The Latex bragger strikes again!

https://latex.artofproblemsolving.com/miscpdf/ujuyolds.pdf?t=1566795355878


We can find all the solutions of the equation $|x^2-14x+29|=4$ by splitting it up into two cases:
$$x^2-14x+29=4$$
$$-x^2+14x-29=4$$

Looking at the first case, where $x^2-14x+29\geq 0$, we begin to notice that in the form of a quadratic $ax^2+bx+c$ $a$ is 1 and $b$ is even, making the scenario the easiest to complete the square.

$$x^2-14x+29=4$$
$$x^2-14x=-33$$
$$x^2-14x+49=-33+49$$
$$(x-7)^2=16$$
$$x-7=\pm 4$$

We can then split this into two cases again:
$$x-7=+4$$
$$x-7=-4$$

$$x=11$$
$$x=3$$


Now that we found the values of $x$ when $x^2-14x+29\geq 0$, we can move on to our second case of when $x^2-14x+29<0$

$$-x^2+14x-29=4$$
$$-x^2+14x-33=0$$
$$(-x+11)(x-3)=0$$

$$x=11, 3$$

Interesting...


Now that we have our final two values of $3, 11$ we have to check these values in case there are many erroneous cases since in the first case we took the square root of both sides.


Plugging in $3$ gives us:
$$|x^2 - 14x + 29|=4$$
$$|(3)^2 - 14(3) + 29|=4$$
$$|9 - 42 + 29|=4$$
$$|9 - 42 + 29|=4$$
$$|-4|=4$$
$$4=4$$

Plugging in $11$ gives us:
$$|x^2 - 14x + 29|=4$$
$$|(11)^2 - 14(11) + 29|=4$$
$$|121-154+29|=4$$
$$|-4|=4$$
$$4=4$$

Now that we have checked our work, we finally get that all the solutions of the equations $|x^2 - 14x + 29|=4$ are $\boxed{\boxed{3, 11}}$


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