Does anyone know how to do quadratic equations for word problems?

Question

Find two consecutive positive integers such that the sum of their squares is 85.

Answer 1

if the two integers are a and a+1 respectively then they will be consecutive integers

the sum of their squares are 85

so a^2 + (a+1)^2 = 85

we need to multiply out the square brackets and combine and factorize

a^2 + a^2 + 2a + 1 = 85

to bring everything on one side

2a^2 + 2a - 84 = 0

divide throughout by 2

a^2 + a - 42 = 0

factorize to give

(a+7)(a - 6) = 0

so the answers are

a= -7 or a = 6

and the two possible sets of answers are

-7, -6 and 6 , 7

the first set we can disregard as positive integers only are required

substitute back into original problem and double check if it fits (which it does QED)

Answer 2

x^2 + (x+1)^2 = 85
x^2 + x^2 + 2x + 1 = 85
2x^2 + 2x + 1 - 85 = 0 /-85
2x^2 + 2x - 84 = 0 / *0,5
x^2 + x - 42 = 0
(x + 7) (x - 6) = 0
x + 7 = 0
x = -7 (not positive)
x - 6 = 0
x = 6

Then the two consecutive values are

x = 6 and x+1 = 7

Answer 3

Let x = the first
then x + 1 = the second

x² +(x+1)² = 85

2x² + 2x - 84 = 0

x² + x - 42 = 0

(x + 7)(x - 6) = 0

x = 6, x + 1 = 7 (x = -6 is a trivial result and we will discard it)
.

Answer 4

let, x, x+1 be the consecutive +ve integers.

(x)^2 +(x+1)^2 = 85
x^2 + x^2 + 2x +1 = 85
2x^2 + 2x + 1 = 85
2x^2 +2x - 84 = 0
2x^2 + 14x - 12x - 82 = 0
2x(x + 7) - 12(x + 7) = 0
x = 6, -7
x = -7 is not possible.
thus, x = 6 and x+1 = 7.

so 6 and 7 are the numbers.

Answer 5

6 and 7- 6x6=36 7x7=49 36+49=85