SOLUTION: find 4 rational numbers such that the product of the 1st 2nd and 4th is 54. The 2nd number is 2 less than the 1st number, the 3rd number is 5 less than the 2nd number and the 4th n

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: find 4 rational numbers such that the product of the 1st 2nd and 4th is 54. The 2nd number is 2 less than the 1st number, the 3rd number is 5 less than the 2nd number and the 4th n      Log On


   



Question 1174021: find 4 rational numbers such that the product of the 1st 2nd and 4th is 54. The 2nd number is 2 less than the 1st number, the 3rd number is 5 less than the 2nd number and the 4th number is 3 less than the 3rd number
Answer by ikleyn(52887) About Me  (Show Source):
You can put this solution on YOUR website!
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find 4 rational numbers such that the product of the 1st 2nd and 4th is 54. The 2nd number is 2 less than the 1st number,
the 3rd number is 5 less than the 2nd number and the 4th number is 3 less than the 3rd number
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           This problem  HAS  NO  solutions in rational numbers.


The equation is  

    x*(x-2)*(x-7)*(x-10) = 154,   

or

    x^4 - 19x^3 + 104x^2 - 140x - 154 = 0.


       +------------------------------------------------------+
       |     The  DIAGNOSIS  (obtained with special software  |
       |         using specialized web-site)  is THIS         |
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Result:


The roots of the polynomial x 

    x^4 - 19x^3 + 104x^2 - 140x - 154

are


x1 = −0.31014

x2 = 2.63744

x3 = 6.47111

x4 = 10.20159


Explanation:

This polynomial has no rational roots that can be found using Rational Root Test.

Roots were found using quartic formulas.


So, the problem is a FAKE.

You, the visitor, simply stole my time.