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Guest Dec 17, 2017
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 #1
avatar+5896 
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1:

 

x + 2  =  \(\sqrt{3x+10}\)

                                           Square both sides of the equation.

(x + 2)2  =  3x + 10

 

(x + 2)(x + 2)  =  3x + 10

 

x2 + 4x + 4  =  3x + 10

                                           Subtract  3x  from both sides and subtract  10  from both sides.

x2 + x - 6  =  0

                                           Factor the left side.

(x + 3)(x - 2)  =  0

                                           Set each factor equal to zero and solve for  x .

x = -3    or    x = 2

 

Plug in  -3  into the original equation to see if it makes it true.

 

-3 + 2  =  \(\sqrt{3(-3)+10}\)      ?

-1  =  \(\sqrt{1}\)      ?

-1  =  1       False.

 

Plug in  2  into the original equation to see if it makes it true.

 

2 + 2  =  \(\sqrt{3(2)+10}\)      ?

4  =  \(\sqrt{16}\)      ?

4  =  4         True.

 

So....

 

(a)     2  is the solution.

(b)    -3  is the extraneous solution, because it does not make the given equation true.

(c)     And this can probably explain what an extraneous solution is better than I can.

hectictar  Dec 17, 2017
edited by hectictar  Dec 17, 2017
 #2
avatar+5896 
+2

2:      x + a  =  \(\sqrt{bx + c}\)

 

For  a , I pick  1 .  For  b , I pick  2 .  And they tell us that  x = 7 .

 

7 + 1  =  \(\sqrt{2(7) + c}\)        Now we just need to solve this for  c .

64  =  14 + c

c  =  50              So...

 

(a)      \(x + 1=\sqrt{2x+50}\)

 

----------

 

(b)      Square both sides to get...

 

(x + 1)2  =  2x + 50

x2 + 2x + 1  =  2x + 50      Subtract  2x  from both sides and subtract  1  from both sides.

x2  =  49                            Take the  ± square root of both sides.

x = 7    or    x = -7

 

Test each potential solution.'

 

7 + 1  =  \(\sqrt{2(7)+50}\)     ?

8  =  \(\sqrt{64}\)      True.

 

-7 + 1  =  \(\sqrt{2(-7)+50}\)

-6  =  \(\sqrt{36}\)     False.

 

So  7  is the solution.  -7  is an extraneous solution.

 

----------

 

If  a = 2 ,  b = 2,  and  x = 7 , the equation is...

 

\(7+2=\sqrt{2(7)+c}\)       And  c  must equal...

 

c  =  67    So......

 

(c)      \(x+2=\sqrt{2x+67}\)

hectictar  Dec 17, 2017
 #3
avatar+5896 
+2

3:

 

(a)     x - 1   =   \(\sqrt{2x+22}\)          and          The extraneous solution is  -3  .

 

 

(b)      bx + c  must be positive or zero because the square root of a negative number is imaginary.

hectictar  Dec 17, 2017

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