how do u solve product of two consective odd whole number is 323?

Answer 1

The easy way is to just use trial and error. Since 20^2 is 400 and 15^2 is 225 the numbers you're looking for should be between those two numbers and it shouldn't take long to come up with 17 and 19 (of course, you can also use -17 and -19 if you want to be cute).

The hard way would be to use the quadratic formula, which would be algorithmic so it would be nice for a computer program but would take much longer for us humans. Note that we won't get an even number for x because two even numbers can't be multiplied to get an odd number.

--- x(x+2) = 323
--- x^2+2x = 323
--- x^2+2x-323 = 0

x = [-2 +- sqrt(4+1292)]/2
= -1 +- sqrt(1296)/2
= -1 +- 36/2 = -1 +- 18 = -19 or 17

If x is -19, the two numbers would be -19 and -19+2 = -17
If x is 17, the two numbers would be 17 and 17+2 = 19

Answer 2

Let the two consecutive odd whole numbers be n and (n + 2).

Since n is odd, n = 2m + 1 (m is a constant whole number)

Therefore,
(n)(n + 2) = (2m +1)(2m + 1 + 2) = 323
==> (2m + 1)(2m + 3) = 323
==> 4mxm + 8m + 3 = 323
==> 4mxm + 8m - 320 = 0
==> mxm + 2m - 80 = 0
==> (m + 10)(m - 8) = 0
==> m = -10 or m = 8
Hence, n = 17 or -19
And the 2 consecutive odd numbers are either 17 & 19 or -19 & -17

Answer 3

Let first consecutive odd number = x
Let second consecutive odd number = x + 2

x (x + 2) = 323
x^2 + 2x = 323
x^2 + 2x - 323 = 0
(x + 19)(x - 17) = 0
x = -19 or x = 17
The two consecutive odd numbers are -19 and -17, and 17 and 19. The product of each pair is 323.

Answer 4

17 times 19 equals 323

Answer 5

First define an "odd" number with variables.

Say "n" is any number. Any number can be odd or even.
But, "2n" will always be even. (check it out, it works)
If that's true, "2n + 1" will always be odd.

So, an unknown odd-number is "2n + 1"


Second, with two consecutive odd numbers, like "5"and "7",
the difference will always be "2". So the other number is:

(2n + 1) + 2

Two consecutive odd number are
(2n + 1) and (2n + 3)

Their product equals 323
(2n + 1)(2n + 3) = 323
4n^2 + 8n + 3 = 323
4n^2 + 8n - 320 = 0
4(n^2 + 2n - 80) = 0
n^2 + 2n - 80 = 0

n = [ -2 +/- sqrt( (2^2) - (4)(1)(-80) ) ] / (2)(1)
= [ -2 +/- sqrt( 4 + 320 ) ] / 2
= [ -2 +/- 18 ] / 2
n = -10 or 8


For n = -10
(2n + 1) = -19 and (2n + 3) = -17

For n = 8
(2n + 1) = 17 and (2n + 3) = 19

Answer 6

If first numer = x
Second number will be = x + 2
x(x+2) = 323
x^2 + 2x - 323 = 0
(x+19)(x-17) = 0
X+19 = 0 or x-17 = 0
x = -19 or x = 17

first set of solutions: first number = 17, second number = 19
second set : first number = -19, second number = -17

Answer 7

n^2+2n-323=0
(n+19)(n-17)=0
so the nosare 17 and 19 or -19 and -17
since the sum says whole nos the choice is 17 and 19

Answer 8

x(x + 2) = 323
x^2 + 2x = 323
x^2 + 2x - 323 = 0

x = (-b ± sqrt(b^2 - 4ac))/(2a)

x = (-2 ± sqrt(2^2 - 4(1)(-323))/(2(1))
x = (-2 ± sqrt(4 + 1292))/2
x = (-2 ± sqrt(1296))/2
x = (-2 ± 36)/2
x = (-38/2) or (34/2)
x = -19 or 17

ANS : -19 and -17 or 17 and 19