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Find all solutions of the equation(answers)

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

Guest Aug 14, 2018
 #1
avatar+91160 
+1

|x^2 - 14x + 29| = 4

 

We have that  either

 

x^2 - 14x + 29   = 4          or that         x^2  - 14x  +  29   = -4           simplify

 

x^2 - 14x  + 25  = 0                            x^2 - 14x + 33  = 0               factor

 

The second equation   factors as  ( x - 11) ( x - 3)  = 0

Setting  both factors to 0  and solving for x produces   x = 11   and x  = 3

 

For the first equation we can complete the square on x  to solve

x^2  - 14x   = -25       take 1/2 of 14  = 7....square it  = 49....add it to both sides

x^2 - 14x + 49 = -25 + 49       factor the left side....simpify the right

(x - 7)^2  = 24       take  both roots

x - 7  = ±√24

x - 7 = ±2√6    add 7 to  both sides

x = 7 ±2√6

 

None of the 4 solutions are extraneous   as we  can see by this graph....the x axis is  intersected  four times  [ 4 real "zeroes"]

 

https://www.desmos.com/calculator/8incrgx666

 

 

cool cool cool

CPhill  Aug 14, 2018
 #2
avatar
+1

Thank you sir!!!

Guest Aug 14, 2018
 #3
avatar+91160 
0

No prob....!!!!

 

 

cool cool cool

CPhill  Aug 14, 2018
 #4
avatar+20153 
0

Find all solutions of the equation(answers)

|x^2 - 14x + 29| = 4

 

\(\begin{array}{|lrclcrcl|} \hline & |x^2 - 14x + 29| &=& 4 \\ \Rightarrow & |(x-7)^2-20| &=& 4 \quad & | \quad \text{substitute} \quad u = (x-7)^2 \\\\ & |u-20| &=& 4 \\ & u_1-20 &=& 4 &\text{ or }& u_2-20 &=& -4 \\ & u_1 &=& 24 && u_2 &=& 16 \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline u &=& (x-7)^2 \\ \pm\sqrt{u} &=& x-7 \\ 7 \pm\sqrt{u} &=& x \\ \mathbf{x} & \mathbf{=} & \mathbf{7 \pm\sqrt{u}} \\ \hline \end{array}\)

 

Solutions:

\(\begin{array}{|rclcl|} \hline x_1 &=& 7+\sqrt{u_1} = 7 + \sqrt{24} &=& 7+2\sqrt{6} \\ x_2 &=& 7-\sqrt{u_1} = 7 - \sqrt{24} &=& 7-2\sqrt{6} \\ x_3 &=& 7+\sqrt{u_2} = 7 + \sqrt{16} &=& 11 \\ x_4 &=& 7-\sqrt{u_2} = 7 - \sqrt{16} &=& 3 \\ \hline \end{array}\)

 

Proof:

\(\begin{array}{|rcll|} \hline |(7+2\sqrt{6})^2 - 14\cdot (7+2\sqrt{6}) + 29| &=& 4 \\ |11.8989794856^2 - 14\cdot 11.8989794856 + 29| &=& 4 \\ |-25 + 29| &=& 4 \\ |4| &=& 4 \\ 4 &=& 4 \ \checkmark \\\\ \hline |(7-2\sqrt{6})^2 - 14\cdot (7-2\sqrt{6}) + 29| &=& 4 \\ |2.10102051443^2 - 14\cdot 2.10102051443 + 29| &=& 4 \\ |-25 + 29| &=& 4 \\ |4| &=& 4 \\ 4 &=& 4 \ \checkmark \\\\ \hline |11^2 - 14\cdot 11 + 29| &=& 4 \\ |121 - 154 + 29| &=& 4 \\ |-33 + 29| &=& 4 \\ |-4| &=& 4 \\ 4 &=& 4 \ \checkmark \\\\ \hline |3^2 - 14\cdot 3 + 29| &=& 4 \\ |9 - 42 + 29| &=& 4 \\ |-33 + 29| &=& 4 \\ |-4| &=& 4 \\ 4 &=& 4 \ \checkmark \\ \hline \end{array}\)

 

laugh

heureka  Aug 15, 2018

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