Solving a very radical equation. An algebraic challenge.

This video is about solving a radical equation. See what happens if you square both sides.

Radical expressions,algebra,algebraic equations,SyberMath,algebraic manipulations,equations,substitution,Challenging Math Problems,Non-routine Math Problems,polynomial equations,an algebra challenge,algebraic identities,non-standard methods,symmetry,math,maths,mathematics,An algebraic challenge,Polynomial equations,an algebraic challenge,an equation,cubic equations,power of substitution,systems of equations,quadratic formula,Solving a very radical equation

[mva]

Sustituting u^3 = 2x - 6 makes for an easier substitution, leading to (sqrt(u^3+8)/2) = u+2 which, after squaring both sides and simplifying leads to u^3-2u^2-u = 0, or u(u-4)(u+2)=0.

[kippie]

A very unexpected solution to the equation. Very clever.

[yusrizalakbar]

more radical equation please, your content is amazing!

[armacham]

I just wanted to share another way to solve it: I moved the cube root by itself ( sqrt(x+1) - 2 = (2x-6)^(1/3) ) and then I substituted u=sqrt(x+1), so u-2 = (2x-6)^(1/3).

if you cube both sides, you get (u-2)^3 = 2x-6

expanded, that is uuu-6uu+12u-8 = 2x-6

if you add 8 to both sides, you get uuu-6uu+12u = 2x+2

because u = sqrt(x+1), uu = x+1

2x+2 = 2(x+1) = 2u^2

so you can write it as: uuu-6uu+12u = 2uu

from there, it is trivial to move everything to one side, factor it, and then solve for x based on the possible values of u

[tarekraja]

Such a complicated equation, very nice and very clever way done. It is amazing. Thank you sir.

[michidayo_]

There are many questions that are different from the ones taught in Japan, which is helpful. thank you.
from Japan.

[-basicmaths]

Just put x+1=t^2 and then transfer the equation in terms of t.
t-2=cubrut√2t^2-8
t^3-6t^2+12t-8=2t^2-8
t^3-8t^2+12t=0,
t(t^2-8t+12)=0
t(t-6)(t-2)=0
t=0, √x+1=0, x+1=0, x=-1
t=6, x=35,
t=2, x=3
x={-1, 3, 35}

[ioannismichalopoulos]

Sorry SyberMath, I have an objection! At 1:40 the solution is just in front of your eyes! I agree with what you say at 2:54 ie that "one of the most powerful methods in mathematics is substitution". Thus, I propose a substitution as follows:
u = (2x-6)^(1/3) <=> x = (u^3 / 2) + 3
Thus, the equation at 1:40 is transformed as follows:
u^2 = u^3 / 2 -4u <=> u^3 -2u^2 - 8u = 0 <=> u(u+2)(u-4)=0
u = 0 <=> x = 3
u = -2 <=> x = -1
u = 4 <=> x = 35

[paultoutounji]

My favorite video ! Love those kind of equations. Hope you will present some more . Thank you very much !!!! 👍😀

[amoswittenbergsmusings]

Much simpler to substitute x=u^2-1. Then the second term becomes cube root ot 2u^2-8 or 2(u+2)(u-2). Isolate the cube root and cube the entire thing. That gives (u-2)^3 =2(u+2)(u-2). The common factor (u-2) stares you in the face and the rest is a simple quadratic, already factored. The advantage of this approach is that the factoring already happens inside the cube root!

[trdastidar]

Take the equation in the form at 1:12, subtract 4 from both sides, and set y = (2x-6)^1/3. It transforms into a nice cubic equation y^3 -2y^2 - 8y = 0. This has 3 roots, y = 0, 4, -2. From this, the 3 x roots are 3, 35, -1. Probably no need to use 2 variables. Using techniques from your other videos :-)

[elifalk]

A much quicker way: reach the point in the aborted method; substitute b=cube root of 2x-6 (note that x-3=b^3/2), and you get the same cubic equation quickly.

[diedoktor]

I accidentally flipped a sign and had to redo the problem at one point but once I did it more carefully it wasn't difficult.
I started with the assumption that 2x-6 would be a perfect cube. Since 2 is even x+1 and 2x-6 both need to be even.
if x+1 is even then x is odd.
It's easy to check 2x-6=0 and see this is a solution.
sqrt(3+1)-cubert(2*3-6)=2
Since we have a solution if we rearrange the problem into cubic form we can reduce it to a quadratic. Isolate the cube root and cube both sides. Factor out the square root and isolate it then square both sides. Multiply both sides by the denominator and combine like terms.
The result is x^3-37x^2+67x+105=0
Factor out x-3.
(x-3)(x^2-34x-35)=0
You can either observe x=-1 is a solution and factor or use the quadratic formula.
17 +- sqrt(1156+140)/2
17 +- 18 =x
x= -1, 35, and 3

[protoroxsinha]

I'm learning a lot of fun things thank you

[neuralwarp]

What happens with the negative square roots of the first term for each solution?

[elkincampos]

Let u=√(x+1), u>=0. But (2*x+2-8). That is equivalent to u>=0 and (u-2)^3=2*u^2-8. u>=0 and u^3-6*u^2+12*u-8=2*u^2-8. u>=0 and
We obtain that u=0, u=2 and u=6 solve the equation.

[ananthiyengar]

Well done sir, I want More such problems

[XBGamerX]

just a clarification:

for the first root, you must put a rule and that is x ≥ 1, so x = -1 is not a real solution. also in the second root it also must be x ≥ 3

so the solutions are x = 3 and x = 35

[tmacchant]

2a^2-b^3=2(x+1)-(2x-6)=8.
a-b=2 => b=a-2.
Substitute b=a-2 to 2a^2-b^3=8.
I got a^3-8a^2+12a=0.

[jarikosonen]

This looks good. Can you make this if there is 2 square roots and 1 cube root summed at the left hand side of the equation?