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Question

The product of two whole numbers is 1122.their difference is 1.find both numbers.
show work plz an ty!!!!!!!!!!!!

Answer 1

Two pairs of consecutive *numbers* have a product of 1122:
33 x 34 and -33 x -34. The whole number answer is 33 x 34 = 1122.

Intuitively the answer has to be close to the sqrt(1122). This is ±33.4962684. So trying the whole numbers on either side, you get the answer.

Alternatively:
Let x be the smaller number and x+1 be the next consecutive number.

Write the product as:
x(x+1) = 1122

Now distribute the x through:
x² + x = 1122

Now subtract 1122 from both sides:
x² + x - 1122 = 0

Now use the quadratric formula to get the solutions for x:
x = 33 or x = -34

Since your problem states they have to be whole numbers (0 and the positive integers), there is only one solution:
33 x 34 = 1122

Answer 2

Start by factoring 1122 into its prime number factors.
Right off you can see that 11 is a factor so
1122=11(102)
102=2(51) so
1122=(11)(2)(51)
but 51=(3)(17) or
1122=(11)(2)(3)(17)
These are all prime numbers so we cannot factor 1122 any further.
Multiply any combination of these primes to get all the factors of 1122.
Since you are looking for two consecutive numbers as factors
start multiplying pairs of prime factors to see if you get factors of 1122 which are close together.
(11)(2)=22, (11)(3)=33 (2)(17)=34,(3)(17)=51.
So you see that 33 and 34 are both factors of1122.
34-33=1.
There you have it.

The two whole numbers are 34 and 33.

Answer 3

Let the numbers be x and y.
Hence, x-y=1.
i.e., x=y+1.
Also, x * y = 1122.
Since, x=y+1,
hence,(y+1) * y =1122
i.e., y^2 + y = 1122
i.e., y^2 +y -1122 = 0.
Solving for y, we get, y = +33 or y = -34.
Since y is a whole number, therefore, y is positive.
Hence, y = 33.
x = y + 1 = 33 + 1 = 34.
Thus, the numbers are 34 and 33.

Answer 4

according to problem, a x b = 1122,

a-b=1

so a=1+b

(1+b) x b = 1122

b+b^2 -1122 = 0

(b+34) (b-33) =0

b= -34 or b = 33

according to problem, the product of two whole number is

maximum.

Answer 5

It's an algebra equation:

x(x+1)=1122 then multiply the first half out
x^2 +x = 1122 then subtract 1122 from both sides to get the equation = to 0
x^2 + x - 1122 = 0

then enter it into the quadratic equation

x= [-b +/- sqrt(b^2-4ac)]/2a
which gives you

x= [-1 +/- sqrt(4489)]/2; sqrt 4489 equals 67
x= (-1 +/- 67)/2;
so your numbers are 33 and 34

33*34=1122

Answer 6

Find the square root of the number 1122, and then find the closest whole numbers to the square root that multiply to give you 1122, and that subtract to give you 1.
The answer is 33 & 34

Answer 7

this is simple let them be x, x+1
x^2 < 1122 and (x+1)^2 > 1122
so x <33.49 x = 33 x+1 = 34
multiply and test(the reason to cross check is if I said product 1101 this is still an) = 1122.
so it matches
so the whole numbers are 33 , 34

(-34 and -33 are not whole numbers so excluded)

Answer 8

I'll show you all the steps including the quadratic formula below:
1. x*(x+1)=1122
2. x^2 + x - 1122 = 0
3. quadratic formula: x = [-1 +/- sqrt(1-4*(-1122))]/2
4. x=[ -1 +/- sqrt(4489)]/2
5. x= 66/2 OR x= -68/2
6. x=33 and therefore x+1 = 34
OR x= -34 and therefore x+1 = -33
7. Solutions: 33 and 34 (whole numbers)

Answer 9

take the square root of the number 1122, and then find the closest whole numbers to the square root that multiply to give you 1122, and that subtract to give you 1.

Answer 10

33 and 34

xy=1122
and x-y=1
so x=y+1
(y+1)y=1122
y^2+y-1122
use the quadratic equation to find that y=33 and x=34