weird symmetric equation

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

weird symmetric equation

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

146 posts

#1 h Apr 27, 2025, 4:29 AM

Y by

Solve the equation: $8x^2-11x+1=(1-x)\sqrt{4x^2-6x+5}$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

54 posts

#2 h Apr 27, 2025, 8:25 AM • 1 Y

Y by ehuseyinyigit

giangtruong13 wrote:

Solve the equation: $8x^2-11x+1=(1-x)\sqrt{4x^2-6x+5}$

The domain of the equation is all real $x$ .
Let $y = \sqrt{4x^2 - 6x + 5}$ , $z = 1 - x$ , then we obtain the system of equations:
$5y^2 - 2yz - 4y^2 = 19 \quad (1)$ $y^2 - 4z^2 = 3 - 2z \quad (2)$ $(2) \times 7 - (1)$ gives:
$7(y^2 - 4z^2) - (5y^2 - 2yz - 4y^2) = 7(3 - 2z) - 19$ $\iff 2y^2 + 2yz - 24z^2 = 2 - 14z$
$\iff y^2 + yz - 12z^2 + 7z - 1 = 0$
$\iff (y - 3z)(y + 4z) + (y + 4z) - (y - 3z) - 1 = 0$
$\iff (y - 3z + 1)(y + 4z - 1) = 0$
Therefore $y + 4z = 1$ or $y - 3z = -1$ . Consider 2 cases:
Case 1: $y + 4z = 1$
$\iff \sqrt{4x^2 - 6x + 5} + 4(1 - x) = 1$
$\iff \sqrt{4x^2 - 6x + 5} = 4x - 3$
$\Rightarrow 4x^2 - 6x + 5 = (3 - 4x)^2$
$\iff 6x^2 - 9x + 2 = 0$
$\iff x = \frac{9 \pm \sqrt{33}}{12}$
Substituting in the values of $x$ , we see that $x = \frac{9 + \sqrt{33}}{12}$ is a solution.
Case 2: $y - 3z = -1$
$\iff \sqrt{4x^2 - 6x + 5} - 3(1 - x) = -1$
$\iff \sqrt{4x^2 - 6x + 5} = 2 - 3x$
$\Rightarrow 4x^2 - 6x + 5 = (2 - 3x)^2$
$\iff 5x^2 - 6x - 1 = 0$
$\iff x = \frac{3 \pm \sqrt{14}}{5}$
Substituting in the values of $x$ , we see that $x = \frac{3 - \sqrt{14}}{5}$ is a solution.
Hence the equation has 2 roots $x = \frac{3 - \sqrt{14}}{5}$ and $x = \frac{9 + \sqrt{33}}{12}$ .

This post has been edited 1 time. Last edited by pooh123, Apr 27, 2025, 8:34 AM
Reason: typo

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a