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The sum of the roots of the equation $(x-4)^2+8|x-4|+15=0$ is

My attempt: Let $|x-4|=y$. So, the equation becomes $y^2+8y+15=0$. So, $y=-3,-5$. Both values should be rejected as $|x-4|$ cannot be negative. But the answer has been given as $16$.

aarbee
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    I think you're working too hard. $(x-4)^2\ge0$, $8|x-4|\ge0$, $15>0$, so the equation can't have any solutions. If someone has asked you to solve this problem, I advise you to ask them to clarify. (But first check to see whether you have copied it down correctly.) – Gerry Myerson Aug 07 '19 at 06:31
  • I have copied the statement correctly. But I see your point, it should be clear at the beginning itself that the equation has no solutions. – aarbee Aug 07 '19 at 06:35
  • Your answer is correct as well. As stated it has no solutions. You can plug in $x=16$ and see that it doesn't solve the equation. – quarague Aug 07 '19 at 06:36
  • @quarague, $16$ is the sum of roots. – aarbee Aug 07 '19 at 06:38

1 Answers1

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Since the answer has been given as $16$, I think that the equation should read

$$(x-4)^2-8|x-4|+15=0.$$

Now let $y:=|x-4|$ and proceed as above.

Fred
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