THE SUM OF A NUMBER AND ITS RECIPROCAL IS 53/14. FIND THE NUMBER show your solution
Answers
Step-by-step explanation:
a + 1/a = 53/14
=> a^2 + 1 / a = 53 / 14
=> 14a^2 + 14 = 53a
=> 14a^2 - 53a = -14
=> a ( 14a - 53 ) = -14
a = -14 or
14a - 53 = -14
=> 14a = -14 + 53
=> 14a = 39
a = 39/14.
Since reciprocal is needed, we take a as -14.
Let that number be x. The reciprocal of the number is 1/x.
Therefore, x + 1/x = 53/14.
If you recall, if you have something like 2 + 1/2, for example, that is the same thing as 2 1/2, which is the same thing as 5/2.
In order to go from 2 1/2 to 5/2, all I did was multipy the denominator of the 1/2 fraction (which is 2), by the whole number (which, in this case, is also 2) and add the digit in the numerator of the fraction (which is 1) to give [(2 x 2 )+ 1)] = 5
In your problem, x + 1/x, similarly, would give you [(x * x) + 1] = (x^2 + 1)/x = 53/14
From here you cross multiply to get 14 (x^2 + 1) = 53 x
If you rearrange and bring everything to the right hand side, you get 14(x^2) - 53x + 14 = 0.
14x^2 - 4x - 49x + 14 = 0
2x(7x - 2) - 7(7x - 2) = 0
(2x - 7)(7x - 2) = 0
x = 7/2, 2/7
Hope this helps