My Math Forum Absolute value term in a quadratic.

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 December 17th, 2015, 09:43 AM #1 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,316 Thanks: 1023 Absolute value term in a quadratic. Example: x^2 + |x - 7| - 9 = 0 Can be solved this way : x^2 + SQRT[(x - 7)^2] - 9 = 0 SQRT[(x - 7)^2] = 9 - x^2 squaring both sides and simplifying: x^4 - 19x^2 + 14x + 32 = 0 Soooo....got rid of absolute term |x - 7| After googling "quadratic with absolute value term", I saw a few ways to handle these, but not the above. Did I simply "invent something stupid"?
 December 17th, 2015, 10:02 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,636 Thanks: 2621 Math Focus: Mainly analysis and algebra That looks fairly normal to me. I'd consider writing $y=x-7$ so that $x^2=y^2+14y+49 = |y|^2\pm 14|y|+49$ and substituting into the original equation allows the use of the quadratic formula to garner solutions. Last edited by v8archie; December 17th, 2015 at 10:48 AM.
 December 17th, 2015, 01:45 PM #3 Global Moderator   Joined: May 2007 Posts: 6,730 Thanks: 689 You have two possibilities: x^2+x-16 (x > 7) or x^2-x-2 (x < 7). Solve for x in each case and use any solution fitting the condition. Note: I found no solution for the first case and 2 for the second. Last edited by mathman; December 17th, 2015 at 10:45 PM. Reason: correct error
 December 17th, 2015, 02:11 PM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,636 Thanks: 2621 Math Focus: Mainly analysis and algebra I do think that is probably the easiest approach.
December 17th, 2015, 04:22 PM   #5
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Quote:
 Originally Posted by mathman You have two possibilities: x^2+x-16 (x > 7) or x^2-x-2 (x < 7).
Ya, that's the "popular" approach; I was finding a one operation
to replace the two operations...no big deal...

December 18th, 2015, 02:47 PM   #6
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Quote:
 Originally Posted by Denis Ya, that's the "popular" approach; I was finding a one operation to replace the two operations...no big deal...
The x > 7 branch has no roots - by inspection, since the polynomial is positive and increasing for that case. This leaves a simple quadratic for x < 7.

 December 18th, 2015, 03:29 PM #7 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,316 Thanks: 1023 Once more: I simply made up an equation. Purpose was not to solve it, but use it as illustration. Merci.
December 19th, 2015, 03:08 PM   #8
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Quote:
 Originally Posted by Denis Once more: I simply made up an equation. Purpose was not to solve it, but use it as illustration. Merci.
You replaced a quadratic equation (two equations actually), easy to handle, into a fourth degree equation. In general - not a good idea.

 December 19th, 2015, 04:18 PM #9 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,316 Thanks: 1023 Agree...but much easier when writing a program...my only purpose. That one was easy to handle, but not all are...

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