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# help

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Solve y^2 - |4y - 2| + 5 = 0.

Nov 18, 2019

#1
+29257
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"Solve y^2 - |4y - 2| + 5 = 0."  I'm going to assume only real solutions are required.

(1) Suppose 4y-2 is positive.  Then we have y2-4y+7=0    but b2-4ac = 16 - 28 <0 so solutions are complex.

(2) Suppose 4y-2 is negative.  Then we have y2+4y+3=0. so y = (-4±√(16-12))/2 or y = -1 and y = -3

Check:

If complex solutions are required as well, just solve the quadratic in (1).

Nov 18, 2019
#2
+109750
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I agree with your real solutions, Alan

I also thought that the complex solutions to  y^2 - 4y + 7 =  0    were answers

But...Wolframalpha  gives  these  answers for the complex solutions :

https://www.wolframalpha.com/input/?i=y%5E2+-++abs+%284y+-+2+%29++%2B+5+%3D0

Do you see how they got those  ???

Nov 18, 2019
#3
+29257
+3

Hmm!  My simplification when |4y-2| is positive leads to incorrect solutions.  i.e. when the solutions are plugged back into the LHS of the original equation the resut isn't zero.   The Wolfram Alpha solutions, on the other hand, do give zero, so they are obviously correct.  I will need to think further about how to obtain these!

Alan  Nov 18, 2019
edited by Alan  Nov 18, 2019
#4
+29257
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Further thought has produced the following:

I guess WolframAlpha used something like this reasoning!

Alan  Nov 18, 2019
#5
+109750
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THX, Alan....I'm gonna' need to ponder this one....!!!!

CPhill  Nov 18, 2019