mabolden14
Answered

Looking for reliable answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.

Find Two consecutive positive integers such that the square of the larger integer is nineteen more than nine times the smaller integer

Sagot :

LeCoolKid

Answer:

9 and 10

Step-by-step explanation:

To start, let's have what we know:

a and b are two consecutive positive integers

a = first integer

b = second integer, 1 greater than variable a

b = a + 1

b^2 = 19 + 9a

To solve this, we would only have to substitute values in

b^2 = 19 + 9a

(a+1)^2 = 19 + 9a

a^2 + 2a +1     = 19 + 9a

       -9a -19    -19 - 9a

a^2 -7a - 18 = 0

a^2 + 2a - 9a - 18

a(a + 2) - 9(a+2) = 0

(a - 9)(a + 2) = 0

So solving this, we would get a = 9 or a = -2. Since this is supposed to be a positive integer, a will be 9.

Solving for b,

b = a + 1

  = 9 + 1

  = 10

Let me know in the comments if you have any questions! If you could mark this answer as brainliest, I would very much appreciate it!

We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.