Welcome :: Homework Help and Answers :: Mathskey.com
Welcome to Mathskey.com Question & Answers Community. Ask any math/science homework question and receive answers from other members of the community.

13,459 questions

17,854 answers

1,446 comments

805,839 users

pls help me :(

0 votes
4) Find the values of x in the number x, x+1 , x+2 such that the reciprocal of the smallest number equals the sum of the reciprocal of the other two numbers. Assume that x is possitive.
asked Dec 9, 2014 in PRECALCULUS by anonymous

1 Answer

0 votes

The values of x are x, x+1, x+2

Reciprocal of the smallest number is 1/x

Sum of reciprocal of the other two number is (1/x+1) + (1/x+2)

1/x = (1/x+1) + (1/x+2)

Multiply each side by (x+1)(x+2)

[(x+1)(x+2)] / x = x+2 + x + 1

Multiply each side by x

(x+1)(x+2) = x(2x+3)

x²+x+2x+2 = 2x²+3

x²+3x+2 = 2x²+3

2x²+3- (x²+3x+2) = 0

2x² + 3 - x² - 3x - 2 = 0

x² - 3x+1 = 0

 

Now the expression is the form of Quadratic Equation then values of x are

x = [-(-3) ± √((-3)²-4*1*(1))] / (2*1)

x = [3 ± √(9-4)] / (2)

x = [3 ± √5] / 2

x = [3 ± 2.2360] / 2

x = [3 + 2.2360] / 2 and [3 - 2.2360] / 2

x = 2.6180 and 0.382

Therefore the values of x are 2.6180 and 0.382.

answered Dec 9, 2014 by Lucy Mentor

Related questions

asked Dec 9, 2014 in PRECALCULUS by anonymous
asked Dec 9, 2014 in PRECALCULUS by anonymous
asked Dec 9, 2014 in PRECALCULUS by anonymous
asked Dec 9, 2014 in PRECALCULUS by anonymous
asked Oct 22, 2014 in CALCULUS by anonymous
asked Jan 19, 2015 in PRECALCULUS by anonymous
asked Dec 19, 2014 in PRECALCULUS by anonymous
asked Dec 9, 2014 in PRECALCULUS by anonymous
asked Dec 9, 2014 in PRECALCULUS by anonymous
asked Nov 28, 2014 in PRECALCULUS by anonymous
asked Nov 21, 2014 in PRECALCULUS by anonymous
asked Jan 19, 2015 in PRECALCULUS by anonymous
asked Nov 21, 2014 in PRECALCULUS by anonymous
asked Oct 22, 2014 in CALCULUS by anonymous
asked Oct 21, 2014 in CALCULUS by anonymous
asked Oct 21, 2014 in CALCULUS by anonymous
asked Sep 26, 2014 in CALCULUS by anonymous
asked Sep 18, 2014 in CALCULUS by anonymous
asked Oct 22, 2014 in CALCULUS by anonymous
...