help with a math problem/// i asked 10 ppl and they dont know... HELP?

Question

The sum of the squares of two (2) consecutive integers is nine (9) greater than eight (8) times the smaller. Find the numbers!?

Answer 1

to denote two consecutive integers, you can call one x and the other x+1
the sum of the squares of these two would be :

x^2 + (x+1)^2

and this equals 9 greater than 8 times the smaller, which can be denoted as:
8x + 9, since x is the smaller ( x+1 is one bigger) and you are multiplying it by 8 and adding 9
so set these two equal to eachother:

x^2 + (x+1)^2 = 8x + 9

now, multiply out (x+1)(x+1) using FOIL

so you get
x^2 + x^2 +2x +1 = 8x + 9

now combine like terms and get everything on one side:
2x^2 - 6x - 8 = 0
now factor:
(2x - 8)(x+1 ) = 0
so x = 4, -1

now, if x = 4 then x+1 would equal 5
recheck the conditions with these numbers:

4^2 + 5^2 = 8*4 +9
16 + 25 = 41
41 = 41
so the numbers are 4 and 5!!
or now you can try the other which was -1
so if x = -1, x+1 = 0
now lets double check these:
(-1)^2 + (0)^2 = 8*(-1) +9
1=1
and these also work

so the numbers could either be
4 and 5
or -1 and 0

if you know algebra, this is a fairly simple problem, if you don't you just have to do alot of guess and check and notice patterns and see if you could get the answer!

I hope this helps!!

Answer 2

The digits are consecutive. So if the smaller one is n, then the other is n+1.

So the statement becomes this equation:
n^2 + (n+1)^2 = 9 + 8n

Expand the brackets:
n^2 + n^2 + 2n + 1 = 9 + 8n

Rearrange and collect like terms:
2n^2 - 6n - 8 = 0

Divide through by 2:
n^2 - 3n - 4 = 0

Factorise:
(n+1)(n-4) = 0

So n is either -1 or 4 (making the other number either 0 or 5).

Check:
-1 & 0:
(-1)^2 + 0^2 = 1
9 + 8*(-1) = 9-8 = 1 So that pair works.

4 & 5:
4^2 + 5^2 = 16 + 25 = 41
9 + 8*4 = 9+32 = 41. So that pair works too.

Answer 3

Let 'x' be the smaller of the two integers. Then (x+1)is the next larger consecutive integer.
So square up the two unknowns.
x^2 + (x+1)^2 = 8x + 9
NB x is the smaller and 8x is x multiplied by 8.
So multiplying out to a quadratic equation.
x^2 + x^2 + 2x + 1 = 8x + 9
Collecting all terms to one side of the 'equals' and equate to 0 (zero).
2x^2 + 2x + 1 -8x - 9 = 0
2x^2 - 6x -8 = 0
take a factor of '2' out, we have:-
x^2 - 3x - 4 = 0
Now factorise this quadratic eq'n.
(x - 4)(x + 1) = 0
Hence to two values of 'x' are +4 & -1.
-1 can be discounted as an impossibility.
So '4' is the smaller integer and its next connsecutive integer is '5'.
This can be verfiieid
4^2 + 5^2 = 16 + 25 = 41
& 8 x 4 = 32 + 9 = 41.
So '4' & '5' are the two integers.

Answer 4

You came to the right place!

the integers are:
x and x+1
x^2 + (x+1)^2 = 9 + 8x
2x^2 + 2x +1 -8x =9
2x^2 - 6x -8 =0
x^2 -3x - 4 = 0
(x-4)(x+1) = 0
The integers are 4 & 5 OR
-1 & 0

Answer 5

let x be the integer
x + 1 is the next integer

sum means addiction
x^2 + (x+1)^2 = 8x + 9

x^2 + x^2 + 2x + 1 = 8x + 9

2x^2 - 6x - 8 = 0

2x^2 -8x + 2x - 8 = 0

(2x^2 - 8x) + (2x - 8) = 0

2x (x - 4) + 2(x - 4) = 0

(2x + 2) (x - 4) = 0

x = -1 or 4


so the two integers are -1 and 0

OR

4 and 5

Answer 6

let x, x+1 be the consecutive numbers.
x^2+(x+1)^2=9+8x
x^2+x^2+2x+1=8x+9
2x^2-6x-8=0
x^2-3x-4=0
(x-4)(x+1)=0
x=4 or -1
check : 4^2+5^2=8(4)+9 ? yes
The numbers are 4 and 5

Answer 7

x^2+(x+1)^2=8x+9
2x^2-6x-8=0
2(x-4)(x+1)=0
x=4 or x=-1
Therefore the numbers could be -1 and 0, or 4 and 5. both work

Answer 8

Call the first number x, the second number x+1.

x^2 + (x+1)^2 =8x+9

You'll FOIL the set of parentheses, then simplify the equation.

Set it equal to zero by subtracting 8x and 9 from both sides.

From here, you can do the quadratic equation.

Answer 9

n^2+(n+1)^2 >8n+9
2n^2 -6n-8 >0
n^2-3n-4>0 so n>4 if you mean positive integers
and in general n>4 or n<-1
There are infinite possibilities

Answer 10

4,5
sum squares=41
4*8=32