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, 2012

Algebra Difficulty 3.1 AMC 10/12 Prove it Canada

Determine the value of $(a+b)^2$ , given that $a^2+b^2 = 24$ and $ab=6$ .
If $(x + y)^2 = 13$ and $x^2 + y^2 = 7$ , determine the value of $xy$ .
If $j + k = 6$ and $j^2 + k^2 = 52$ , determine the value of $jk$ .
If $m^2 + n^2 = 12$ and $m^4 + n^4 = 136$ , determine all possible values of $mn$ .

Solution

Expanding, $(a+b)^2=a^2+2ab+b^2=(a^2+b^2)+2ab$ .

Since $a^2+b^2=24$ and $ab=6$ , then $(a+b)^2=24+2(6)=36$ .
Expanding, $(x+y)^2=x^2+2xy+y^2=(x^2+y^2)+2xy$ .

Since $(x+y)^2=13$ and $x^2+y^2=7$ , then $13=7+2xy$ or $2xy=6$ , and so $xy=3$ .
Expanding, $(j+k)^2=j^2+2jk+k^2=(j^2+k^2)+2jk$ .

Since $j+k=6$ and $j^2+k^2=52$ , then $6^2=52+2jk$ or $2jk=-16$ , and so $jk=-8$ .
Expanding, $(m^2+n^2)^2=m^4+2m^2n^2+n^4=(m^4+n^4)+2m^2n^2$ .

Since $m^2+n^2=12$ and $m^4+n^4=136$ , then $12^2=136+2m^2n^2$ or $2m^2n^2=8$ or $m^2n^2=4$ , and so $mn=\pm2$ .

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Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.