If #x^2 + y^2 = 73 # and #xy =24#, what is the value of #(x+y)^2# ?

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Answer 1 · 2016-07-12T19:45:54

#(x+y)^2= 11^2#

#(x+y)^2=x^2+2x y+y^2#

so

#(x+y)^2 = 73+2 xx 24 = 121 = 11^2#

We can even ask for the #x,y# values

From #(x+y)^2=11^2->x+y = pm11#

Now we have

#(s+x)(s+y) = s^2+(x+y)s+xy#

then solving

#s^2pm11s+24=0# we will obtain #x,y# as roots.

Taking as referent

#s^2+11s+24=0# we have

#x = -8, y = -3# Analogously for

#s^2-11s+24=0# we have

#x = 8, y = 3#