practice problems pending

1) Determine x, y, z such that

2x^2 + y^2 + 2z^2 - 8x + 2y - 2xy + 2xz - 16z + 35 = 0

S1)
We will try to make it as sum of squares. And by making each one zero we will solve for it.
First we will complete the square in y. Why?
Because y^2 has coefficient of 1 (x^2,z^2 have 2) and it has only 1 mixed term(x has 2: xz,yz).
So group those terms:
y^2 + 2y - 2xy = y^2 + 2y(1-x) = (y + 1 - x)^2 - (1-x)^2

Now the remaining expression is:
2x^2 + 2z^2 - 8x + 2xz - 16z + 35 - (1-x)^2 = 0
=> x^2 -6x + 34 + 2z^2 + 2xz - 16z = 0

Now group 'x' terms since its coefficient is 1 now:
x^2 + 2.x.(z - 3) = (x + z - 3)^2 - (z-3)^2

Now the remaining expression is:
2z^2 - 16z + 34 - (z-3)^2 = 0
=> z^2 - 10z + 25 = 0 = (z-5)^2

Finally:
(z-5)^2 + (x + z - 3)^2 + (y + 1 - x)^2 = 0
=> z = 5, x = -2, y = -3 Answer.

Q2.


S2.
= |sqrt(x-1) + 1| + |sqrt(x-1) - 1|
First term is positive so
= sqrt(x-1) + 1 + |sqrt(x-1) - 1|

Now for it to be constant
sqrt(x-1) - 1 <= 0
=> sqrt(x-1) < =1
square both sides:
x-1 <= 1 => x <= 2
Also for sqrt(x-1) to be real x >= 1

So 1 <= x <= 2
Answer.


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