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# Math9

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Solve for x in the given equation $$\dfrac{\sqrt{x}}{\sqrt{3}x+\sqrt{2}} = \dfrac{1}{2\sqrt{6}x+4}$$

tertre  Mar 12, 2017

#3
+82870
+5

sqrt(x) / [ sqrt(3)*x + sqrt(2) ] = 1 / [ 2*sqrt(6)*x + 4]

sqrt(x) / [ sqrt(3)*x + sqrt(2) ] = 1 / [(2(*sqrt(6)*x + 2) ]   multiply both sides by 2

2sqrt (x) / [sqrt(3)*x + sqrt(2)]  = 1 / [ sqrt(6)*x + 2]  cross multiply

(2sqrt (x))  [ sqrt(6)*x + 2]  =  sqrt(3)*x + sqrt(2)

Factor  sqrt (2)  out of sqrt  sqrt(6)*x + 2

[2sqrt (x) ] ( [sqrt (2)] * [sqrt(3) *x + sqrt(2)] )  = [ sqrt (3)*x + sqrt(2) ]

2sqrt(2)* sqrt(x) [ sqrt (3)*x + sqrt(2)] = [sqrt(3)*x + sqrt(2)]

2sqrt(2)*sqrt(x) [ sqrt (3)*x + sqrt(2)] - [sqrt(3)*x + sqrt(2)]  = 0

[ sqrt (3)*x + sqrt(2)] [ 2sqrt(2)* sqrt(x) - 1 ]  = 0

Set both factors to 0  and either

sqrt (3) * x + sqrt (2) = 0  →   x  = -sqrt (2)/ sqrt(3)

But this gives a non-real result to the original problem

Or

2sqrt(2) * sqrt(x)   - 1  = 0

sqrt(8) * sqrt (x) = 1

sqrt (8x) = 1     square both sides

8x   = 1     divide both sides by 8

x = 1/8

CPhill  Mar 12, 2017
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#1
+1702
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Anyone there! Help!!!!!!!!!!!!!!!!!!!!!!!!

tertre  Mar 12, 2017
#2
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Solve for x:
x/(3 x^2 + 2) = 1/(24 x^2 + 16)

Cross multiply:
x (24 x^2 + 16) = 3 x^2 + 2

Expand out terms of the left hand side:
24 x^3 + 16 x = 3 x^2 + 2

Subtract 3 x^2 + 2 from both sides:
24 x^3 - 3 x^2 + 16 x - 2 = 0

The left hand side factors into a product with two terms:
(8 x - 1) (3 x^2 + 2) = 0

Split into two equations:
8 x - 1 = 0 or 3 x^2 + 2 = 0

Add 1 to both sides:
8 x = 1 or 3 x^2 + 2 = 0

Divide both sides by 8:
x = 1/8 or 3 x^2 + 2 = 0

Subtract 2 from both sides:
x = 1/8 or 3 x^2 = -2

Divide both sides by 3:
x = 1/8 or x^2 = -2/3

Take the square root of both sides:
x = 1/8 or x = i sqrt(2/3) or x = -i sqrt(2/3)

x/(3 x^2 + 2) ⇒ 1/(8 (2 + 3 (1/8)^2)) = 8/131
1/(24 x^2 + 16) ⇒ 1/(16 + 24 (1/8)^2) = 8/131:
So this solution is correct

x/(3 x^2 + 2) ⇒ -(i sqrt(2/3))/(2 + 3 (-i sqrt(2/3))^2) = ∞^~
1/(24 x^2 + 16) ⇒ 1/(16 + 24 (-i sqrt(2/3))^2) = ∞^~:
So this solution is incorrect

x/(3 x^2 + 2) ⇒ (i sqrt(2/3))/(2 + 3 (i sqrt(2/3))^2) = ∞^~
1/(24 x^2 + 16) ⇒ 1/(16 + 24 (i sqrt(2/3))^2) = ∞^~:
So this solution is incorrect

The solution is:
Answer: |x = 1/8

Guest Mar 12, 2017
#3
+82870
+5

sqrt(x) / [ sqrt(3)*x + sqrt(2) ] = 1 / [ 2*sqrt(6)*x + 4]

sqrt(x) / [ sqrt(3)*x + sqrt(2) ] = 1 / [(2(*sqrt(6)*x + 2) ]   multiply both sides by 2

2sqrt (x) / [sqrt(3)*x + sqrt(2)]  = 1 / [ sqrt(6)*x + 2]  cross multiply

(2sqrt (x))  [ sqrt(6)*x + 2]  =  sqrt(3)*x + sqrt(2)

Factor  sqrt (2)  out of sqrt  sqrt(6)*x + 2

[2sqrt (x) ] ( [sqrt (2)] * [sqrt(3) *x + sqrt(2)] )  = [ sqrt (3)*x + sqrt(2) ]

2sqrt(2)* sqrt(x) [ sqrt (3)*x + sqrt(2)] = [sqrt(3)*x + sqrt(2)]

2sqrt(2)*sqrt(x) [ sqrt (3)*x + sqrt(2)] - [sqrt(3)*x + sqrt(2)]  = 0

[ sqrt (3)*x + sqrt(2)] [ 2sqrt(2)* sqrt(x) - 1 ]  = 0

Set both factors to 0  and either

sqrt (3) * x + sqrt (2) = 0  →   x  = -sqrt (2)/ sqrt(3)

But this gives a non-real result to the original problem

Or

2sqrt(2) * sqrt(x)   - 1  = 0

sqrt(8) * sqrt (x) = 1

sqrt (8x) = 1     square both sides

8x   = 1     divide both sides by 8

x = 1/8

CPhill  Mar 12, 2017
#4
+6351
+1

Hey CPhill, I was trying to follow the Guest's work and it looks like the first step was square every term. I was just wondering is that really allowed??

hectictar  Mar 13, 2017
#5
+82870
0

To be honest, hectictar, I wondered about that, too.....maybe the Guest knows something that we don't......!!!!

CPhill  Mar 13, 2017
#6
+82870
+5

After looking at this problem again....I see  that it's really very easy to solve

sqrt(x) / [ sqrt(3)*x + sqrt(2) ] = 1 / [ 2*sqrt(6)*x + 4]

sqrt(x) / [ sqrt(3)*x + sqrt(2) ] = 1 / [ (2)[sqrt(6)*x + 2 ] ]   multiply both sides by 2

2sqrt(x) / [ sqrt(3)*x + sqrt(2) ] = 1 / [ sqrt(6)*x + 2]

Factor  sqrt (2)  out of  sqrt(6)*x + 2    =   sqrt(2) [ sqrt(3)*x + sqrt(2)]

2sqrt(x) / [ sqrt(3)*x + sqrt(2) ] = 1 /  (sqrt(2) [ sqrt(3)*x + sqrt(2)] )

Multiply both sides by sqrt(2)

sqrt(2)*2sqrt(x) / [ sqrt(3)*x + sqrt(2) ] = 1 /  [ sqrt(3)*x + sqrt(2)]

Since the denominators  are the same, we can solve for the numerators

sqrt (2)* 2sqrt(x) = 1

2sqrt(2)sqrt(x)  = 1

sqrt(8)sqrt(x)  = 1

sqrt (8x)  = 1    square both sides

8x  = 1  →   x  = 1/8

CPhill  Mar 13, 2017

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