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Algebra Level 2

$(x+1)^2+(y-1)^2+(z-3)^2=4(y-z)$

Find x,y,z.

x=3,y=z=4 x=-1,y=3,z=1 x=y=z=0 x=0,y=1,z=3

6 solutions

Prateek Gupta
Oct 31, 2014

easy observation sum of squares can't be -ve for any real x , y , z so y - z can't be -ve or zero which is satisfied by only 1 option

Exactly how I did it, didn't need to do any substitution...

Nicky Cleaver - 6 years, 7 months ago

Soumen Bhowmik
Oct 30, 2014

The equation can be written as (x+1)^2 + (y-3)^2 + (z-1)^2 = 0 So every square element should be zero: x+1 = 0; x = -1 y-3 = 0; y = 3 z-1 = 0; z = 1

ang hirap talaga pag medyo matalino

Geraldine Lobetana - 6 years, 7 months ago

Harsh Rathore
Nov 3, 2014

Just substitute the options in given equations

Joshua Villanueva
Oct 31, 2014

Test and Trial (An easy way if multiple choices are offered):

Plugging in the correct answer:

x = -1

y = 3

z = 1

This should work!

Parayus Mittal
Dec 24, 2014

Bring all the figure on RHS the open the square consisting of y&z the simplify try to complete the identity hence for each and every variable Discriminant=0 therefore like roots possible. To make equation=0 x,y&z should be simultaneously zero.

Aadi Naik
Nov 3, 2014

There is no x term in the RHS, so either x=0 or x+1=0 =>x=-1 Considering x=-1, y²-2y+z²-6z+10=4(y-z) => y²-6y+z²-2z+1+9=0 =>(y-3)²+(z-1)²=0 is y-3=0 and z-1=0

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