Question

The product of four consecutive positive integers is 840. Find the numbers.

Answer 1

Let x be the average of the four consecutive positive integers. x is a fraction that has 0.50 as its decimal part.

Then the numbers are:    x-1.5,  x-0.5,  x+0.5, x+1.5

Product:   (x-1.5)(x+1.5) (x-0.5)(x+0.5) = 840
                 (x² - 2.25) (x² - 0.25) = 840
                 x⁴ - 2.50 x² -  839.4375
                 x² =  [2.50 +- √(2.50² + 4*839.4375) ] /2
                     =  [ 2.50 +- 58] /2
                     =  30.25
                 x = 5.5

So the numbers are = 4, 5, 6, 7
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It can be done by factorization also:

840 = 2 * 2 * 10 * 21 = 2*2*2*3*5*7
                                = *4*5*6*7

             by rearranging the factors.

Answer 2

Answer:

Step-by-step explanation

Question:-

The product of four consecutive positive integers is 840. Find the Numbers

Solution :-

Let x be the average of the four consecutive positive integers. x is a fraction that has 0.50 as its decimal part.

Then the numbers are: x-1.5, x-0.5, x+0.5, x+1.5

Product: (x-1.5)(x+1.5) (x-0.5)(x+0.5) = 840

(x² - 2.25) (x² - 0.25) = 840

x⁴ - 2.50 x² - 839.4375

x² = [2.50 +- √(2.50² + 4*839.4375) ] /2

= [ 2.50 +- 58] /2

= 30.25

x = 5.5

So the numbers are = 4, 5, 6, 7

==============

It can be done by factorization also:

840 = 2 * 2 * 10 * 21 = 2*2*2*3*5*7

= *4*5*6*7

by rearranging the factors.

Thanks _!!