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

harrychess
Member
Registered: 2014-04-04
Posts: 33

### Special Functions

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

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

ShivamS
Member
Registered: 2011-02-07
Posts: 3,648

### Re: Special Functions

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)

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

harrychess
Member
Registered: 2014-04-04
Posts: 33

### Re: Special Functions

how did you do it?

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

ShivamS
Member
Registered: 2011-02-07
Posts: 3,648

### Re: Special Functions

Do you know how to solve absolute value equations?

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

harrychess
Member
Registered: 2014-04-04
Posts: 33

### Re: Special Functions

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

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

ShivamS
Member
Registered: 2011-02-07
Posts: 3,648

### Re: Special Functions

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

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

harrychess
Member
Registered: 2014-04-04
Posts: 33

### Re: Special Functions

Wait, isnt it root 5?

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

ShivamS
Member
Registered: 2011-02-07
Posts: 3,648

### Re: Special Functions

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)

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