find 3 consecutive integers such that the square of the first, increased by twice the sum of the other two is 51

Answer 1

You are looking for three numbers x, y, and z that fit the equation x²+2(y+z) = 51.

However, we know the numbers are consecutive integers, so we can say:

1. y = x+1
2. z = x+2

This reduces our equation to be a one-variable equation:

1. x²+2(x+1+x+2) = 51
2. x²+2(2x+3) = 51
3. x²+4x+6 = 51
4. x²+4x-45 = 0

From here, we need to factor the quadratic.

1. We know that the constants must be two integers that multiply into -45, and add up to 4.
2. If you know your multiplication table well, you'll know 9 and 5 multiply to 45.
3. We want the product to be negative, so one of the constants needs to be negative and the other positive.
4. Since they add up to positive 4, we know the greater of the two (9) should be positive. This gives us:
5. (x+9)(x-5) = 0

So our two solutions are x = -9 and x = 5. Plug these in to the original equation and we get:

1. y = -8 or 6
2. z = -7 or 7
3. x = -9 gives us (-9)²+2(-8-7) = 81+2(-15) = 81-30 = 51
4. x = 5 gives us 5²+2(6+7) = 25+2(13) = 25+26 = 51

So, your answer could be (-9,-8,-7) or (5,6,7)