Solving a quartic equation using an unusual idea. An algebraic challenge.

This video is about a quartic equation that can be solved without the formula. Two methods shown.

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[randomjin]

You can actually "cheese" through the problem once you noticed that 182 = 13² + 13. You can then re-write this equation " *in 13* " (yep, sounds silly, but treat "13" as a variable) - it will be "quadratic in 13". It ends up like this: 13² + (2x²+1)13 + (x⁴-x) = 0, leading to 13 = (-(2x²+1)±√4x²+4x+1)/2, so 13 = (-(2x²+1)±(2x+1))/2 and this yields just two quadratics in *x*

[alnitaka]

The second method is really the quartic formula in disguise. Assume the equation is x^4+px^2+qx+r=0, and set the polynomial equal to (x^2+ax+b)(x^2-ax+c) as you say above. This gives b+c-a^2=p, a(c-b)=q, and bc=r. This means that (c+b)^2-(c-b)^2 = (c+b+c-b)(c+b-c+b) = 2c(2b) = 4bc = 4r. But c+b= p+a^2, and c-b = q/a. This means that (p+a^2)^2-(q/a)^2 = 4r. Multiply by a^2 and get a^2(p+a^2)^2-q^2 = 4a^2r. Simplifying this equation and letting y = a^2 yields y^3+2py^2+(p^2-4r)y-q^2=0, which is the cubic resolvent, the central part of the quartic formula.

[adandap]

The first approach is neat. You can streamline the argument a bit if you realise that if functions f and f inverse are equal, it can only be at a point where they both equal x. So you can just set x^2 + 13 = x and solve right away. Then the other two roots can be got by dividing the quartic by that quadratic. (BTW, I think there's a little consistency problem in the first approach. You argued that x - 13 > 0 allows you to take a positive square root, but then get complex values for x.)

[gilbertogarbi]

I enjoy youŕ classes. Congratulations.

[rijadmuric]

You should finish the second method off too bro, you know we ain't gonna do it "at home" 🤣 btw awesome 👏🏿👏🏿👏🏿

[mxsjncv]

I think it is "good" to mention that 182=2x91 so we can real quick assign 1, -1, 2, - 2, 91, -91, 182, -182 and see that there are no rational roots.

[-basicmaths]

X^4+26x^2-x+182=0
For real values of x the above equation is greater than zero.
x is complex number.
x^4+26x^2+182=x
x^4+27x^2+182=x^2+x
(x^2+13)(x^2+14)=x(x+1)
x^2+13=x
x^2-x+13=0
x={1±(√1-52)}/2

[GravityMaster]

11:05 How do you check if polynomials have real solutions? Are you referring to the descartes rule of signs+ some other tricks?

[martylewinter]

Both methods are interesting. Good presentation.

[davidbrisbane]

@12:30 You could have just jumped to this equation immediately by noticing that the right hand side of the original equation is always greater than zero and so there are four complex roots so it factors as two quadratic equations whose discriminates are complex, which when multiplied out, has no x^3 term.

[strikerstone]

An explanation to second method : if u input c+b-a²=26 = eqn1, a(c-b)=-1=eqn2 and bc=182=eqn3 in wolfram you'll get abc = (-1, 13, 14)(1, 14, 13) and 4 other complex sets which we dont care, so try eliminating b first then c, cause a is easiest to find, by substituting b=182/c in eqn2 you'll see (a)c² + c +(-182a) = 0, which is a quadratic eq where c = (-1±√(1+728a²))/(2a), putting that in eqn1, you will quickly notice √(1+728a²) must be √729 cuz its perfect square so a must be 1 and later you can find b and c

[seegeeaye]

the eqution is written as X^2 (X^2 + 26) = x - 182. The LHS is a U-shaped curve touching at (0, 0), the RHS is a straight line cutting the y-axis at --182 with slope = 1, they do not intersect,
we therefore know before solving, the equattion has two pairs of complex numbers

[banglacartoon]

You can use greaffe's root square method

[vinayakjoshi]

I did exactly the same as the 1st method. Your videos are fun.

