Which are the x,y,z values so that sqrt(25x^2 -20x+8)+sqrt(3y^2 -6y+19)+sqrt(16z^2 -16z+40)<12



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Posted on (Answer #1)

We'll try to write the folowing expressions in this way:




=(sqrt 3 y-sqrt 3)^2+16


We've put the expressions above in this manner of writting for highlighting binomial expressions.

sqrt [(5x-2)^2+4]>sqrt 4=2

sqrt[(sqrt 3 y-sqrt 3)^2+16]>sqrt 16=4

sqrt[(4z-2)^2+36]>sqrt 36=6

We'll add the expressions above:

sqrt [(5x-2)^2+4]+sqrt[(sqrt 3 y-sqrt 3)^2+16]+sqrt[(4z-2)^2+36]>2+4+6=12

But the utterance says the contrary, meaning that the sum is smaller than 12, so the conclusion must be that:

(5x-2)^2=0, 5x=2, x=2/5

(sqrt 3 y-sqrt 3)^2=0, sqrt 3 y=sqrt 3, y=1

(4z-2)^2=0, 4z=2, z=2/4=1/2


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