab title="HINT 2" _builder_version="4.1"]Now, solve for $r$. Since $2^r+44=76$, $2^r$ must equal $32$,  and 32 can be written as \( 2^5 \) .So from here we have r=5.[/et_pb_tab][et_pb_tab title="HINT 3" _builder_version="4.1"]Similarly now, solve for $s$. $5^3+6^s=1421$ can be simplified to $125+6^s=1421$ which simplifies further to $6^s=1296$=\(6^4\) , which gives s=4.[/et_pb_tab][et_pb_tab title="HINT 4" _builder_version="4.1"]Lastly, $prs$ equals $2*5*4$ which equals $40$. So, the answer is 40.[/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" min_height="12px" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

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