yume420
contestada

Find twe positive even consecutive intecers such
that the square of the smaller integer is 10 more
than the larger integer.

Respuesta :

Answer:

The numbers are 4 and 6

Step-by-step explanation:

Let:

  • x - the smaller even consecutive integer  
  • x+2 - the larger even consecutive integer

we know that

[tex]x^{2} =(x+2)+10[/tex]

solve for x

[tex]x^{2} -x-12=0[/tex]

We can use the quadratic formula for quadratics in the form [tex]ax^{2} +bx+c=0[/tex]

[tex]x= \frac{-b(+/-)\sqrt{b^{2} -4ac}}{2a}[/tex]

to give us an answer of 4 and -3

Since the question states positive and even we use 4 and that gives us 6 as the other answer
4 and 6

clarch19

Answer:

4 and 6

Step-by-step explanation:

x = smaller integer

x+2 = larger integer

x² = 10 + (x + 2)

x² = x + 12

x² - x - 12 = 0

factors are:

(x-4)(x+3) = 0

x-4=0;  x = 4

x+3=0;  x = -3

Since the question states the numbers are even and positive we use only the 4 and the next consecutive even integer is 6

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