Respuesta :
Let's call the three integers [tex] x,\ x+1,\ x+2 [/tex]
The product of the second and third integers is thus
[tex] (x+1)(x+2) = x^2+x+2x+2 = x^2+3x+2 [/tex]
We know that this quantity equals fifteen times the first integer:
[tex] x^2+3x+2 = 15x \iff x^2-12x+2=0 [/tex]
The solutions of this equation are
[tex] x_{1,2} = \dfrac{-b\pm\sqrt{\Delta}}{2a} [/tex]
So, the determinant [tex] \Delta [/tex] has to be a perfect square.
This is not the case, because
[tex] \Delta = b^2-4ac = 144-8 = 136 [/tex]
which is not a perfect square.
So, it is not possible to find three consecutive integers with that property.
Answer:
hope this helps have an amazing day :)
Step-by-step explanation:
Let's call the three integers
The product of the second and third integers is thus
We know that this quantity equals fifteen times the first integer:
The solutions of this equation are
So, the determinant has to be a perfect square.
This is not the case, because
which is not a perfect square.
So, it is not possible to find three consecutive integers with that property.
