If $x$ and $y$ are real numbers, the minimum possible value of the expression $(x+3)^{2}+2(y-2)^{2}+4(x-7)^{2}+(y+4)^{2}$ is
(A) 172
(B) 65
(C) 136
(D) 152
(E) 104
Solution 1

We expand the given expression to obtain

$$
\left(x^{2}+6 x+9\right)+2\left(y^{2}-4 y+4\right)+4\left(x^{2}-14 x+49\right)+\left(y^{2}+8 y+16\right)
$$

We expand further to obtain

$$
x^{2}+6 x+9+2 y^{2}-8 y+8+4 x^{2}-56 x+196+y^{2}+8 y+16
$$

We simplify to obtain

$$
5 x^{2}-50 x+3 y^{2}+229
$$

We remove a common factor of 5 from the first two terms

$$
5\left(x^{2}-10 x\right)+3 y^{2}+229
$$

and then complete the square to obtain

$$
5\left(x^{2}-10 x+5^{2}-5^{2}\right)+3 y^{2}+229
$$

This gives

$$
5(x-5)^{2}-125+3 y^{2}+229
$$

or

$$
5(x-5)^{2}+3 y^{2}+104
$$

Since $(x-5)^{2} \geq 0$ for all real numbers $x$ and $3 y^{2} \geq 0$ for all real numbers $y$, then the minimum value of $5(x-5)^{2}+3 y^{2}+104$ (and hence of the original expression) is $5(0)+3(0)+104$ or 104. We note that this value is actually achieved when $x=5$ (which gives $(x-5)^{2}=0$ ) and $y=0$ (which gives $3 y^{2}=0$ ).

## Solution 2

We expand the given expression to obtain

$$
\left(x^{2}+6 x+9\right)+2\left(y^{2}-4 y+4\right)+4\left(x^{2}-14 x+49\right)+\left(y^{2}+8 y+16\right)
$$

We expand further to obtain

$$
x^{2}+6 x+9+2 y^{2}-8 y+8+4 x^{2}-56 x+196+y^{2}+8 y+16
$$

The terms involving $x$ are

$$
x^{2}+6 x+9+4 x^{2}-56 x+196=5 x^{2}-50 x+205=5 x^{2}-50 x+125+80=5(x-5)^{2}+80
$$

The terms involving $y$ are

$$
2 y^{2}-8 y+8+y^{2}+8 y+16=3 y^{2}+24
$$

Since $(x-5)^{2} \geq 0$ for all real numbers $x$, then the minimum value of $5(x-5)^{2}+80$ is 80 .

Since $y^{2} \geq 0$ for all real numbers $y$, then the minimum value of $3 y^{2}+24$ is 24 .

Since the minimum value of $(x-3)^{2}+4(x-7)^{2}$ is 80 and the minimum value of $2(y-2)^{2}+(y+4)^{2}$ is 24 , then the minimum value of $(x-3)^{2}+2(y-2)^{2}+4(x-7)^{2}+(y+4)^{2}$ is $80+24=104$.