if a number is subtracted from twice its reciprocal the result is 7/3. find the numbers.
help me..
let the number be x
the reciprocal of the number is 1/x
twice the reciprocal is 2/x
2/x - x = 7/3
multiply each term by 3x
6 - 3x^2 = 7x
3x^2 + 7x - 6 = 0
(3x - 2)(x + 3) 0
x = 2/3 or x = -3
TY
To find the numbers, let's start by expressing the problem as an algebraic equation.
Let's assume the number is "x". The reciprocal of "x" is "1/x".
According to the problem, when we subtract a number from twice its reciprocal, the result is 7/3. Mathematically, this can be written as:
2 * (1/x) - x = 7/3
To solve this equation, let's first get rid of the fraction by multiplying everything by 3:
3 * [2 * (1/x) - x] = 7
Next, simplify the equation:
6 * (1/x) - 3x = 7
To eliminate the fraction, let's multiply both sides of the equation by "x":
6 - 3x^2 = 7x
Rearranging the terms:
3x^2 + 7x - 6 = 0
Now, we can solve this quadratic equation using factoring, completing the square, or the quadratic formula. Let's use factoring:
(3x - 2)(x + 3) = 0
Now, set each factor equal to zero and solve for "x":
3x - 2 = 0 or x + 3 = 0
When 3x - 2 = 0, we get:
3x = 2
x = 2/3
When x + 3 = 0, we get:
x = -3
So, the two numbers that satisfy the given condition are 2/3 and -3.