What positive number satisfies the condition that twice the number minus three times the reciprocal of the number is equal to 1?

Solve:

"twice the number minus three times the reciprocal of the number is equal to 1."
                                                     3(1)
Let the number be n.  Then 2n - = 1
                                                       n

Mult all 3 terms by n to elim. the fractions:

2n^2 - 3 = n.  Rearranging this, we get 2n^2 - n - 3 = 0.

We need to find the roots (zeros or solutions) of this quadratic equation.

Here a=2, b= -1 and c= -3.  Let's find the discriminant b^2-4ac first:

disc. = (-1)^2 - 4(2)(-3) = 1 + 24 = 25.

That's good, because 25 is a perfect square.
                -(-1) plus or minus 5         1 plus or minus 5
Then x = =
                            2(2)                                  4

x could be 6/4 = 3/2, or -5/4.

You must check both answers in the original equation.  If the equation is true for one or the other or for both, then you have found one or more solutions.

Dis the correct  answer