Solve Real Roots of $(x-3)^4+(x-7)^4=24832$

[anemone]

Solve for real roots of the equation $(x-3)^4+(x-7)^4=24832$.

 

[MarkFL]

Here is my solution:

Spoiler

Expanding, dividing through by 2, and writing in standard form, we obtain:

$$x^4-20x^3+174x^2-740x-11175=0$$

Let's define:

$$f(x)=x^4-20x^3+174x^2-740x-11175$$

Using the rational roots theorem, we find:

$$f(-5)=0$$

$$f(15)=0$$

And so, carrying out the division, we find:

$$f(x)=(x+5)(x-15)\left(x^2-10x+149\right)$$

The discriminant of the quadratic factor is negative, hence the only real roots are:

$$x=-5,\,15$$

 

[kaliprasad]

Spoiler

We can convert this quartic equation to quadratic by putting (x-5) = t (x-5 is mean of x-3 and x-7)
So we get
$(t+2)^4 +(t-2)^4 = 24832$
$2(t^4 + 6 t^2 (-2)^2 + 16) = 24832$
or $t^4 + 24 t^2 = 12400$
$t^4 + 24 t^2 – 12400 = 0$
or $(t^2 – 100)(t^2 + 124) = 0$
so $t^2$ = 100 or t = + or – 10 or x = -5 or 15

 

[anemone]

Thanks to MarkFL and kaliprasad for participating and provided the good method with correct answers and kali, I remember you once used the same trick to crack my other challenge problem!(Sun)