I've seen it before - 1

Algebra Level 3

If $x, y, z \in\mathbb{R}$

$x^2+11z=-40.25$ $y^2-9x=-28.25$ $z^2+7y=5.75$

Then the value of $|x|+|y|+|z|$ is

Note

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3 solutions

Chew-Seong Cheong
Oct 25, 2014

Add the three equations together we have:

$x^2 - 9x + y^2 +7y + z^2 +11z = -62.25$

$(x-4.5)^2 - 20.25 + (y+3.5)^2 -12.25 +(z+5.5)^2 - 30.25 = - 62.25$

$(x-4.5)^2 + (y+3.5)^2 +(z+5.5)^2 - 62.25 = - 62.25$

$(x-4.5)^2 + (y+3.5)^2 +(z+5.5)^2 = 0$

Since the three squared quantities are always positive, the above is true only when:

$(x-4.5)^2 = 0$ , $\quad (y+3.5)^2 = 0 \quad$ and $\quad (z+5.5)^2 = 0$

$\Rightarrow \quad x=4.5$ , $\quad y=-3.5 \quad$ and $\quad z=-5.5$

$\Rightarrow \quad |x|+|y|+|z| = 4.5+3.5+5.5 = \boxed{13.5}$

Good approach using the trivial identity.

Note that you have only found a necessary condition. You should verify that the initial equations are satisfied by these values.

Calvin Lin Staff - 6 years, 7 months ago

Maybe you meant non-negative

"Since the three squares are always positive"

But other than that, the solution is elegant!

Joel Tan - 6 years, 7 months ago

Same way!!

Dev Sharma - 5 years, 6 months ago

Ali Alyahyai
Oct 25, 2014

X2 - 9x+20.25+20+y2+7y+12.25+16+z2+11z+30.25-36=0

(x-4.5)2+(y+3.5)2+(z+5.5)2=0

X=4.5 y=-3.5 z=-5.5

4.5+3.5+5.5=13.5

Abhinav Rawat
Sep 19, 2015

Done exactly the same as Chew-Seong