Special Functions

harrychess · 2014-05-16 11:54:23

Find all solutions of the equation |x^2 - 14x + 29| = 4. Discuss whether or not your solution generates extraneous solutions.

ShivamS · 2014-05-16 11:59:21

x = 3
x = 11
x = - 2sqrt(6) + 7
x = 2sqrt(6) + 7

No extraneous solutions.

Last edited by ShivamS (2014-05-16 12:16:50)

harrychess · 2014-05-16 12:29:15

how did you do it?

ShivamS · 2014-05-16 12:30:41

Do you know how to solve absolute value equations?

harrychess · 2014-05-16 12:39:17

dont you set if its positive or its negative? 2 cases?

ShivamS · 2014-05-16 12:40:34

Yes. Set it like that and complete the square on the quadratic polynomial and then solve it.

harrychess · 2014-05-16 13:03:24

Wait, isnt it root 5?

ShivamS · 2014-05-16 13:33:13

Hi harrychess;

From |x^2 - 14x + 29| = 4, we can create two equations.

+(x^2 - 14x + 29) = 4 which is equivalent to simply x^2 - 14x + 29 = 4
or
-(x^2 - 14x + 29) = 4 which is equivalent to simply -x^2 + 14x - 29 = 4

Those two are simply quadratic equations which we can solve through a number of ways, such as completing the square, using the quadratic formula, partial factoring, factoring etc. By doing that, we can have the four solutions.

To complete the solution, just use the quadratic formula on x^2 - 14x + 29 = 4 and -x^2 + 14x - 29 = 4

In this example a=1,b=−14,c=25. Just plug in the values for a, b, and c into the quadratic formula.

Last edited by ShivamS (2014-05-16 13:38:44)

Math Is Fun Forum

#1 2014-05-16 11:54:23

Special Functions

#2 2014-05-16 11:59:21

Re: Special Functions

#3 2014-05-16 12:29:15

Re: Special Functions

#4 2014-05-16 12:30:41

Re: Special Functions

#5 2014-05-16 12:39:17

Re: Special Functions

#6 2014-05-16 12:40:34

Re: Special Functions

#7 2014-05-16 13:03:24

Re: Special Functions

#8 2014-05-16 13:33:13

Re: Special Functions

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