What is wrong with my solving of this equation

Question

Equation : $(6-x)^4 + (8-x)^4 = 16$

Here is my solution

I am a high school student. According to this question's solution, the other two roots are 6 and 8. However, I am only getting 2 of the 4 roots. It would be very helpful if I know what I was wrong in this case, and how to get the other two roots.

Answer 1

That question is ugly: why use $6-x$ and $8-x$ instead of $x-6$ and $x-8$ when you're squaring it anyway?

So, you get:$$(x-6)^4+(x-8)^4=16$$

Now say $u=x-7$, then you get:

$(u+1)^4+(u-1)^4=16$, or $2u^4+12u^2+2=16$

This, you can solve to $u^2$ and then back to $x$.

As your method was right, this should give you the same results :-)

Answer 2

The approach mentioned in the comment by eyeballfrog and discussed by Dominique is likely the quickest way to solve the equation, since it exploits the symmetry of the function, on the left side of the equation, about $ \ x \ = \ 7 \ \ , \ $ which arises because each term is individually symmetric about their own "axes" at $ \ x \ = \ 6 \ \ $ and $ \ x \ = \ 8 \ \ . $

Another approach that might be used to save some writing is to re-arrrange the equation as $ \ (6-x)^4 + (8-x)^4 - 16 \ = \ 0 \ $ and apply the substitution $ \ u \ = \ 6 - x \ \ . \ $ We will only need to perform one "binomial expansion" in this way, producing $$ \ u^4 \ + \ (u \ + \ 2)^4 \ - \ 16 \ \ = \ \ u^4 \ + \ u^4 \ + \ 4·2·u^3 \ + \ 6·2^2·u^2 \ + \ 4·2^3·u \ + \ 2^4 \ - \ 16 $$ $$ = \ \ 2·u^4 \ + \ 8·u^3 \ + \ 24·u^2 \ + \ 32·u \ \ = \ \ 2·u \ · \ ( \ u^3 \ + \ 4·u^2 \ + \ 12·u \ + \ 16 \ ) \ \ = \ \ 0 \ \ . $$

This makes the coefficients far smaller than they become by "multiplying out" the two "binomial-fourths" in the original equation, so that in making use of the Rational Zeroes Theorem, it is easier to see that $ \ u \ = \ -2 \ $ is a zero of the cubic polynomial factor (the Rule of Signs tells us that it has no positive real zeroes). From here, synthetic or polynomial division shows that the equation can be factored as $$ 2·u·(u \ + \ 2) \ · \ ( \ u^2 \ + \ 2·u \ + \ 8 \ ) \ \ = \ \ 0 \ \ , $$ the remaining quadratic factor being irreducible over the real numbers. The zeroes of that factor are then $$ u \ \ = \ \ -\frac22 \ \pm \ \frac{\sqrt{4 \ - \ 32}}{2} \ \ = \ \ -1 \ \pm \ i·\sqrt7 \ \ = \ \ 6 \ - \ x \ \ , \ $$ confirming your result. [If one has already found "by inspection" of the original equation that $ \ x \ = \ 6 \ $ and $ \ x \ = \ 8 \ $ are two of its roots, then one could carry out a polynomial division $ \ ( \ 2·u^4 + 8·u^3 + 24·u^2 + 32·u \ ) \ / \ (u^2 \ + \ 2u) \ $ more immediately to obtain the second quadratic factor. [A similar procedure follows if we had chosen $ \ u \ = \ 8 - x \ $ instead.]

Answer 3

The reason you only got $2$ roots is you divided by functions of x on both sides of the equation. To illustrate: consider you have a quadratic $(x+5)(x+10)=(x+5)$. There are two solutions. Suppose we take $(x+5)$ from the right to the left hand side of the equation; $(x+5)(x+9)=0$. Now we see the solutions are $x\in\{-5,-9\}$. Suppose at the first step we decided instead to divide both sides by $(x+5)$ then unwittingly both sides have been divided by $0$ which is an invalid operation (if we repeated the process we get $1=0$, a contradiction). Nevertheless, doing so gives the constrained solution set $x=\{-9\}$. Until you know you're not dividing by $0$ you shouldn't divide by anything. If you're forced to divide by some function $f(x)$ you immediately invent the constraint $f(x)\neq 0$ on your solution-set out of thin air.