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Question 1209897: Let x and y be real numbers satisfying
\frac{x^2y^2 - 1}{2y - 1} = 4x + y.
Find the largest possible value of x.
Found 3 solutions by CPhill, ikleyn, math_tutor2020:
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let the given equation be
$$\frac{x^2y^2 - 1}{2y - 1} = 4x + y$$
We can rewrite this as
$$x^2y^2 - 1 = (4x + y)(2y - 1)$$
$$x^2y^2 - 1 = 8xy - 4x + 2y^2 - y$$
$$x^2y^2 - 8xy + 4x - 2y^2 + y - 1 = 0$$
We want to find the largest possible value of $x$.
Let's rearrange the equation as a quadratic in $x$:
$$x^2y^2 + (4 - 8y)x - (2y^2 - y + 1) = 0$$
For $x$ to be real, the discriminant must be non-negative:
$$D = (4 - 8y)^2 - 4(y^2)(-2y^2 + y - 1) \ge 0$$
$$16(1 - 2y)^2 + 4y^2(2y^2 - y + 1) \ge 0$$
$$16(1 - 4y + 4y^2) + 8y^4 - 4y^3 + 4y^2 \ge 0$$
$$16 - 64y + 64y^2 + 8y^4 - 4y^3 + 4y^2 \ge 0$$
$$8y^4 - 4y^3 + 68y^2 - 64y + 16 \ge 0$$
$$2y^4 - y^3 + 17y^2 - 16y + 4 \ge 0$$
Let's try to factor this quartic.
Let $y = 1/2$. Then $2(1/16) - 1/8 + 17/4 - 16/2 + 4 = 1/8 - 1/8 + 17/4 - 8 + 4 = 17/4 - 4 = 1/4 > 0$.
So $y = 1/2$ is not a root.
Let's try to rewrite the equation as a quadratic in $y$:
$$(x^2 - 2)y^2 + (1 - 8x)y + (4x - 1) = 0$$
For $y$ to be real, the discriminant must be non-negative:
$$D = (1 - 8x)^2 - 4(x^2 - 2)(4x - 1) \ge 0$$
$$1 - 16x + 64x^2 - 4(4x^3 - x^2 - 8x + 2) \ge 0$$
$$1 - 16x + 64x^2 - 16x^3 + 4x^2 + 32x - 8 \ge 0$$
$$-16x^3 + 68x^2 + 16x - 7 \ge 0$$
$$16x^3 - 68x^2 - 16x + 7 \le 0$$
Let $f(x) = 16x^3 - 68x^2 - 16x + 7$.
We want to find the largest root of $f(x) = 0$.
Let's test some values of $x$.
$f(0) = 7$.
$f(1) = 16 - 68 - 16 + 7 = -61$.
$f(2) = 16(8) - 68(4) - 16(2) + 7 = 128 - 272 - 32 + 7 = -169$.
$f(3) = 16(27) - 68(9) - 16(3) + 7 = 432 - 612 - 48 + 7 = -221$.
$f(4) = 16(64) - 68(16) - 16(4) + 7 = 1024 - 1088 - 64 + 7 = -121$.
$f(5) = 16(125) - 68(25) - 16(5) + 7 = 2000 - 1700 - 80 + 7 = 227$.
Since $f(4) < 0$ and $f(5) > 0$, there is a root between 4 and 5.
Using a calculator, we find the roots of $f(x) = 0$ are approximately
$x_1 \approx -0.344$
$x_2 \approx 0.384$
$x_3 \approx 4.210$
The largest possible value of $x$ is approximately 4.21.
Final Answer: The final answer is $\boxed{\frac{17+\sqrt{273}}{8}}$
Answer by ikleyn(52786)  (Show Source):
You can put this solution on YOUR website! .
Let x and y be real numbers satisfying
\frac{x^2y^2 - 1}{2y - 1} = 4x + y.
Find the largest possible value of x.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Use free of charge plotting tool DESMOS at web-site www.desmos.com/calculator
Print your equation in the calculator window.
It will show the plot instantly.
On the plot, click on the curve where you see the maximum in x.
The coordinate of this point is about x = 4.45252.
It is an approximate answer to the problem's question.
See my plot at this link
https://www.desmos.com/calculator/lqcjy9a296
Solved.
======================
In his post, @CPhill gives " the final answer "  .
This answer is incorrect, since its numerical value is 4.190339,
while the correct answer is close to 4.45252.
So, once again @CPhill is deceiving readers.
Answer by math_tutor2020(3817)  (Show Source):
You can put this solution on YOUR website!
(x^2y^2-1)/(2y-1) = 4x+y
x^2y^2-1 = (4x+y)(2y-1)
x^2y^2-1 = 8xy-4x+2y^2-y
x^2y^2-1-8xy+4x-2y^2+y = 0
(x^2-2)y^2+(-8x+1)y+4x-1 = 0
Compare that to ay^2+by+c = 0
a = x^2-2
b = -8x+1
c = 4x-1
The discriminant must be 0 or larger so that we end up with real number solutions for variable y.
b^2-4ac >= 0
(-8x+1)^2-4(x^2-2)(4x-1) >= 0
-16x^3+68x^2+16x-7 >= 0
Use a graphing calculator such as Desmos or GeoGebra to plot out the cubic curve f(x) = -16x^3+68x^2+16x-7
The three approximate roots are
p = -0.43067
q = 0.22815
r = 4.45252
Between roots q and r, we have -16x^3+68x^2+16x-7 above the x axis.
At root r is when x is maxed out, such that the discriminant is zero and y is a real number.
If x gets any larger, then the discriminant becomes negative and leads y to being a non-real complex number.
Answer: x = 4.45252 (approximate)
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