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One positive integer is 1 less than twice another. The sum of their squares is 461. Find the integers.

Sagot :

ManpreetCheema

Answer:

10 and 19

Step-by-step explanation:

  • In Algebra, an even number = 2x, because anything multiplied by 2 is even.
  1. There are two numbers in the question, I'll define the first one as 2x
  2. the other number is 1 less than twice of the first, so I'm going to define it in algebra as 2(2x) - 1, which is just 4x - 1
  3. the square of the first is [tex]4x^2[/tex]
  4. the square of the second is [tex]16x^2-8x+1[/tex]
  5. add them together: [tex]20x^2-8x+1[/tex]
  6. this equation = 461, so [tex]20x^2-8x+1 = 461[/tex]
  7. simplify to [tex]20x^2-8x-460 = 0[/tex]
  8. I wont explain how to solve this here, since it is very complicated, if you don't know search up solving quadratics.
  9. the solutions are 5 or -4.6, it cannot be -4.6 since the question says positive, so the solution for x = 5
  10. substitute x into our equations for the two numbers in steps 1 and 2
  11. the first number = 10, the second = 19
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