Welcome to our community

Be a part of something great, join today!

Solve for real roots

  • Thread starter
  • Admin
  • #1

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,755
Solve for real roots of the equation $(x-3)^4+(x-7)^4=24832$.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Here is my solution:

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

\(\displaystyle x^4-20x^3+174x^2-740x-11175=0\)

Let's define:

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

Using the rational roots theorem, we find:

\(\displaystyle f(-5)=0\)

\(\displaystyle f(15)=0\)

And so, carrying out the division, we find:

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

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

\(\displaystyle x=-5,\,15\)
 

kaliprasad

Well-known member
Mar 31, 2013
1,322
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
 
  • Thread starter
  • Admin
  • #4

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,755
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)