- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,808

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

- Thread starter anemone
- Start date

- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,808

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

- Admin
- #2

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

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:

\(\displaystyle x=-5,\,15\)

- Mar 31, 2013

- 1,333

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

- Feb 14, 2012

- 3,808