[meetwhole]

Can you explain 2 things for me?

First, when I put x^2+13=(x-13)^(1/2) in my calculator it says there are no solutions for the equation. How is that so, even though it's derived from the initial equation? Is it because the 'x' on the right is caught under the square root symbol?

Second, after you got to (x-y)(1+x+y)=0, how did you come to determine that x-y=0? I heard something earlier about the inverse functions being symmetric where x=y or something like that. Can you elaborate?
Thank you. This is a great video!

Edit: I understand the second part lol.

[UCDMconJoseMande]

in 3:23 you have (x²+13)^2 = x-13 being the right hand side non negative for x greater than 13, we expect to get real solutions from then on. But the function on the left hand side has derivative grater than the function on the right hand side, and, since its value at x=13 is greater than 0 (the value of the line at 13), there can be no solutions.

[ramaprasadghosh]

x^4 +26 x^2 - x + 182 = 0
can be rewritten as
(x^2 +13)^2 - x +13 = 0
or x^4 + 2*13*x^2 - x + 13*14 = 0
Coefficient of x^3 being zero, this can be rewritten as
(x^2 +bx + 13)( x^2 - bx + 13+1) = 0
comparing coefficients of x one gets
13*b - (13+1)b = -1 i.e. b =1
comparing coefficients of x^2 one gets
13+1 +13 -b*b = 13*2 i.e b*b =1
(which is in compliance of b=1)
Hereby, given equation can be rewritten as
(x^2 +x + 13)( x^2 - x + 13+1) = 0

[NadiehFan]

If you want to factor a quartic into two quadratics there are smarter ways than what you do in your second method. The idea is to transform the quartic into a difference of two squares where one square is the square of a quadratic polynomial whereas the other square is the square of a linear polynomial. Then, when we use the difference of two squares identity we get a product of two quadratics and we have achieved what we wanted.

The equation is

x⁴ + 26x² − x + 182 = 0

As you already noted we have x⁴ + 26x² + 169 = (x² + 13)² so we can write the equation as

(x² + 13)² − (x − 13) = 0

Now, the first term is already a square of a quadratic polynomial, but (x − 13) is not a square of a linear polynomial. But what we can do here is introduce a parameter t and add 2(x² + 13)t + t² because (x² + 13)² + 2(x² + 13)t + t² = (x² + 13 + t)² so the first term will then remain a square of a quadratic polynomial regardless of the value of t. But of course if we add 2(x² + 13)t + t² = 2tx² + 26t + t² we also need to subtract this again, so the equation then becomes

(x² + 13 + t)² − (2tx² + x + t² + 26t − 13) = 0

Now, for any value of t unequal to zero 2tx² + x + t² + 26t − 13 is a quadratic polynomial in x, and this will be a perfect square, i.e. the square of a linear polynomial in x, if the discriminant of this quadratic is zero, that is, if t satisfies

1² − 4·2t·(t² + 26t − 13) = 0

This is a cubic equation in t, but rather than first expanding the left hand side it is a good idea to start looking for rational solutions right away. It is clear that the left hand side is equal to 1 for t = 0 and negative for t = 1 so there must be a root on the interval [0, 1]. We may also note that 26t − 13 = 0 for t = ½ and for t = ½ we therefore get 1 − 4·2·½·¼ = 0, so t = ½ is indeed a solution of this equation in t. With t = ½ our quartic equation becomes

(x² + 13 + ½)² − (x² + x + ¼) = 0

and indeed x² + x + ¼ = (x + ½)² is a perfect square, so we have

(x² + 13½)² − (x + ½)² = 0

Applying the difference of two squares identity this can be written as

(x² − x + 13)(x² + x + 14) = 0

and we have factored the quartic into two quadratics.

[shanmugasundaram]

Could you please solve the quartic equation 2x^4+12x^3+2x^2-72x-63=0 the equation you derived in solving a geometrical problem.

[jimmyzhang]

In method 2 at 12 minutes, I notice 182=13x14, Let b=13, c=14 . Is it any better?