An algebra problem by Dev Sharma

Before giving away my solution I wanna give you some information which is used in this problem.

The function $f(x) = ax^2 + bx+c$ has minimum which is equal to $\dfrac{4ac-b^2}{4a}$ and it occurs at $\dfrac{-b}{2a}$ if $a>0$ and vice versa.

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So, $2x^2 -6x +6$ and $y^2 -2y+7$ has minimum values as the coefficient of $x^2$ in both cases is $>0$ . So,

$2x^2 - 6x + 6 \geq \dfrac{4\times 2 \times 6 - 36}{4 \times 2} = \dfrac{3}{2}$ and equality holds at $x= \dfrac{3}{2}$ .

Similarly, $y^2 - 2y + 7 \geq 6$ and equality holds at $y= 1$ .

Then, $\left(2x^2 - 6x + 6 \right) \times \left( y^2 - 2y + 7 \right) \geq 9$ and equality hold when $x= 1.5$ and $y=1$ . But, $\left(2x^2 - 6x + 6 \right) \times \left( y^2 - 2y + 7 \right) = 9$ . So, $x= 1.5$ and $y=1$ .

Therefore, $x+y = \boxed{2.5}$ .

We can rewrite the given equation as

$2(x^{2} - 3x + 3)(y^{2} - 2y + 7) = 2((x - \frac{3}{2})^{2} + \frac{3}{4})((y - 1)^{2} + 6) = 9$

$\Longrightarrow 2(x - \frac{3}{2})^{2}(y - 1)^{2} + 12(x - \frac{3}{2})^{2} + \frac{3}{2}(y - 1)^{2} = 0.$

Since both $(x - \frac{3}{2})^{2}, (y - 1)^{2}$ are $\ge 0$ we require that both equal $0.$ This implies that $x + y = \frac{3}{2} + 1 = \boxed{2.5}.$

Let $f\left( x \right) ={ 2x }^{ 2 }-6x+6\quad and\quad f\left( y \right) ={ y }^{ 2 }-2y+7$

Given, $f\left( x \right) f\left( y \right) = 9$

So, $f\left( y \right) =\frac { 9 }{ f\left( x \right) }$

Now, For $y$ to be real, $\triangle \ge 0$ Which means, $f\left( x \right) \le \frac { 3 }{ 2 }$

=> $f\left( x \right) -\frac { 3 }{ 2 } \le 0$

=> $2{ \left( x-\frac { 3 }{ 2 } \right) }^{ 2 }\le 0\quad$ and so $x=\frac { 3 }{ 2 }$

Now, $f\left( \frac { 3 }{ 2 } \right) =\frac { 3 }{ 2 } \quad =>\quad f\left( y \right) =\frac { 9 }{ f\left( x \right) } =6$

Which gives $y=1$ and hence $x+y=2.5$

let $\begin{cases} 2x^2-6x+6=a\Longrightarrow 2x^2-6x+6-a=0\\ y^2-2y+7=\dfrac{9}{a}\longrightarrow y^2-2y+7-\dfrac{9}{a}=0\end{cases}$ note that x,y must be real hence both equation must have a discriminant $\geq$ 0. we get that: $\begin{cases} 6^2-4*2(6-a)\geq 0\Longrightarrow a\geq 1.5\\ 2^2-4(7-\dfrac{9}{a})\geq 0\Longrightarrow a\leq 1.5\end{cases}$ we see that a must be 1.5, putting this in individual equations, $\begin{cases}2x^2-6x+6=1.5\Longrightarrow x= 1.5\\ y^2-2y+7=\dfrac{9}{1.5}\longrightarrow y = 1\end{cases}$ $x+y=1.5+1=\boxed{2.5}$

We write the 2 equations: (1) 2x^2 - 6x + 6 = t, and (2) y^2 -2y + 7 = 9/t. Solve each one by the quadratic formula, and get: (3) 4x = 6 +/- sqrt(36 - 8(6 - t), and (4) 2y = 2 +/- sqrt $ - 4(7 - 9/t). Since we are wanting a single solution fot x + y, both discriminants must vanish, which happens when t = 1.5. Then x = 1.5, y = 1, and x + y = 2.5.Ed Gray

If one solves for x in terms of y, then:

2(x^2 - 3x + 3) = 9/(y^2 - 2y +7);

or x^2 - 3x = 9/[2*(y^2 - 2y +7)] - 3;

or x^2 - 3x + 9/4 = 9/[2*(y^2 - 2y +7)] - 3 + 9/4;

or (x - 3/2)^2 = 9/[2*(y^2 - 2y +7)] - 3/4;

or x - 3/2 = (+/-)sqrt{9/[2*(y^2 - 2y +7)] - 3/4};

or x = 3/2 (+/-) (y-1)/2 * sqrt[-3/(y^2 - 2y + 7)].

The quadratic equation y^2 - 2y + 7 is strictly positive for all real y since (-2)^2 - 4(1)(7) < 0 (by the discriminant test). Hence x can only be real iff y = 1, which yields x = 3/2 and x + y = 5/2 (or 2.5).