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.
