If x, y and z are all non-zero numbers, what is the minimum value of x

Question

If x, y and z are all non-zero numbers, what is the minimum value of  x^4 + z^2 -4yz + 4y^2 -4x^2y + 2x^2z - 2  ?

A. -2
B. -1
C. 0
D. 1
E. 2

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IanStewart · Accepted Answer

The standard way to do a question like this is to find a factorization involving a square (and perhaps other terms), and then, by noticing that squares can never be less than zero in value, finding the minimum or maximum value of the expression by making the square exactly equal to zero. yashikaaggarwal has done that perfectly above, so I'll just point out a different way one could solve almost instantly. If you notice that when x, y and z are all exactly equal to zero, the value of the expression is -2, then it must be true that when x, y and z are infinitesimally different from zero (so are all equal, say, to 0.00000001) the value of the expression will be negligibly different from -2. So its minimum value clearly must be less than -1, and only answer A can be right.

yashikaaggarwal · Answer

X^4+Z^2+4Y^2-4YZ-4X^2Y+2X^2Z-2
Compare it with (a+b+c)^2
Where A = -x^2
B = 2y
C = -Z
{X^4+Z^2+4Y^2-4YZ-4X^2Y+2X^2Z}-2
=> (-X^2+2y-Z)^2-2
For minimum value put (-X^2+2y-Z)^2 = 0
=> (-X^2+2y-Z)^2 = 0
=> 0-2 = -2

IMO A

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Kinshook · Answer

Asked: If x, y and z are all non-zero numbers, what is the minimum value of  x^4 + z^2 -4yz + 4y^2 -4x^2y + 2x^2z - 2  ?

z^2 - 4yz + 4y^2 + x^2 (x^2 - 4y + 2z) - 2
(z - 2y)^2 + x^4 + 2x^2(z - 2y) - 2
((z - 2y) + x^2)^2 - 2

Min value = - 2
Since min ((z - 2y) + x^2)^2 = 0 ; when z = 2y & x = 0

IMO A

Lipun · Answer

x^4 + z^2 - 4yz + 4y^2 - 4x^2y + 2x^2z - 2

=    - 2

=  (x^2 - 2y +z)^2 - 2

Since the first term is a square, therefore, the minimum value can only be zero.
=> Min value of the expression = -2

saumyagupta1602 · Answer

min val x^4= 1 for x=+1 or -1 expression has only x^2 so x=1 or -1 doesn't matter
for z^2=1 z=1 or -1 
taking x^4, x^2 and z^2=1 we get 1+1 -4yz + 4y^2 -4y+2z-2
checking for z=-1; 4y^2-8y+2 so case 1 y=1 ie 4-8+2=-2 or case 2 y=-1 ie 4+8+2=14 , min so y=1 and -2
checking for z=1 ; 4y^2-2 so y=+1or -1 gives Y^2=1 and val= 2
min(-2,2)=-2 answer

ShankSouljaBoi · Answer

Yu have taken x= 0 ; however, it is given in the stem that x y and z cant assume zero values.

BlueRocketAsh · Answer

x^4+z^2−4yz+4y^2−4x^2y+2x^2z−2
= (x^4+z^2+4y^2−4x^2y+2x^2z−4yz)−2
= (x^2 - 2y + z)^2 - 2
Here all the variable are inside a square term. Minimum value of the square term is 0
Therefore minimum value of the expression is -2

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If x, y and z are all non-zero numbers, what is the minimum value of x

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If x, y and z are all non-zero numbers, what is the minimum value of x

Prep Toolkit

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