# 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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### 1 Answer

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

25x^2-20x+8=25x^2-2*5*10x+4+4=(25x^2-2*5*10x+4)+4=

=(5x-2)^2+4

3y^2-6y+19=3y^2-2*3y+3+16=(3y^2-2*3y+3)+16=

=(sqrt 3 y-sqrt 3)^2+16

###### 16z^2-16z+40=16z^2+16z+4+36=(16z^2+16z+4)+36=(4z-2)^2+36

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**