Solve for x: \(\dfrac{\sqrt{3x}-4\sqrt{3}}{\sqrt{x}-\sqrt{2}}=\dfrac{2\sqrt{2x}+\sqrt{2}}{\sqrt{6x}-2\sqrt{3}}\)
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
Add 13 to both sides:
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
Add 2 to both sides:
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
+118667 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)
+129845 √ [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
