solve this... see details?

Question

find three consecutive integers such that the sum of the squares of the 2nd and third integers exceeds the square of the 1st by 21.

Answer 1

2, 3 & 4

2 x 2 = 4
3 x 3 = 9
4 x 4 = 16
3 x 3 + 4 x 4 = 16 + 9
= 25
25 - 4 = 21

Answer 2

x, y, z in that order

z - y = 1
y - x = 1

z - x = 2


"the sum of the squares of the 2nd and third integers exceeds the square of the 1st by 21":

(y^2 + z^2) - x^2 = 21

The key is to get this equation in terms of one variable:

From above,
z = 1 + y
x = y - 1

Substitute and combine terms:
(y^2 + (1 + y)^2) - (y - 1)^2 = 21
y^2 + 1 + 2y + y^2 - y^2 + 2y - 1 = 21

y^2 + 4y = 21
y^2 + 4y - 21 = 0
(y + 7)(y - 3) = 0

y = -7 or 3

z - y = 1 so, z = -6 or 4
y - x = 1 and x = -8 or 2

There are two answers:

-8, -7 , -6 or 2, 3, 4

Answer 3

let three consecutive integers are x, x+1, x+2
according to question (x+1)^2 + (x+2)^2= x^2 +21
x^2 +2x+1+x^2 +4x+4=x^2 +21
x^2 +6x-16=0
x^2 +8x-2x-16=0
x(x+8)-2(x+8)=0
(x+8)(x-2)=0
x+8=0, x-2=0
x= -8 , x=2
so possible integers are 2,3,4 or -8,-7,-6

Answer 4

once you've x^2 = 9, the merely element preventing you from having a answer is that exponent. to have the ability to remove it and to that end isolate x, that is sensible to take the sq. root of both aspect of the equation to get x = ± ?9, meaning x = 3 or -3. there is not any longer something incorrect with doing it this way or doing it the following way: x^2 = 9 x^2 - 9 = 0 (x+3)(x-3) = 0 x = -3, 3 yet back, in the first case you're merely one step away. So why no longer merely remove the exponent and get the answer. I even ought to chortle on the guy who stated to apply the quadratic formulation. ought to someone somewhat take x^2 = 9, and regulate it to x^2 - 9 = 0, and then plug a=a million, b=0, c= -9 into the quadratic formulation? it ought to artwork, yet geez, what a waste of time.

Answer 5

let the first integer be x
therefore x^2 + 21 = (x+1)^2 + (x+2)^2
(I am going to write x^2 as x2)
or, x2 + 21 = x2 + 2x + 1 + x2 + 4x + 4
or, 21 = x2 + 6x + 5
or, x2 + 6x - 16 = 0
or, x2 + 8x - 2x - 16 = 0
or, x (x + 8) - 2 (x + 8) = 0
or, x = - 8 or 2
therefore, the integers are -6, -7 and -8 or 2,3 and 4.

Answer 6

let x= second no.
let x-1= first no.
let x+1= third no.

(x+1)^2+x^2=(x-1)^2+21
x^2+2x+1+x^2=x^2-2x+1+21
2x^2+2x+1=x^2-2x+22
x^2+4x-21=0
(x+7)(x-3)=0
x+7=0 x-3=0
x=-7 x=3
-6,-7,-8 and 2,3,4
where -6 is the third no -7 is the second -8 is the first
where 4 is the third no. 3 is the second and 2 is the first

Answer 7

There are two sets of answers to this question since it is quadratic:

(x+1)^2 + (x+2)^2 = x^2 +21
x^2 +2x +1 + x^2 +6x +9 = x^2 +21

simplified:
x^2 + 6x - 16 = 0

factored:
(x+8)(x-2) = 0

so x = -8 and 2

two sets:
-8, -7, -6
and
2,3,4

Answer 8

(x+1)^2 + (x+2)^2 = x^2 + 21
x^2 + 6x - 16 = 0
(x+8)(x-2) = 0
x = -8
x = 2

Therefore, the numbers are 2, 3, 4

Answer 9

im suppose to ask u a question a while ago if ur sure ur question is clearly stated since i have solved -8,-7,-6 and sets2,3,4... if ur question is restated like now... but ur question a while ago... i gor an imaginary number since there is no such thing as square root of a negative value... but i guess everybody else answered same as mine and they answered first too... i was waiting for u to restate it a while ago and i found ur restated question just now...

Answer 10

let the no. be (n-1),n,(n+1)

Now : n2 + (n+1)2 = (n-1)2 + 21

this gives n = 3 & -7

therefore no. are : 2, 3 ,4 or -8,-7,-6