can anyone solve this?

Question

the area of a square is 45 more than the perimeter. find the length of a side.

and

the product of two positive consecutive even integers is 224. find the integers.

Answer 1

If the side length of a square is "s", then the area is s^2 and the perimeter is just 4s. Since the area has a value that's 45 more than the perimeter, this means:
s^2 = 4s + 45, or
s^2 - 4s - 45 = 0
Solve this for s.

If N is an even integer, then the even integer that comes after that is N+2. So N(N+2) = 224. Multiply this out like before and solve for N, keeping in mind that we want to find the positive solution. Then use N to write N and N+2.

Answer 2

Let side of square be x so preimeter is sum of all sides. As all sides are equal. Perimeter = 4 × x = 4x
Area fo square = (Length)^2 = x^2
So as per statement,
x^2 = 4 x + 45
or x^2 – 4 x – 45 = 0
Product of Coefficient of 1st and 3rd term = 1 × 45 = 45
Find factors of 45 such that the sum of difference in factors is – 4, such factors are + 5 and – 9
so the expression can be written as,
x^2 + 5 x – 9 x – 45 = 0
= x (x + 5) – 9(x – 5) = 0 ----------- I assume that you know this step
(x – 9)(x + 5) = 0 i.e either of thse bracket's value is zero
(x – 9) = 0 or (x + 5) = 0
i.e. x = 9 or x = – 5, but value of side can not be – ve.
so x = side = 9

the product of two positive consecutive even integers is 224. find the integers
Let smaller number be x so the next even number = (x + 2)
as per statement,
x × (x + 2) = 224
i.e x^2 + 2 x = 224 or x^2 + 2 x – 224 = 0
similar to previous find factors of 224
= 2 × 2 × 8 × 7 = 14 ×16 -------- the factors having difference of 2
so x^2 + 2 x – 224 = 0
x^2 + 16 x – 14 x – 224 = 0
x(x + 16) – 14(x – 16) = 0
(x – 14)(x + 16) = 0
x = 14 or x = – 16
But +ve integer is required so
x = 14 and other number = 14 + 2 = 16

Answer 3

Solution of 1st question...:-
Let the side of square be "a"
the area of square will be "a2 “
and the perimeter will be “ 4a “
now it is given that the area of the square is 45 more than its perimeter....
then....a2 – 4a = 45
 a2 – 4a – 45 = 0
 a2 - 9a + 5a – 45 = 0
 a(a – 9) + 5(a – 9) = 0
 (a – 9) (a + 5) = 0
 a = 9, a = -5
since avalue of “a” can not be negative because side can not be positive
so the value of “a” which is the length of the square will be 9.
Ans: 9


Solution of 2nd question:
Let first no be “ x “
Then the second no. be “ x + 2 “
The product will be “ x(x + 2)
Then x(x + 2) = 224
So x2 + 2x – 224 = 0
x2 + 16x – 14x – 224 = 0
x(x +16) – 14(x + 16) = 0
(x + 16) (x – 14) = 0
x = 14,
x = -16
for x = 14, two integers will be 14 and 16
for x = -16, two integers will be -16 and -14
ans: 14 and 16 or -16 and -14
All d Best...

Answer 4

Area = x^2
Perimeter = 4x

Area = Perimeter + 45 Substitute above into this equation

x^2 = 4x+ 45 Subtract 4x and 45 from both sides
x^2 - 4x - 45 = 0 Factor
(x -9)(x+5) = 0 Set each term = 0

x - 9 = 0
x = 9 ANSWER

Check
Area 9(9) = 81
Perimeter 4(9) = 36
81 - 36 = 45

or
x + 5 = 0
x = - 5 The side of a square can't be a negative number
-----------------------------------------------------------------------------
product of two positive consecutive even integers is 224
x and x+2 are the numbers
x(x+2) = 224
x^2 +2 x = 224 Subtract 224 from both sides
x^2 + 2x - 224 = 0
(x - 14)(x + 16) = 0

x - 14 = 0
x = 14
x + 2 = 16
The numbers are 14 and 16 ANSWER

The term x+16 = 0 will be a negative number so use the other term above
x + 16 = 0
x = - 16

Answer 5

Problem 1:
Area= Perimeter +45
Area since it is a square is s^2 (s meaning side)
Perimeter is s+s+s+s (4s)

Therefore:
s^2= 4s+45
s^2-4s-45=0
(s-9)(s+5)=0
set each equal to 0
s-9 = 0
s=9

and s+5 = 0
s= -5

so your two possibilities are side can = 9, -5
but you can't have negative sides so your side is 9.


Problem 2:
Actually for the second one you can't just say the first number is x and the second is x+2 because x can be an odd number and x+2 can be an odd number too.

so to write it properly. the first number should be 2k and the second would be this number plus 2 (2k+2) because that's the definition of consecutive even number.

so (2k)(2k+2)=224
4k^2+4k=224
4k^2+4k-224=0
divide both sides by 0
k^2+k-56=0
factor
(k+8)(k-7)=0
k=-8 or k=7
but it can't be -8 because then our first number would be -16 which is a negative number.

therefore k must be 7.

our first number is 2k = 14
the next consecutive even number is 16.

Answer 6

Let side of the square be s.
Area=s.s
Perimeter=4.s
Given,Area=4.s+45
s.s=4.s+45
s.s-4.s-45=0
s.s-9.s+5.s-45=0(by resolving into factors)
s(s-9)-5(s-9)=0
(s-9)(s-5)=0
If a(b)=0,a=0 or b=0 or a=b=0
Thus,
s-9=0 or s-5=0
s=9 or s=5
Therefore, length of a side is either 5units or 9units.

Let the two positive consecutive even numbers be 2x and 2x+2.
2x(2x+2)=224
4x.x+4.x=224
x.x+x=224/4
x.x+x=56
x(x+1)=56
x(x+1)=7(8)
Thus, x=7
Therefore the integers are14(2x) and 16(2x+2).

Answer 7

Question 1
let side be L
Area = L²
P = 4L
L² = 4L + 45
L² - 4L - 45 = 0
(L - 9)(L + 5) = 0
L = 9

Question 2
2x (2x + 2) = 224
4x² + 4x - 224 = 0
x² + x - 56 = 0
(x + 8)(x - 7) = 0
x = 7
Integers are 14 and 16

Answer 8

the area of a square is L^2, where L is the length of one side
the perimeter is 4L

4L+ 45= L^2

using quadratic eq, L=9

the other one is:

x times (x+2) = 224

or x(x+2) = 224

X^2 + 2X = 224. use quadratic eq.

x=4

Answer 9

Let the side of square to be 'a'.

a^2 = 45 + 4a
a^2 - 4a - 45 =0
a^2 - 9a + 5a - 45 = 0
a(a - 9) + 5(a - 9)=0
(a + 5)(a - 9)=0
a = -5 or a = 9

a = 9 { since distance can't be negative thus eliminating a=-5

Answer 10

if side length=L
perimeter=4L
Area=L^2
so
L^2-4L=45
L^2-4L-45=0
(L+5)(L-9)=0
L=-5(wrong answer)
L=9
so the length of the side is 9
-------------------------
if number2 are 2n & 2n+2
so
2n(2n+2)=224
n(n+1)=56
n^2+n-56=0
(n+8)(n-7)=0
n=-8 wrong
n=7
so the numbers are 2*n=14 & 2n+2=16