IMG0 Determine the value of (a+b)2, given that a2+b2=24 and ab=6. If (x+y)2=13 and x2+y2=7, determine the value of xy. If j+k=6 and j2+k2=52, determine the value of jk. If m2+n2=12 and m4+n4=136, determine all possible values of mn.
Solution
Expanding, (a+b)2=a2+2ab+b2=(a2+b2)+2ab. Since a2+b2=24 and ab=6, then (a+b)2=24+2(6)=36. Expanding, (x+y)2=x2+2xy+y2=(x2+y2)+2xy. Since (x+y)2=13 and x2+y2=7, then 13=7+2xy or 2xy=6, and so xy=3. Expanding, (j+k)2=j2+2jk+k2=(j2+k2)+2jk. Since j+k=6 and j2+k2=52, then 62=52+2jk or 2jk=−16, and so jk=−8. Expanding, (m2+n2)2=m4+2m2n2+n4=(m4+n4)+2m2n2. Since m2+n2=12 and m4+n4=136, then 122=136+2m2n2 or 2m2n2=8 or m2n2=4, and so mn=±2.
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