A Fascinating Radical Equation | An Algebra Challenge

[@Shobhamaths]

Cubing on both sides
x^3=(1-2x-9x^2) /(x^2-2x-9)
Let t=2x+9
x=(t-9) /2, substituting
a=t-9
a^5-4a^4-36a^3+16a^2+144a-32=0
a=2;
then x=1

[@tunneloflight]

First two (of 5) roots: x = 1, x = -phi^2
trinomial remaining to resolve = x^3 - (phi+2)*x^2+(4*phi-7)*x+(2-phi).
resolve: (2-phi), 2+root(3), 2-root(3)

[@tunneloflight]

So full solution: x = 1, -phi-1, 2-phi, 2+root(3), 2-root(3).
in ~ numbers: x = 1, -2.618, 0.31966, 3.73205, 0.26795
ordered: 3.73205, 1, 0.31966, 0.26795, -2.618

[@kassuskassus6263]

Five real solutions x=1; x=(-3+or-sqrt5)/2; x=2+or-sqrt3

[@paulortega5317]

First comment. You can reduce down first and avoid dealing with a quintic polynomial
The initial equation can be expressed as:
x³ = ((x - 1)² - 10x²)/((x - 1)² -10)
x³(x - 1)² - 10) = (x - 1)² - 10x²
x³(x - 1)² - 10x³ = (x - 1)² - 10x²
x³(x - 1)² - (x - 1)^2 = 10x³ - 10x²
(x - 1)²(x³ - 1) = 10x²(x - 1)
(x - 1)(x³ - 1) = 10x²
x⁴ - x³ - 10x² - x + 1 = 0
Second comment. Any time you see a format like x⁴ + ax³ + bx² +ax + 1 = 0 you know you can use u = x + 1/x to simplify it. As you did. In this case a = -1 and b = -10

[@rabotaakk-nw9nm]

x=1 => x-1=0 !!! 😬 =>
(x-1)²(x³-1)-10x²(x-1)=0
(x-1)((x-1)(x³-1)-10x²)=0
(x-1)(x⁴-x³-10x²-x+1)=0
=> x-1=0 => x1=1…………

[@DhirajKumar-g]

Hey !can try this problem with at least two different methods , the program looks like -: x + 3x-y/(x²+y²) =3 and y -(x+3y)/(x²+y²). Solve for x and y

[@Comedy_boy_30]

I am also wondering with this equations to solve but can't .One method to solve this is the video made on it by the channel prime newton may be you know about this channel. I give this equations to my teacher but he could not able to solve this problem😢😢😢 so bad. Plz share this to different persons who have wide knowledge about algebra

[@DhirajKumar-g]

I know shashwat 😂

[@RashmiRay-c1y]

Cubing both sides, we get x^5-1 -2x^4+2x-9x^3+9x^2=0 > (x-1)[x^4-x^3-10x^2-x+1]=0. So, x=1 or x^4-x^3-10x^2-x+1=0 > x^2+1/x^2 -(x+1/x) = 10 > (x+1/x)^2 - (x+1/x) = 12 > x+1/x= -3,4 > x^2+3x+1=0 or x^2-4x+1=0. Thus the valid solutions are, x=1,1/2(√5-3), -1/2(√5+3), 2+√3, 2-√3.

[@walterwen2975]

An Algebra Challenge: x = ³√[(1 - 2x - 9x²)/(x² - 2x - 9)], x ϵ R; x =?
x ≠ 0; x³(x² - 2x - 9) = 1 - 2x - 9x², x⁵ - 2x⁴ - 9x³ = 1 - 2x - 9x²
x⁵ - 2x⁴ - 9x³ + 9x² + 2x - 1 = 0, (x⁵ - 2x⁴ + x³) - (10x³ - 10x²) - x² + 2x - 1 = 0
x³(x² - 2x + 1) - 10x²(x - 1) - (x² - 2x + 1) = x³(x - 1)² - 10x²(x - 1) - (x - 1)² = 0
(x - 1)[x³(x - 1) - 10x² - (x - 1)] = (x - 1)(x⁴ - x³ - 10x² - x + 1)
x⁴ - x³ - 10x² - x + 1 = (x⁴ + 2x² + 1) - x³ - x - 12x²
= (x² + 1)² - x(x² + 1) + (3x)(- 4x) = (x² + 1 - 4x)(x² + 1 + 3x)
x⁵ - 2x⁴ - 9x³ + 9x² + 2x - 1 = (x - 1)(x² - 4x + 1)(x² + 3x + 1) = 0
x - 1 = 0, x² - 4x + 1 = 0 or x² + 3x + 1 = 0
x = 1, x² - 4x + 4 = (x - 2)² = (√3)², x = 2 ± √3, or x = (- 3 ± √5)/2
Answer check:
x = ³√[(1 - 2x - 9x²)/(x² - 2x - 9)]; Convert to the following Quintic Equation
x⁵ - 2x⁴ - 9x³ + 9x² + 2x - 1 = (x - 1)(x² - 4x + 1)(x² + 3x + 1)
x = 1: ³√[(1 - 2x - 9x²)/(x² - 2x - 9)] = ³√[(1 - 2 - 9)/(1 - 2 - 9)] = 1 = x; Confirmed
x = 2 ± √3: x² - 4x + 1 = 0; (x - 1)(x² - 4x + 1)(x² + 3x + 1) = 0; Confirmed
x = (- 3 ± √5)/2: x² + 3x + 1 = 0; (x - 1)(x² - 4x + 1)(x² + 3x + 1) = 0; Confirmed
Final answer:
x = 1, x = 2 + √3, x = 2 - √3, x = (- 3 + √5)/2 or x = (- 3 - √5)/2

[@RealQinnMalloryu4]

(x^3 )➖ (1)2x ➖ 9x^3/(x^2)^2➖( 2x )^3➖ 9 ={x^9 ➖ 1}2x ➖ 9x^2/{x^4 ➖ 4x^2} ➖ 9={x^8*2x} ➖ 9x^2/4x^2 ➖ (9)^2= 2x^8 ➖ (9x^2)^2/4x^2 ➖ (9)^2={2x^8 ➖ 81x^4}/{4x^2 ➖ 81}=79x^4/87x^2 79^1x^2^2/3^29x^2 1^1x^1^1/3^29^1x^2/3^1^1x^2 3x^2 (x ➖ 3x+2).

[@Quest3669]

X= 1; 2+- √3; (-3+-√5)/2