The difference between a number and its reciprocal is 3/2 find the number
The Correct Answer and Explanation is:
Let’s define the number as x.
The reciprocal of x is 1/x. According to the problem, the difference between a number and its reciprocal is 3/2. Therefore, the equation can be written as:x−1x=32x – \frac{1}{x} = \frac{3}{2}x−x1=23
To solve for x, first multiply both sides of the equation by x to eliminate the denominator:x2−1=32xx^2 – 1 = \frac{3}{2} xx2−1=23x
Next, multiply the entire equation by 2 to get rid of the fraction:2×2−2=3x2x^2 – 2 = 3x2x2−2=3x
Now, move all terms to one side to set the equation to zero:2×2−3x−2=02x^2 – 3x – 2 = 02×2−3x−2=0
This is a quadratic equation. We can solve it using the quadratic formula, which is:x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}x=2a−b±b2−4ac
For our equation 2×2−3x−2=02x^2 – 3x – 2 = 02×2−3x−2=0, the coefficients are:
- a=2a = 2a=2
- b=−3b = -3b=−3
- c=−2c = -2c=−2
Substitute these values into the quadratic formula:x=−(−3)±(−3)2−4(2)(−2)2(2)x = \frac{-(-3) \pm \sqrt{(-3)^2 – 4(2)(-2)}}{2(2)}x=2(2)−(−3)±(−3)2−4(2)(−2)
Simplify the terms:x=3±9+164x = \frac{3 \pm \sqrt{9 + 16}}{4}x=43±9+16x=3±254x = \frac{3 \pm \sqrt{25}}{4}x=43±25x=3±54x = \frac{3 \pm 5}{4}x=43±5
Now, solve for the two possible values of x:
- x=3+54=84=2x = \frac{3 + 5}{4} = \frac{8}{4} = 2x=43+5=48=2
- x=3−54=−24=−12x = \frac{3 – 5}{4} = \frac{-2}{4} = -\frac{1}{2}x=43−5=4−2=−21
So, the two possible values for x are 2 and -1/2.
To verify, we can check both values:
- If x = 2, the reciprocal is 1/2, and their difference is 2−1/2=3/22 – 1/2 = 3/22−1/2=3/2, which matches the given condition.
- If x = -1/2, the reciprocal is -2, and their difference is −1/2−(−2)=3/2-1/2 – (-2) = 3/2−1/2−(−2)=3/2, which also satisfies the condition.
Thus, the two possible numbers are 2 and -1/2.
