# How do you solve (2-x)^4+(3-x)^4-(5-x)^4 = 0 ?

Aug 5, 2016

Use a numeric method to find approximations to the zeros:

${x}_{1} \approx 3.49773$

${x}_{2} \approx - 10.5327$

${x}_{3 , 4} \approx 3.5175 \pm 1.3997 i$

#### Explanation:

Note that:

${\left(a + b\right)}^{4} = {a}^{4} + 4 {a}^{3} b + 6 {a}^{2} {b}^{2} + 4 a {b}^{3} + {b}^{4}$

Hence:

${\left(2 - x\right)}^{4} = 16 - 32 x + 24 {x}^{2} - 8 {x}^{3} + {x}^{4}$

${\left(3 - x\right)}^{4} = 81 - 108 x + 54 {x}^{2} - 12 {x}^{3} + {x}^{4}$

${\left(2 - x\right)}^{4} + {\left(3 - x\right)}^{4} = 97 - 140 x + 78 {x}^{2} - 20 {x}^{3} + 2 {x}^{4}$

${\left(5 - x\right)}^{4} = 625 - 500 x + 150 {x}^{2} - 20 {x}^{3} + {x}^{4}$

${\left(2 - x\right)}^{4} + {\left(3 - x\right)}^{4} - {\left(5 - x\right)}^{4} = {x}^{4} - 72 {x}^{2} + 360 x - 528$

So the roots of the original equation are the zeros of:

$f \left(x\right) = {x}^{4} - 72 {x}^{2} + 360 x - 528$

By the rational root theorem, any rational zeros of this quartic are factors of $528$. That gives us rather a lot of factors to check, and none works. It is somewhat quicker to find approximations to the zeros using a method like Durand-Kerner.

Here's an example C++ program for this quartic... Using this program I found approximations:

${x}_{1} \approx 3.49773$

${x}_{2} \approx - 10.5327$

${x}_{3 , 4} \approx 3.5175 \pm 1.3997 i$