How do I solve for x in the following equation...?

Question

I am stumped as to how to solve for x in:

3/2 = x - (49/2x)

Answer 1

First get rid of the fractions by multiplying by the greatest common denominator:
3/2 = x - (49/2x)
2x*3/2 = 2x*x - 2x*(49/2x)
3x = 2x^2 - 49
0 = 2x^2 - 3x - 49

Then use the quadratic formula from there.

I hope this helps!

Answer 2

3/2 = x - (49/2x)

Adding 49/2x on both sides
3/2 + 49/2x = x

Taking LCM on the LHS
(3x + 49)/2x = x

Multiplying both sides by 2x
(3x + 49) = 2x^2

Subtracting 3x + 49 from both sides
0 = 2x^2 - 3x - 49

or

2x^2 - 3x - 49 = 0

Now solve for that equation:

a = 2, b = -3, c = -49

x = { -b ± Sq Root ( b^2 - 4ac ) } ÷ 2a
= { 3 ± Sq Root ( 9 + 392 ) } ÷ 4
= { 3 ± Sq Root (401) } ÷ 4
Therefore, x = {3 + sqrt(401)}/4 or {3 - sqrt(401)}/4
In decimal form, x evaluates to 5.7562 and -4.2562

Hope this helps!

Answer 3

Multiply the whole thing by 2x. You'll get 6x/2=2x(squared) -49. Then subtract the 3x (6/2=3) from both sides and you'll have 2x(squared) - 3x -49. Use the quadratic formula from there to find x.

Answer 4

First, you have to get rid of your brackets. so make the equation: 3/2= x-49/2x

then multiply everything in the equation by 2x so you don't have anymore x's in the denomenator.
you will have
3/2(2x)=x(2x)-49/2x (2x)
then
3x= 2x^2 - 49
move everything to one side to make the equation equal zero;
2x^2+3x-49
solve by factoring
hint:you need two numbers that multiply to give you 98 and add to give you 3

Answer 5

3/2 = x - (49/2x)
3 = 2x - (49/x)
3x = 2x^2 -49
2x^2 - 3x -49 =0

x=5.7562 or x=-4.2562