My Math Forum Absolute value term in a quadratic.

 Algebra Pre-Algebra and Basic Algebra Math Forum

 December 17th, 2015, 10:43 AM #1 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,624 Thanks: 954 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, 11:02 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,512 Thanks: 2514 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 11:48 AM.
 December 17th, 2015, 02:45 PM #3 Global Moderator   Joined: May 2007 Posts: 6,641 Thanks: 625 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 11:45 PM. Reason: correct error
 December 17th, 2015, 03:11 PM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,512 Thanks: 2514 Math Focus: Mainly analysis and algebra I do think that is probably the easiest approach.
December 17th, 2015, 05:22 PM   #5
Math Team

Joined: Oct 2011

Posts: 13,624
Thanks: 954

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, 03:47 PM   #6
Global Moderator

Joined: May 2007

Posts: 6,641
Thanks: 625

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, 04:29 PM #7 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,624 Thanks: 954 Once more: I simply made up an equation. Purpose was not to solve it, but use it as illustration. Merci.
December 19th, 2015, 04:08 PM   #8
Global Moderator

Joined: May 2007

Posts: 6,641
Thanks: 625

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, 05:18 PM #9 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,624 Thanks: 954 Agree...but much easier when writing a program...my only purpose. That one was easy to handle, but not all are...

,

### absolute term in quadratic eq

Click on a term to search for related topics.
 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post corleoneee Calculus 5 October 22nd, 2013 01:31 AM shiseonji Algebra 2 September 24th, 2013 09:36 AM David_Lete Calculus 2 January 31st, 2011 08:13 PM benoit Linear Algebra 0 December 1st, 2010 03:34 AM forcesofodin Real Analysis 4 August 14th, 2010 06:50 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top