The product of two consecutive positive integers is 75 more than nine times the second integer. What are the integers?

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neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

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If one positive integer is x, then the other positive consecutive integer is  x+1.

Their product is x(x+1) .

 Again given that product x(x+1) = 75 more than  9 times the  second integer.

So x(x+1) = 75+9(x+1)  is the required equation. We get the solution from this.

x^2+x = 9x+75+9

x^2 +x-9x -84 = 0

x^2-8x-84 = 0

(x-14)(x+6) = 0

x-14 = 0 or x+6 = 0

x =14 or x = -6.

x = 14 is valid as the positive integer.

Tally:

75+9(x+1) = 75+9*15 = 75+135 = 210

 

 

krishna-agrawala's profile pic

krishna-agrawala | College Teacher | (Level 3) Valedictorian

Posted on

Let:

First integer= x

Then:

Next integer = x +1

As per the given conditions:

x*(x + 1) = 9(x + 1) + 75

Simplifying this equation we get:

==> x^2 + x = 9x + 9 + 75

==> x^2 - 8x - 84 = 0

==> x^2 - 8x + 16 - 100= 0

==> (x -  4)^2 10^2 = 0

==>(x - 4 + 10)(x - 4 - 10) = 0

==>(x + 6)(x - 14) = 0

x = -6, and 14

As x is a positive number the correct value can only be:

x = 14

Answer:

the integers are 14 and 15.

 

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

Let's note the first integer as x and the secon consecutive integer is x+1.

Now, we'll write mathematically the condition of enunciation:

- the product of 2 consecutive integers: x(x+1)

- is: =

- 75 more: 75 +

- nine times the second integer: 9(x+1)

Now, let's join them:

x(x+1) = 75 + 9(x+1)

We'll remove the brackets:

x^2 + x = 75 + 9x + 9

We'll combine like terms:

x^2 + x = 84 + 9x

We'll subtract 84 + 9x both sides:

x^2 + x - 84 - 9x = 0

We'll combine like terms:

x^2 - 8x - 84 = 0

We'll apply the quadratic formula:

x1 = [8 + sqrt(64 + 336)]/2

x1 = (8 + 20)/2

x1 = 14

or

x1 = (8-20)/2

x1 = -6

Since the integer has to be positive, we'll accept just x1 = 14.

The second consecutive integer is x2 = 14+1

x2 = 15.

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