+0

0
293
3

Solve for x:   $$\dfrac{\sqrt{3x}-4\sqrt{3}}{\sqrt{x}-\sqrt{2}}=\dfrac{2\sqrt{2x}+\sqrt{2}}{\sqrt{6x}-2\sqrt{3}}$$

Mar 8, 2019

#1
+2

Solve for x:
3 (sqrt(x) - 4) (sqrt(2) sqrt(x) - 2) = -sqrt(2) (sqrt(2) - sqrt(x)) (2 sqrt(x) + 1)

Add sqrt(2) (2 sqrt(x) + 1) (sqrt(2) - sqrt(x)) to both sides:

sqrt(2) (sqrt(2) - sqrt(x)) (2 sqrt(x) + 1) + 3 (sqrt(x) - 4) (sqrt(2) sqrt(x) - 2) = 0

sqrt(2) (sqrt(2) - sqrt(x)) (2 sqrt(x) + 1) + 3 (sqrt(x) - 4) (sqrt(2) sqrt(x) - 2) = 26 + (-13 sqrt(2) - 2) sqrt(x) + sqrt(2) x:
26 + (-13 sqrt(2) - 2) sqrt(x) + sqrt(2) x = 0

Simplify and substitute y = sqrt(x).
26 + (-13 sqrt(2) - 2) sqrt(x) + sqrt(2) x = 26 + (-13 sqrt(2) - 2) sqrt(x) + sqrt(2) (sqrt(x))^2
= sqrt(2) y^2 + (-2 - 13 sqrt(2)) y + 26:
sqrt(2) y^2 + (-2 - 13 sqrt(2)) y + 26 = 0

The left hand side factors into a product with two terms:
(y - 13) (sqrt(2) y - 2) = 0

Split into two equations:
y - 13 = 0 or sqrt(2) y - 2 = 0

y = 13 or sqrt(2) y - 2 = 0

Substitute back for y = sqrt(x):
sqrt(x) = 13 or sqrt(2) y - 2 = 0

Raise both sides to the power of two:
x = 169 or sqrt(2) y - 2 = 0

x = 169 or sqrt(2) y = 2

Divide both sides by sqrt(2):
x = 169 or y = sqrt(2)

Substitute back for y = sqrt(x):
x = 169 or sqrt(x) = sqrt(2)

Raise both sides to the power of two:

x = 169               x = 2 is extraneous

Mar 8, 2019
#2
+2

I'll get you started :)

$$\dfrac{\sqrt{3x}-4\sqrt{3}}{\sqrt{x}-\sqrt{2}} =\dfrac{2\sqrt{2x}+\sqrt{2}}{\sqrt{6x}-2\sqrt{3}}\\ \dfrac{\sqrt{x}+\sqrt{2}}{\sqrt{x}+\sqrt{2}}\cdot\dfrac{\sqrt{3x}-4\sqrt{3}}{\sqrt{x}-\sqrt{2}} =\dfrac{2\sqrt{2x}+\sqrt{2}}{\sqrt{6x}-2\sqrt{3}}\cdot \dfrac{\sqrt{6x}+2\sqrt{3}}{\sqrt{6x}+2\sqrt{3}}$$

When you expand the each side out you will have no irational terms on the bottom so it will then be much easier.

If you get stuck you can let us know (with an explanation)

Mar 8, 2019
edited by Melody  Mar 8, 2019
#3
+2

√ [3x] - 4√3               2√[2x ]  + √2

__________   =       ____________

√x  - √2                      √[6x] - 2√3

√ [3x] - √48              √[8x ]  + √2

__________   =       ____________     cross multiply

√x  - √2                      √[6x] - √12

( √ [3x] - √48 ) (  √[6x] - √12) =  (√x  - √2) ( √[8x ]  + √2)     simplify

3x√2  - 12√[2x] - 6√x + 24   =  2x√2 -  4√x + √[2x] - 2

x√2 - 13√[2x]  - 2√x + 26  = 0

√2 √x √x -  13√2 √x  - 2√x + 26 = 0      factor as

√2√x [ √x  - 13 ] - 2 [ √x - 13]  = 0

[ √x - 13]  [ √[2x] - 2 ]  = 0

So either

√x - 13  = 0                                   or         √[2x] - 2  = 0

√x = 13   square both sides                       √[2x]  = 2        square both sides

x = 169                                                          2x   = 4

x = 2

Reject the second solution....it makes the original denominators = 0

So

x = 169   Mar 8, 2019