Math problem that i cant figure out?

Question

Use an algebraic solution to find four consecutive positive integers such that the product of the first and fourth is four less than twice the first multiplied by the fourth.

The four consecutive integers, in order of least to greatest are...

Answer 1

Set up your integers like this:

x
x+1
x+2
x+3

Then convert your word problem into an algebraic problem:

"product of first and fourth" = x(x+3)

"4 less than twice the first multiplied by the fourth" = 2x(x+3) - 4

So, your problem to solve is:

x(x+3) = 2x(x+3) - 4

x² + 3x = 2x² + 6x - 4

0 = x² + 3x - 4

0 = (x+4)(x-1)

So, x = 1 or -4

Therefore your 4 consecutive integers are {1,2,3,4}

{-4,-3,-2,-1} would've worked too, but your problem said four consecutive POSITIVE integers.

.

Answer 2

Let the first integer be x. The second is x+1, etc. From this process, we get these expressions for our integers:

x, x+1, x+2, x+3

This is simply straightforward algebra. The product of the first and fourth integers is x(x+3) = (1) x^2 + 3x. Twice that is 2(x^2+3x) = 2x^2+6x. Four less than this last expression is (2) 2x^2+6x-4.

Now we simply set expression (1) equal to expression (2). So x^2 + 3x = 2x^2+6x-4. We collect terms by transposing them to the right side of the equation in order to minimize pesky negative numbers.

That gets us 0 = x^2+3x-4, which is easily factorable into 0 = (x+4)(x-1). Setting these factors equal to zero yields only one positive integer, which is 1. This is our first integer. We get the fourth simply by adding 3 to it.

Now, simply back substitute the appropriate numbers into our equation to assure that the requirements for our numbers have been met. This I leave as an exercise for you to do.

Answer 3

Since they are all consecutive, we can call the "first" integer N, and the second (N+1)...and so on. Then, all we need to do is set up the equation (N)(N+3)+4=2(N)(N+4) and solve for N.

Answer 4

n, n + 1, n + 2, n + 3 are 4 consecutive integers starting with n
n(n+3) + 4 = 2n(n + 3)
n^2 + 3n +4 = 2n^2 +6n
n^2 + 3n - 4 = 0
n = (-3 +/- sqrt(9 + 16))/2
n = (-3 + 5)/2 = 2/2 = 1
1,2,3,4

1*4 + 4 = 2*4
QED

Answer 5

1, 2, 3, 4

Answer 6

your integers will be x x+1 x+2 x+3

1st*4th=2*1st*4th-4

x(x+3)=2(x(x+3))-4
x^2+3x=2x^2+6x-4
0=x^2+3x-4
0=(x+4)(x-1)
x=1,-4

use 1 for a positive integar.

so your integers are 1,2,3,4

Answer 7

1,2,3,4 is the answer
use x-2,x-1,x and x+1 as consecutive numbers

Answer 8

i do no longer understand approximately "front end estimation," yet right this is one thank you to get the respond: First, you already know the thank you to calculate your contemporary ordinary, suited? --you're taking the given 5 rankings, upload them up, and divide via 5: one hundred + ninety + 80 two + ninety six + seventy 8 = 446 / 5 = 89.2 you have one extra attempt to take, so your very final ordinary score would be one hundred + ninety + 80 two + ninety six + seventy 8 + X / 6 --- you do no longer understand X yet, yet you do understand that there will be six rankings to ordinary. Your *optimal* obtainable ordinary score will take place in case you get one hundred on the suited attempt, suited? in case you plug in one hundred for X above and do the arithmetic -- the respond is ninety one.