answer this algebra question.. thank you :]?

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 second number be x then first number is (x – 1)
Square of 2nd No = x^2
twicw of first No. = 2 × (x – 1) = 2 x – 2
Sum of these No. = x^2 + 2x – 2 = 116
or x^2 + 2 x + 1 – 3 = 116
or x^2 + 2 x + 1 = 119
(x + 1)^2 = 119
(x + 1) = rt of 119 ---------- This is not integer
rt 119 = 10.908
x = 9.908
Numbers are 8.908 and 9.908

Answer 2

Ok 1st integer = A,
Second integer = A+1

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

2A + A^2 + 2A + 1 = 116 <-----Expanded the binomial

A^2 + 4A = 115

A^2 + 4A -115 = 0

Hmmmm... This isn't going to come out cleanly...

Are you sure you wrote this right?

Here's some examples proving this won't work:

If A = 9 and B = 10: 18 + 100 = 118
If A = 8 and B = 9, 16 + 81 = 97
If A = 10 and B = 11, 20 + 121 = 141
If A = -13 and B = -12, -26 + 144 = 118
If A = -12 and B = -11, -24 + 121 = 97
If A = -14 and B = -13, -28 + 169 = 141

That's all the consecutive integers that get close to 116, and none of them work. Sorry but I'm either dumb, or this problem is. Most likely the latter.

Answer 3

Let x = 1st integer, x + 1 = 2nd integer.

Equations:

Find the 1st integer:
2x + (x + 1)^2 = 116
2x + x^2 + x + x + 1 = 116
x^2 + 4x - 115 = 0
x = 8.9087121146357

Find the 2nd integer:
= 8.9087121146357 + 1
= 9.9087121146357

Answer: 1st integer = 8.9087121146357, 2nd integer = 9.9087121146357

Proof( sum is 116):
= 2(8.9087121146357) + 9.9087121146357^2
= 17.8174242292716 + 98.1825757707284
= 116

I think that if you copied the problem correctly the sum should have been 118. The answer could have been two fold:
= 2(9) + 10^2
= 18 + 100
= 118

OR 1st integer = -13 and 2nd integer = - 12

= 2(- 13) + (- 12)^2
= - 26 + 144
= 118

That is why I had no choice but to do the problem almost by trial and error method. With sum of 116 it will never result in two consecutive integers. I think you miswrote the problem.

Answer 4

Let's assume one integer is x and the second is x+1 since they're consecutive. Your equation would then look like this:

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

If you factor it out, you get:

x^2+2x+1+2x=116

Combine like terms and subtract 116 on both sides to make the right side equal 0:

x^2+4x-115=0

Since this equation doesn't factor, we'll try the Quadratic Formula. That gives us:

x is approximately 8.908

AND

x is approximately -12.908.

Now, for the x+1 part:

8.908 + 1 = 9.908

AND

-12.908 + 1 = -11.908

Thus, you have two solutions: x = 8.908 and x+1 = 9.908

as well as: x = -12.908 and x+1 = -11.908

But these answers are not integers. Perhaps you should go back and check the original problem - you might have copied it incorrectly.

Good luck in solving it! :)

Answer 5

Let:
x = first integer
x +1 = second integer
2x + (x+1)^2 = 116
2x + x^2 + 2x + 1 = 116
x^2 + 4x = 115
By completing square
x^2 + 4x + 4 = 115 + 4
(x + 2)^2 = 119
x + 2 = (119)^(1/2) which will not yield an integer.

If x = first integer
x - 1 = second integer
2x + (x - 1)^2 = 116
2x + x^2 - 2x + 1 = 116
x^2 = 115
x = (115)^(1/2), again not an integer

No valid integer for this problem.

Answer 6

Let the smaller integer be I. the bigger one is I+1.
The square of the bigger one is (I+1)^2.
Twice the first integer is 2I.
Their sum (I+1)^2 +2I = 116.
or I^2+4I-115=0
That doesn't give you integers. Are you sure the sum wasn't 118?

Answer 7

Let us take the integers as I integer= x-1
ii integer = x
iii = x=1

Square of second = x2
is added TWICE the i integer = 2^ x -1 = 2x-2

Now,
square of ii int. + twice the i integer = 116
2x + 2x-2 = 116
4x -2 = 116
4x = 116 + 2
4x= 118
x= 118/ 4
x=29.5

i = 29.5 -1 = 28.5
ii = 29.5
iii= 29.5+1= 30.5

Answer 8

Let x=1st Integer
x+1= 2nd integer
(x+1)(x+1)+(2x)=116
x2+2x+1+2x=116
x2+4x-115=0
then factor it