Twice the sum of a number and its reciprocal is same as 33/16? please help po thanks :>
To solve this equation, let's break it down step by step.
Let's assume the number in question is represented by the variable "x".
The sum of a number and its reciprocal can be calculated as x + 1/x.
According to the problem, twice the sum of a number and its reciprocal is equal to 33/16:
2(x + 1/x) = 33/16
To simplify the equation, we need to eliminate the fraction. We can do this by multiplying both sides of the equation by 16:
16 * 2(x + 1/x) = 16 * (33/16)
Now, simplify:
32(x + 1/x) = 33
Expand the equation:
32x + 32(1/x) = 33
Multiply both sides by x to eliminate the fractions:
32x^2 + 32 = 33x
Rearrange the equation to set it to zero:
32x^2 - 33x + 32 = 0
This equation is a quadratic equation. We can solve it using factoring, completing the square, or the quadratic formula. In this case, factoring may not be the easiest method.
Applying the quadratic formula:
x = (-b ± √(b^2 - 4ac))/(2a)
For our equation, a = 32, b = -33, and c = 32.
Plug these values into the quadratic formula:
x = (33 ± √((-33)^2 - 4 * 32 * 32))/(2 * 32)
Simplify further:
x = (33 ± √(1089 - 4096))/(64)
x = (33 ± √(-3007))/(64)
Since the discriminant (√(b^2 - 4ac)) is negative, the equation does not have real solutions. However, if you are looking for complex solutions, you can further simplify the expression:
x = (33 ± i√3007)/(64)
So, the solutions to the equation are:
x = (33 + i√3007)/(64) and x = (33 - i√3007)/(64)
Please note that in the context of the problem, if you are looking for a real number solution, there is no such solution.
2 (x+1/x)
HATDOOOOOOG
sum -- ___ + ___
twice the sum -- 2(___ + ___)
now recall that 1/x is the reciprocal of x
so now just fill in the blanks and finish the equation