easy algebra..?

Question

pls show the representation,equation, solution and the final answer.. thank you.

1. when the square of the second of two consecutive integers is added to twice the first integer, the sum is 116. Find the integers.

Answer 1

Let the integers by x and (x + 1)

The word problem can be represented in this algebraic form:

(x + 1)^2 + 2x = 116
x^2 + 4x + 1 = 116
x^2 + 4x - 115 = 0

Using the quadratic formula:

x = [-4 +/- sqrt(16 + 460)]/2
x = [-4 +/- sqrt 476]/2
x = 2(-2 +/- sqrt 119)/2
x = -2 +/- sqrt 119


WOAH! I'm sure there's a mistake in your question (I cross - checked my answer 3 times as I am clearly not getting integers)

Answer 2

If the first integer is x the next consecutive integer would have to be x+1 and the second consecutive would be (x+2)
Now they ask for the square of the second of two consecutive integers, so that would be (x+2)^2
is added to twice the first integer (2x) : (x+2)^2+2x
the sum is 116: (x+2)^2+2x=116
x^2+4x+4+2x=116
x^2+6x+4=116
x^2+6x-112=0
factors of 112: 2,56, 4,28, 8, 14
8 and 14 are six apart, so that looks like it will work out
(x-8)(x+14)=0
x={8, -14)
Your possible solutions could be
8, 9, 10 or -14, -13, -12

Answer 3

Call the numbers X and x+1

(x+1)^2 +2x=116

x^2+4x-115=0

These numbers don't quite add up...

If you meant 118 instead of 116, then the answer will be 9,10 and -13,-12 using the quadratic equation.

Answer 4

Let integers be x and (x + 1)
2x + (x + 1)² = 116
2x + x² + 2x + 1 = 116
x² + 4x - 115 = 0
Doesn`t factorise so must be error in question.

Answer 5

I worked this problme all out and something is not right about it. Did you type something wrong? Every time I check my answers I get 118 not 116.

Answer 6

my bad, the equation is

((n+1)^2)+2n=116

but 116 doesn't seem to fit it, read what Greg answered..

Answer 7

2x+(x+1)^2= 116 i think its this. solve it.

Answer 8

2a+(a+1)^2=116
2a+(a*a+1)=116
4a+1=116
4a=115
a=28.75

OR...

8.908712115