Find a number such that the sum of two times the number and six times its reciprocal is 79/5.
To find the number, let's call it "x," we need to set up an equation based on the given information. We are told that the sum of two times the number (2x) and six times its reciprocal (6/x) is equal to 79/5.
So, we can write the equation as:
2x + 6/x = 79/5
To solve this equation, we need to combine the terms on the left side and get a common denominator on the right side:
(2x * x + 6 * 5) / x = 79/5
Simplifying the equation further:
(2x^2 + 30) / x = 79/5
Cross-multiplying to eliminate the fractions:
5(2x^2 + 30) = 79x
Expanding and rearranging the equation:
10x^2 + 150 = 79x
Bringing all the terms to one side to set up a quadratic equation:
10x^2 - 79x + 150 = 0
Now, we can solve this quadratic equation for x by factoring, completing the square, or using the quadratic formula. Completing the square or using the quadratic formula will give us the exact value of x. Let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
For the equation 10x^2 - 79x + 150 = 0, the values of a, b, and c are:
a = 10, b = -79, and c = 150
Plugging these values into the quadratic formula:
x = (-(-79) ± √((-79)^2 - 4 * 10 * 150)) / (2 * 10)
Simplifying:
x = (79 ± √(6241 - 6000)) / 20
x = (79 ± √241) / 20
Therefore, the solutions for x will be:
x = (79 + √241) / 20 and x = (79 - √241) / 20
These are the two possible values for the number x that satisfy the given equation.
2x + 6/x = 79/5
10x^2 + 30 = 79x
10x^2 - 79x + 30 = 0
(5x-2)(2x-15) = 0
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