Solve (3x-1)^1/2 - (2x-1)^1/2 =1?

Answer 1

√(3x-1) - √(2x-1) =1

So √(3x-1) = 1 + √(2x-1)

Square both sides

3x - 1 = 1 + 2√(2x-1) + 2x - 1

ie x - 1 = 2√(2x-1)

Square both sides again

(x - 1)² = 4(2x - 1)
So x² - 2x + 1 = 8x - 4

ie x² -10x + 5 = 0

So x = (10 ± √[(-10)² - 4*1 *5])/2
= 5 ± √(20)
= 5 ± 2√5

Check 5 + 2√5
√(3x-1) - √(2x-1) = 1

5 - 2√5
√(3x-1) - √(2x-1) ≠1
In fact in this case √(3x-1) + √(2x-1) = 1

So discarding the extraneous solution

x = 5 + 2√5

Answer 2

It's 1.

Answer 3

3x - 5 = 11 + 2(x-6) Foil first: 2(x-6) = 2x - 12 Insert this into the equation: 3x - 5 = 11 + 2x - 12 Get all x on one area via including the different of 2x: 3x - 2x - 5 = 11 - 12 upload like aspects: 3x - 2x = 1x. 11-12 = -a million so x - 5 = -a million stability the equation via including the different to the two aspects: x = 4

Answer 4

So (3x-1)^1/2 - (2x-1)^1/2 =1, now multiply both sides by (3x-1)^1/2 + (2x-1)^1/2; then (3x-1)^1/2 + (2x-1)^1/2 = (3x-1) - (2x-1) = x; now add LHSides and RHsides and square both sides; thence equation: x^2-10x+5=0, hence x1=5+2sqrt(5), x2=5-2sqrt(5); yes, Kellenraid is right BUT LOOK x2 is not our solution; the answer is x= 5+2sqrt(5);

Answer 5

(3x - 1)^(1/2) - (2x - 1)^(1/2) = 1
(3x - 1)^(1/2) = (2x - 1)^(1/2) + 1
square both sides
3x - 1 = ((2x - 1)^(1/2) + 1)((2x - 1)^(1/2) + 1)
3x - 1 = (2x - 1) + 2(2x - 1)^(1/2) + 1
3x - 1 = 2x - 1 + 2(2x - 1)^(1/2) + 1
3x - 1 = 2x + 2(2x - 1)^(1/2)
x - 1 = 2(2x - 1)^(1/2)
x - 1 = (4(2x - 1))^(1/2)
square both sides
(x - 1)^2 = 4(2x - 1)
(x - 1)(x - 1) = 8x - 4
x^2 - 2x + 1 = 8x - 4
x^2 - 10x + 5 = 0

x = (-b ± sqrt(b^2 - 4ac))/(2a)

x = (-(-10) ± sqrt((-10)^2 - 4(1)(5)))/(2(1))
x = (10 ± sqrt(100 - 20))/2
x = (10 ± sqrt(80))/2
x = (10 ± sqrt(16 * 5))/2
x = (10 ± 4sqrt(5))/2
x = 5 ± 2sqrt(5)

If you were to plug in 5 - 2sqrt(5), it will give you a complex value.

ANS : x = 5 + 2sqrt(5)

Answer 6

(3x - 1)^1/2 - (2x - 1)^1/2 = 1
(3x - 1)^1/2 = 1 + (2x - 1)^1/2
Squaring,
3x - 1 = 1 + 2x - 1 + 2(2x - 1)^1/2
x - 1 = 2(2x - 1)^1/2
Squaring again,
x^2 -2x + 1 = 8x - 4
x^2 -10x + 5 = 0
x = (10 ± √(100 - 20))/2
x = 5 ± √(25 - 5)
x = 5 ± 2√5

Answer 7

Rewrite it
sqr(3x-1) = 1 + sqr(2x-1) then squaring both sides
3x-1 = 1 + 2sqr(2x-1) + (2x-1) then collecting terms
x-1 = 2sqr(2x-1) then squaring
x^2-2x+1 = 4(2x-1) = 8x-4
x^2 - 10x + 5 = 0
And since I cannot find a simple factoring, I use the quadratic formula to get
x = 5 +- 2sqr(5)

Answer 8

Your first step is to set them to square root:
√3x-1-√2x-1=1

Take the √2x-1 and add it from both sides:
√3x-1=1+√2x-1

Your next step is the square both sides:
(√3x-1)^2=(1+√2x-1)^2
3x-1=(1+√2x-1)(1+√2x-1)
3x-1=1+2√2x-1+2x-1
3x-1=2√2x-1+2x

Subtract 2x from both sides:
x-1=2√2x-1

You need to square both sides again:
(x-1)^2=(2√2x-1)^2
x^2-2x+1=4(2x-1)
x^2-2x+1=8x-4

Subtract 8x and add 4 from both sides:
x^2-10x+5=0

Now, use the quadratic formula:
x=-b±√b^2-4ac/2a
a=1,b=-10,c=5
x=-(-10)±√(-10)^2-4(1*5)/2(1)
x=10±√100-4(5)/2
x=10±√100-20/2
x=10±√80/2
x=10±√16*5/2
x=10±4√5/2
x=5±2√5
The answer is 5+2√5

Answer 9

√(3x-1)² =1+ √(2x-1)²
3x - 1 = 1+ 2x - 1
3x - 1 = 2x
3x - 2x = 1
x = 1
The answer is 1.
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Answer 10

ray2_moot's answer will work. Normally, since both the universal gravity constant and the mass of the Earth are constant, the two are multiliplied together to obtain a new constant called the geocentric gravitational constant. It's 3.986004418 x 10^14 m^3/sec^2.

(abhishek j gave your the formula for the FORCE of gravity as you move further away from the Earth. However, if you know that F=ma, then you can obtain the acceleration by dividing out the mass