Posted by purpletree on Sunday, March 29, 2009 at 12:54pm.
Find two consecutive integers, the sum of whose squares is 1 more than twelve times the larger.
(Please provide the steps/explanation for solving this problem)

Quadratic Equations  Reiny, Sunday, March 29, 2009 at 3:58pm
two consecutive integers: x and x+1
<the sum of whose squares is 1 more than twelve times the larger..>
x^2 + (x+1)^2 = 12(x+1) + 1
expand and solve as a quadratic equation.
(see if you can come up with 6,7 or 1,0 )

Quadratic Equations  purpletree, Monday, March 30, 2009 at 9:22am
Thanks for the help. :)
Answer This Question
Related Questions
 Maths  The sum of the squares of three consecutive positive integers is 365 ...
 math  1.twice a number is 6 2.four added to a number gives ten 3.twenty five ...
 math  Given two numbers whose sum is 56. Six times the larger is 72 more than 5...
 Math please help  Three consecutive odd intergers are such that the sum of the ...
 math please help  Three consecutive odd intergers are such that the sum of the ...
 math  The difference of the cubes of two consecutive odd positive integers is ...
 Algebra  The difference of the cubes of two consecutive odd positive integers ...
 Algebra  The difference of the cubes of two consecutive odd positive integers ...
 algebra!  The difference of the cubes of two consecutive odd positive integers ...
 Algebra  The difference of the cubes of two consecutive odd positive integers ...
More Related Questions