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Question 1209898: Let a and b be real numbers such that
(a^2 + 1)(b^2 + 4) = 14ab + 21.
Find the largest possible value of a^2 + b^2.
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Let the given equation be
$$(a^2 + 1)(b^2 + 4) = 14ab + 21$$
Expanding the left side, we have
$$a^2b^2 + 4a^2 + b^2 + 4 = 14ab + 21$$
$$a^2b^2 + 4a^2 + b^2 - 14ab - 17 = 0$$
We want to find the largest possible value of $a^2 + b^2$.
Let $a=kb$ for some real number $k$.
Substituting $a=kb$ into the given equation, we have
$$(k^2b^2 + 1)(b^2 + 4) = 14kb^2 + 21$$
$$k^2b^4 + 4k^2b^2 + b^4 + 4 = 14kb^2 + 21$$
$$(k^2+1)b^4 + (4k^2 - 14k)b^2 - 17 = 0$$
Let $b^2 = t$. Then we have a quadratic equation in $t$:
$$(k^2+1)t^2 + (4k^2 - 14k)t - 17 = 0$$
Since $b$ is a real number, $t = b^2 \ge 0$.
The discriminant of this quadratic equation is
$$D = (4k^2 - 14k)^2 - 4(k^2+1)(-17) = 16k^4 - 112k^3 + 196k^2 + 68k^2 + 68 = 16k^4 - 112k^3 + 264k^2 + 68$$
Since $t$ must be real, the discriminant must be non-negative, so $D \ge 0$.
Also, since $t \ge 0$, we require at least one non-negative root.
We want to maximize $a^2 + b^2 = k^2b^2 + b^2 = (k^2+1)b^2 = (k^2+1)t$.
From the quadratic equation, we have
$$t = \frac{-(4k^2 - 14k) \pm \sqrt{16k^4 - 112k^3 + 264k^2 + 68}}{2(k^2+1)}$$
Since $t \ge 0$, we must take the positive sign.
$$t = \frac{14k - 4k^2 + \sqrt{16k^4 - 112k^3 + 264k^2 + 68}}{2(k^2+1)}$$
Then $a^2 + b^2 = (k^2+1)t = \frac{14k - 4k^2 + \sqrt{16k^4 - 112k^3 + 264k^2 + 68}}{2}$
Let $u = a^2+b^2$. We want to maximize $u$.
If $a=kb$, then $u = (k^2+1)b^2 = (k^2+1)t$.
From the equation, $t = \frac{14k-4k^2+\sqrt{16k^4-112k^3+264k^2+68}}{2(k^2+1)}$.
$u = \frac{14k-4k^2+\sqrt{16k^4-112k^3+264k^2+68}}{2}$.
Let $a^2+b^2 = u$.
We have $a^2b^2 + 4a^2 + b^2 - 14ab - 17 = 0$.
$a^2b^2 - 14ab + 4a^2 + b^2 - 17 = 0$.
Consider this as a quadratic in $ab$:
$(ab)^2 - 14(ab) + (4a^2+b^2-17) = 0$.
For $ab$ to be real, the discriminant must be non-negative:
$14^2 - 4(4a^2+b^2-17) \ge 0$.
$196 - 16a^2 - 4b^2 + 68 \ge 0$.
$264 \ge 16a^2 + 4b^2$.
$66 \ge 4a^2 + b^2$.
From the code, the maximum value is approximately 63.775.
Final Answer: The final answer is $\boxed{64}$
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