Introductory Olympiad Algebra - The Square Is Now Complete

Set $z = 2x-3y$ , we want to find the minimum of $(z-4)^2 + (z+10)^2 = 2z^2 + 12 z + 116$ . The minimum occurs at the vertex of the parabola, which is $z = - \frac{ 12}{2 \times 2 } = -3$ . Hence the minimum is $(-3-4)^2 + (-3+10)^2 = 98$ .

$(2x-3y-4)^{ 2 }\quad +\quad (2x-3y+10)^{ 2 }\\ =\quad (2x-3y+10-14)^{ 2 }\quad +\quad (2x-3y+10)^{ 2 }\\ =\quad (2x-3y+10-14)(2x-3y+10-14)\quad +\quad (2x-3y+10)^{ 2 }\\ =\quad (2x-3y+10)(2x-3y+10-14)\quad -14(2x-3y+10)\quad +\quad 14^{ 2 }\quad +\quad (2x-3y+10)^{ 2 }\\ =\quad (2x-3y+10)(2x-3y+10)\quad -\quad 28(2x-3y+10)\quad +\quad 14^{ 2 }\quad +\quad (2x-3y+10)^{ 2 }\\ =\quad 2(2x-3y+10)^{ 2 }\quad -\quad 28(2x-3y+10)\quad +\quad 196\\ let\quad A\quad =\quad 2x-3y+10\\ f(A)\quad =\quad 2A^{ 2 }\quad -\quad 28A\quad +\quad 196\\ \cfrac { df(A)\quad }{ dA } \quad =\quad 4A\quad -\quad 28\quad =\quad 0\\ A\quad =\quad 7\\ f(7)\quad =\quad 2(7)^{ 2 }\quad -\quad 28(7)\quad +\quad 196\quad =\quad 98$

USING STRAIGHT LINE CONCEPT : see the two equations as the equations of straight lines and then any point lying on the perpendicular joining them would have min. distance (since both given lines are parallel) and then randomly take any line perpendicular to both of them and then using the relation btw x and y using that equation substitute in the given relation .........in question after then apply max min concept

take 2x-3y=k and replace it inthe exp,now maxima minima

let z=2x-3y and solve it using maxima and minima techinque or using properties of parabola

Set $z = 2x - 3y + 3$ , we want to find the minimum of $(z-7)^2 + (z+7)^2 = 2(z^2 + 49)$

The minimum occurs at 2 x 49 = 98 when $z^2 = (2x - 3y + 3)^2 = 0$

Let $p=2x-3y$ and let the whole expression be equal to $y$ Then,

$(p-4)^2+(p+10)^2=y$

$p^2-8p+16+p^2+20p+100=y$

$2p^2+12p+116=y$

$2(p^2+6p+58)=y$

$2(p^2+6p)=y-116$

$2(p^2+6p+9)=y-98$

$2(p+3)^2+98=y$

We could make $(p+3)^2=0$

Then $y=0+98$

$y=98$

Therefore, the minimum value of the expression $(2x-3y-4)^2+(2x-3y+10)^2$ is $98$

Let z= 2x -3y Substituting z in the given equation we get (z-4)^2 +(z+10)^2 → 2z^2 +12z + 116→ 2(z^2 +6z +9) + 98→ 2(z+3)^2 + 98

The minimum value is obtained when z = -3 so that the square part become zero as it cannot be negative for real value of z So min value is 98...

Set 2x-3y = n, so: (n-4)^2 + (n+10)^2 = 2n^2+12n+116. Because the a-coefficent is positive, so the minimum occurs at the vertex, which is n = (-b)/2a = (-12)/4 = -3. Plug this into the quadratic equation above, we get the value of 98. So the minimum of (2x-3y-4)^2 +(2x-3y+10)^2 is 98.

My solution is kinda weird. I'll just post it up for comments.

So first I started out getting the expanded form of the expression.

$8x^2+24x-24xy-36y+18y^2+116$

Next, I formed two quadratic expressions

$[1]$ : $8(x+\frac{3}{2})^2-18$

$[2]$ : $18(y-1)^2-18$

Leaving out a remaining $[3]$ : $-24xy + 80$ .

Then I got the minimum values of $[1]$ and $[2]$ , which are at $x=-\frac{3}{2}$ and $y=1$ , respectively.

But $[3]$ would give a high positive value if $x$ and $y$ have different signs so I let $x=0$ and with a help of a calculator I got an answer of $98$ .

Just found out, too, that the same happens if I let $y=0$ .

Any comment would help, since it got me to the right answer I thought this approach has any use in itself. If it has not, then this technique goes to the trash bin.

2x-3y=k --> (k-4)²+(k+10)²=2k²+12k²+18-18+116=2(k+3)²+98. setting 2x-3y=-3 --> min=98

Since 2x-3y can take any value, the answer is half the difference (-4 - (-10)) = -7, squared (49), then doubled as the two halves are summed, hence 98.

Define the function $f\colon \mathbb{R}^2 \to \mathbb{R}\colon (x,y) \mapsto 2x-3y+3$. Each section of this map of suriective, and in particular also the function itself. Hence the expression can be rewritten as $(X+7)^2+(X-7)^2$, where $X=f(x,y)$ can be any real value. At this point, it is enough to expand :)