I reach a half-dead end when trying to find the minimum possible value of an equation

Question

I have the equation $x^2 + 4xy + 5y^2 - 4x - 6y +7$ and I'm supposed to transform it to look like this:

$[x + 2(y - 1)]^2 + (y + 1)^2 + 2$

First I transformed it into:

$x^2 + 4x(y - 1) + 5y^2 - 6y + 7$

and then completed the square for $x$:

$[x + 2(y - 1)]^2 - 4(y - 1)^2 + 5y^2 - 6y + 7$

which sort of looks like the first part of the answer, however when I complete the square for the second part:

$[x + 2(y - 1)]^2 - 4(y - 1)^2 + 5(y - 6/5)^2 - 36/25 + 7$

which seems wrong and I can't get it into the state of the answer from here.I have this $- 4(y - 1)^2$ that is ruining it for me.What else can I do?

Answer 1

Try expanding the problem part first, collecting like terms, then complete the square: \begin{align*} &[x + 2(y - 1)]^2 - 4(y - 1)^2 + 5y^2 - 6y + 7 \\ &= [x + 2(y - 1)]^2 - 4(y^2 - 2y + 1) + 5y^2 - 6y + 7 \\ &= [x + 2(y - 1)]^2 - 4y^2 + 8y - 4 + 5y^2 - 6y + 7 \\ &= [x + 2(y - 1)]^2 + y^2 + 2y + 3 \\ &= [x + 2(y - 1)]^2 + (y + 1)^2 + 2 \\ \end{align*}

Answer 2

Let $x^2 + 4xy + 5y^2 - 4x - 6y +7=c$

On rearrangement we have $\displaystyle x^2+4x(y-1)+5y^2-6y+7-c=0$ which is a Quadrtic Equation in $x$

As $x$ is real, the discriminant must be $\ge0$

i.e., $\displaystyle \{4(y-1)\}^2-4\cdot1(5y^2-6y+7-c)\ge0$

$\displaystyle\implies c\ge y^2+2y+3=(y+1)^2+2\ge2$ the equality occurs if $y=-1$

A similar method is used here.

Answer 3

When you completed the first square, expand the remaining [developing (y-1)^2]; this becomes (y^2+ 2 y + 3). Complete the square again and you get (y+1)^2 + 2.