Find three consecutive positive integers such that the square of their sum exceeds the sum of their squares by 214.
A3, 4, 5
B4, 5, 6
C5, 6, 7
D6, 7, 8
Answer & Solution
Correct answer: C. 5, 6, 7
Let the integers be $x-1$, $x$, and $x+1$. Their sum is $3x$, so the condition gives $(3x)^2=(x-1)^2+x^2+(x+1)^2+214$. Expanding, $9x^2=3x^2+216$, hence $6x^2=216$ and $x^2=36$. Since the integers are positive, $x=6$, so the numbers are $5,6,7$.
Related questions
The study of inequalities is described as useful in solving problems in science and:A firm needing at least 6 units and at most 10 would write that as a:Which sign makes a statement an inequality rather than an equation?Unless stated otherwise, inequalities in that chapter are solved over the set of:The set of all values of the variable that make an inequality true is called its:Solving 4x plus 3 less than 6x plus 7 gives:When x is a real number in that inequality, the solution set is written as:When x is an integer in that inequality, the largest solution is: