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Dec 17, 2017

#1
+7354
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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.

Dec 17, 2017
edited by hectictar  Dec 17, 2017
#2
+7354
+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)      $$x + 1=\sqrt{2x+50}$$

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(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.

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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)      $$x+2=\sqrt{2x+67}$$

.
Dec 17, 2017
#3
+7354
+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.

Dec 17, 2017