SOLUTION: 1) The sum of the squares of two consecutive integers is 613. What are the integers? Let x be one integer. Therefore x +1 is the next integer. x^2+(x+1)^2= 613 2)The length

Question 204189: 1) The sum of the squares of two consecutive integers is 613. What are the integers?
Let x be one integer. Therefore x +1 is the next integer.
x^2+(x+1)^2= 613
2)The length of a rectangle is four more than the width. If the area of the rectangle is 621 m2, find the width and length of the rectangle.
Let w be the width. Therefore w + 4 is the length.
W(w +4) = 621

Found 2 solutions by RAY100, Theo:
Answer by RAY100(1637) (Show Source): You can put this solution on YOUR website!
Very good setup
.
1) x^2 +(x+1)^2 =613
.
x^2 +x^2 +2x+ 1 =613
.
2x^2 +2x -612 =0
.
x^2 +x -306=0
.
(x+18)(x-17) =0
.
x = 17, -18
.
Integers,,17,18,,,,,,,check,,17^2 +18^2 =613,,,ok
.
also -18,,-17,,,,,,,check, (-17)^2 +(-18)^2 = 613,,,ok
.
.
2) w(w+4) =621
.
w^2 +4w -621=0
.
(w+27)(w-23) = 0
.
w = 23, -27 (not realistic)
.
Dim"s, w=23,,,L=27
.
check , 23*27 = 621,,,ok
.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
problem 1:
x^2 + (x+1)^2 = 613
becomes:
x^2 + x^2 + 2x + 1 = 613
becomes:
2x^2 + 2x = 612
becomes:
2x^2 + 2x - 612 = 0
becomes:
x^2 + x - 306 = 0
solve using quadratic equation:
a = 1
b = 1
c = -306
x =
becomes:
x =
becomes:
x =
becomes:
x =
becomes:
x = (-1 +- 35)/2
-----
x = 34/2 = 17
x = -36/2 = -18
-----
if x = 17, then x + 1 = 18 and 17^2 + 18^2 = 613
if x = -18, then x+1 = -17 and (-18)^2 + (-17)^2 = 613
-----
both values are good.
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problem 2:
w * (w+4) = 621
w^2 + 4w = 621
w^2 + 4w - 621 = 0
(w-23)*(w+27) = 0
w = 23 or w = -27
w can't be negative so w = 23 has to be the answer if it checks out.
23 * 27 = 621 so the answer is good and w = 23 and L = w+4 = 27.
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