A Non-Standard Radical Equation

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[snejpu]

Wow, nice similar. I had quite similar, but I didn't know how to find a solution of the cubic equation, so I proved there is only one solution using a derivative, then I estimated it to be between -3 and -2 and then I used Newton's method. I'm not accustomed to engaging a formula for the root of a cubic equation.

[pk]

The second value of b leads to an extraneous solution . The value of x that it yield leads to a positive value on the left hand side of the original equation and a negative value on the right hand side of the original equation . This extraneous solution happens because in part of your solution you raise (b+1) to the fourth power .

[raystinger]

I just used x=u³ in the expression and got:
4root(u³ + 15) = u + 1
Then elevated both sides to the 4th power:
u³ + 15 = u^4 + 4u³ + 6u² + 4u + 1
=> u^4 + 3u³ + 6u² + 4u - 14 = 0
=> (u - 1)(u³ + 4u² + 10u + 14) = 0
And it was at that point I thought I was doing everything the hard way, because that 3rd degree polynomial didn't have "nice" roots, so I gave up and started watching the video - that's when I discovered this problem isn't "nice" at all. ¯\_(ツ)_/¯

[math]

this one was complex, great job bro, watching a master

[srijanbhowmick]

6:55 Here's an easy way to handle the cubic term, just substitute b = a - 1 in the cubic polynomial, you will get ( a^3 + a^2 + 5a + 4 ) . Now, since *a* is a 4th root, hence a >= 0 which means a^3 + a^2 + 5a + 4 >= 0 + 0 + 0 + 4 = 4 > 0, therefore the cubic term never equal to 0 hence the only solution is x = 1 :)

[GirishManjunathMusic]

⁴√(x + 15) = ³√x + 1

Substitute x = u³:
⁴√(u³ + 15) = u + 1

Raise to 4th power:
u³ + 15 = u⁴ + 4u³ + 6u² + 4u + 1

Move all terms to LHS and negate:
u⁴ + 3u³ + 6u² + 4u – 14 = 0.

As the sum of the coefficients is zero, u – 1 is a factor.
Dividing through,
(u – 1)|(u⁴ + 3u³ + 6u² + 4u – 14) = (u³ – 4u² + 10u – 14)
(u – 1)(u³ – 4u² + 10u – 14) = 0
u = 1 or u³ – 4u² + 10u – 14 = 0.

However, this cubic has no nice factorisation, so we'll have to use the cubic formula or substitution to solve it. Wolfram Alpha suggests substituting 'u = v + 4/3'. I'm not going to solve it through here, but there's only one real solution to the cubic with a really messy exact form. So lets take u = 1. Now, x = u³ so:
x = 1.

[kprdqvg]

The only real solution of b^3+4b^2+10b+14=0 is b=-2.29906. Thus b+1 is negative whilst (15+b^3)^1/4 is positive. Therefore this value of b does not satisfy the original equation. The only solution (real) is b=1 > x=1.

[PatKinson]

X is also equal to approx -12.167 ... the other value (-12.167)^1/3 = -2.3 + 1= -1.3 = absolute value = 1.3, calculation was for the right side proof.
now let us proof check the left side: (-12.167 + 15)^1/4 = (2.833)^1/4 = approx. 1.297... absolute(1.3) approx. 1.297
-2.3 plugs into the cubic equation converges close to zero... u^3+4u^2+10u +14=0.... (-2.3)^3 = -12.167

[wfzvdsb]

> 0+31/4-1=27/4>0, therefore only x = 1

[hawkeyexenotics]

Cardano´s solution formula is so complicated, so very few students challenging for entrance exams can
figured out non-integer solutions.

[adamsulc]

If you graph both functions, they intersect only once. Therefore I think, there is only one real solution.

[bgubtzw]

Левая часть неотрицательна.
Значит и правая часть тоже.
squb x+1>=0. x>=-1.
Только один корень x=1

[leecherlarry]

your 2nd solution *x=-12.1521* (approx. value) results in *1.29906* for the left-hand side of your problem statement (positive!) and *-1.29906* for the right-hand side (negative!).
in addition, wolframalpha shows 1 solution only, enter
<Power[x + 15, (4)^-1] == Power[x, (3)^-1] + 1> on the website.

I am confident that there is only 1 solution, namely x=1:
*𝚁𝚎𝚍𝚞𝚌𝚎[𝚂𝚞𝚛𝚍[𝟷𝟻 + 𝚡, 𝟺] == 𝚂𝚞𝚛𝚍[𝚡, 𝟹] + 𝟷, 𝚡]*

[manojsurya]

Woah!one of the solution is very radical

[johanrichter]

The complicated solution must be a false root. a must be non-negative but b+1 would be negative.

[JojoDrs_]

Me: looks to complicated, so I just plug in 1 and look if it's right
Also me: lol

[carloshuertas]

I got one of the solutions. Another great explanation, SyberMath!

[chessdev]

I only found x=1 as a solution and I think I did it right(its good not to find the other solution since its ugly and perhaps don't satisfy the equation)

[tomasstride]

I think may be the second solution does not actuallly satify the equation.It satifies the negative fourth root rather than the positive. But I could be wrong.

[mathocean]

Could you please show us in detail how to get second solution of b?