Number of real roots of the equation $(6-x)^4+(8-x)^4 = 16$

Question

Find number of real roots of the equation $(6-x)^4+(8-x)^4 = 16$

$\bf{My\; Try::}$ Let $f(x) = (6-x)^4+(8-x)^4\;,$ and we have to find real values of $x$ for

which $f(x) = 16$. Now we will form Different cases.

$\bf{\bullet \; }$ If $x<6$ or $x>8\;,$ Then $f(x)>16$

$\bf{\bullet\; }$ If $6 <x<8\;,$ Then $f(x)<16$.

$\bf{\bullet \; }$ If $x=6\;,x=8\;,$ Then $f(x) = 16$

So Solutions of the above equation are $x=6$ and $x=8$

Can we solve it using Derivative or Algebraic way

Thanks

Answer 1

looking at the solution of $(6-x)^4 + (8-x)^4 = 16$ geometrically as the solution of the simultaneous equations $(6-x)^4 + (8-y)^4 = 16$ and $y = x.$ one is "squashed" circle centered at $(6,8)$ with radius $2$ and the line $y = x$ which is unit distance from the center. i expect the line to cut the circle at two points. the $x$-coordinates of those points are your two solutions. i will confirm this when i get time to graph these.

i see from the comments that there are two solutions.

Answer 2

expanding and factorizing we obtain $2\, \left( x-6 \right) \left( x-8 \right) \left( {x}^{2}-14\,x+56 \right) =0$ from here you will get all solutions.

Answer 3

You can exploit the symmetry of the situation to simplify things if you set $y=7-x$ so that the equation becomes $(y-1)^4+(y+1)^4=16$ or $2y^4+12y^2+2=16$ or $y^4+6y^2-7=0$

This factorises as $(y^2-1)(y^2+7)=0$ with real solutions $y=\pm 1$ corresponding to $x=6, x=8$

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Note this symmetric trick is bound to work, because it gives a symmetric function in $y$, which is inevitably a function of $y^2$ (no odd order terms) - and the order is such that it can't be worse than quadratic in $y^2$, which can therefore be solved.

Answer 4

HINT:

Write $8-x = 2 a$, $\ x-6=2b$. Then $\ a+b=1$ and $a^4+b^4=1$. Draw some level curves.

$\bf{added:}$

Answer 5

I guess by Algebraic you mean this way

$$(6-x)^4+(8-x)^4 = 16$$ put $7-x=t$ (reason behind this substitution is not just because the later is easy to expand using binomial formula but the magic it does). $$(t-1)^4+(t+1)^4 = 16$$ $$(t^4-4t^3+6t^2-4t+1)+(t^4+4t^3+6t^2+4t+1)=16$$

$$2(t^4+6t^2+1)=16$$

$$t^4+6t^2+1=8$$ Well, its quartic but it's easy to find it's roots

put $p=t^2$ $$ p^2+6p-7=0$$ $$\implies p=1,p-7$$ Now $p=-7$ will not yield any real solution so, we ignore that $$\implies t=\pm1$$

$$x=7-t\implies x=6,8$$

Hence, only two real roots, other two being complex.