Question

Four times the sum of three consecutive odd integers is seven hundred sixty-five less than three times the product of the larger two numbers. What are the three integers?
this was my initial answer to the question:
(x-2), x, (x+2)

4((x+2) + x+ (x-2))=3((x+2)(x)) – 765

4(3x)=3x2+6x-765

4x=x2+2x-255

X2-2x-255=0

X2-17x+15x-255=0

X(x-17)+15(x-17)=0

(x+15)(x-17)=0

(x-17)=0

X=17
however it was sent back to me with:
'You should end up with 2 values for x. You then need to write the two sets of three integers.'

I am unsure as to what i am supposed to do with this now :/
Thanks

Answer 1

Answer:

The consecutive odd integers are (15, 17, 19) or (-17, -15, -13).

Step-by-step explanation:

Let the three consecutive odd integers be: $x-2,x\ and \ x+2$

The condition given is:

$4[x-2+x+x+2]=3x(x+2)-765$

Solve this for x as follows:

$4[x-2+x+x+2]=3x(x+2)-765\\4\times 3x=3x^{2}+6x-765\\3x^{2}-6x-765=0\\x^{2}-2x-255=0\\x^{2}-17x+15x-255=0\\x(x-17)+15(x-17)=0\\(x-17)(x+15)=0$

• If $(x-17)=0$ then the value of x is 17.

       The odd numbers are:

          $x-2=17-2=15\\x=17\\x+2=17+2=19$

• If $(x+15)=0$ then the value of x is -15.

       The odd numbers are:

         $x-2=-15-2=-17\\x=-15\\x+2=-15+2=-13$

Thus, the consecutive odd integers are (15, 17, 19) or (-17, -15, -13).