the sum of the squares of two consecutive intergers is 8 more than 7times the sum of the two intergers.?

Answer 1

Let x and x + 1 denote the consecutive integers.
We know:
x^2 + (x + 1)^2 = 8 + 7(x + x + 1)

Answer 2

Let x = smaller number. Since the numbers are consecutive, we can show that the larger number is equal to (x + 1). So our two numbers are:

x = smaller number
(x + 1) = larger number

x² + (x + 1)² = 7(x + x + 1) + 8
x² + x² + 2x + 1 = 14x + 7 + 8
2x² + 2x + 1 = 14x + 15
2x² - 12x - 14 = 0
2(x² - 6x - 7) = 0 --- Divide both sides by 2...
x² - 6x - 7 = 0
(x - 7)(x + 1) = 0
x = 7 or x = -1

CHECK:
x² + (x + 1)² = 7(x + x + 1) + 8
(7)² + (7 + 1)² = 7(7 + 7 + 1) + 8
49 + 64 = 113
113 = 113 --- x = 7 Checks out...

x² + (x + 1)² = 7(x + x + 1) + 8
(-1)² + (-1 + 1)² = 7((-1) + (-1) + 1) + 8
1 + 0 = (-7) + 8
1 = 1 --- x = -1 also checks out...

ANSWER: There are two possibilitiies: (7 and 8) or (-1 and 0).

Answer 3

Call the first number x. Since they're consecutive the second number must be x + 1. We're told that x^2 + (x+1)^2 = 7(x + x+1) + 8.

Now we can simplify by getting rid of the parentheses: x^2 + x^2 + 2x + 1 = 14x + 7 + 8.

Next, we can combine like terms on the same side of the equation: 2x^2 + 2x + 1 = 14x + 15.

Now, subtract 14x + 15 from each side: 2x^2 +2x -14x +1 -15 = 0.

Combine the terms again: 2x^2 -12x -14 = 0. This is a quadratic equation. Solve it by plugging into x = (-b +/-sqrt(b^2 - 4ac))/2a, where a is the coefficient of x^2, b of x, and c is the constant. In this case a = 2, b = -12 and c = -14.

So x = (12 +/- sqrt(144 - 4*2*-14))/2*2. Solve for the inner parentheses: x = (12 +/- sqrt(256)/2*2.

Now solve the square root: x = (12 +/- 16)/2*2.

Therefore, x = (28 or -4)/4 so x = 7 or -1. I'd say 7 is the better bet. Let's plug it in to check.

7^2 + 8^2 = 49 + 64 = 113. 7+8 = 15. 15*7 = 105. 105 + 8 = 113.

Alternatively, you could look at it like this: 7*7 + 8*8 = 7*7 + 7*8 +8, which is obviously true.

The integers are 7 and 8.

Answer 4

Just write the arithmetic diecttly from the sentence, noting that we are dealing with consecutive integers, x and (x+1).


x^2 + (x+1)^2 = 8 + 7 * ( x + (x+1))

x^2 + x^2 + 2x + 1 = 8 + 7 (2x +1)

2x^2 + 2x + 1 = 8 + 14 x + 7

2x^2 - 12x - 14 =0

x^2 - 6x - 7= 0

(x-7)(x+1)=0

So, x= -1 and 7

Therefore the answers (plural) are -1 and 0, or 7 and 8.

{NOTE: -1 is a perfectly good integer!}

Done.

Check your work:
(-1)^2 + 0 = 1 and 1= 8 + (7 * ((-1)+0))= 8 - 7

and

7^2 + 8^2= 113 and 8 + 7*(8+7)= 8 + 105

So, OK!

Answer 5

7 and 8

Answer 6

x^2 + (x + 1)^2 = 7(x + (x + 1)) + 8
x^2 + (x + 1)(x + 1) = 7(x + x + 1) + 8
x^2 + (x^2 + x + x + 1) = 7(2x + 1) + 8
x^2 + (x^2 + 2x + 1) = 14x + 7 + 8
x^2 + x^2 + 2x + 1 = 14x + 15
2x^2 + 2x + 1 = 14x + 15
2x^2 - 12x - 14 = 0
2(x^2 - 6x - 7) = 0
x^2 - 6x - 7 = 0
(x - 7)(x + 1) = 0
x = 7 or -1

Since you didn't state positive or negative

ANS : 7 and 8 or -1 and 0

Answer 7

let the two integers be
x,x+1
We have the equation
x^2+[x+1]^2=7[x+x+1]+8
2x^2+2x+1=14x+15
2x^2-12x-14=0
x^2-6x-7=0
x^2-7x+x-7=0
x[x-7]+1[x-7]=0
[x+1][x-7]=0
x=-1,7
ignore -1
the reqired numbers are 7,8
verify
7^2+8^2=113
[7+8]*7+8=113
ok