SOLUTION: find three consecutive numbers such that twice the square of the first number is 38 more than the product of the other two numbers.

Question 112530: find three consecutive numbers such that twice the square of the first number is 38 more than the product of the other two numbers.
Found 2 solutions by ilana, checkley71:
Answer by ilana(307) (Show Source): You can put this solution on YOUR website!
Because we need 3 consecutive numbers, I would call them x, x+1, x+2. This way we will be solving for x, the smallest number. So, using the clues about the numbers, we get the equation 2x^2 = (x+1)(x+2) + 38.
Simplifying gives 2x^2 = x^2 + 3x + 2 + 38
2x^2 = x^2 + 3x + 40
Now we solve this quadratic equation. We could subtract the right side of the equation from both sides.
x^2 - 3x - 40 = 0
We could factor this to (x-8)(x+5)=0
So x=8 or x=-5. You just need 1 set of numbers, so I would say 8, 9, 10.
You could check this: 2*(8)^2=128 and 9*10+38 = 128, so this set of numbers works.

Answer by checkley71(8403) (Show Source): You can put this solution on YOUR website!
LET X, (X+1) & (X+2) BE THE 3 CONSECUTIVE NUMBERS.
2X^2=38+(X+1)(X+2)
2X^2=38+X^2+3X+2
2X^2-X^2-3X-38-2=0
X^2-3X-40=0
(X-8)(X+5)=0
X-8=0
X=8 ANSWER. 8+1=9 8+2=10.
X+5=0
X=-5 ANSWER. -5+1=-4 -5+2=-3.
PROOF
2(8)^2=38+(9*10)
2*64=38+90
128=128
----------------------
2(-5)^2=38+(-4*-3)
2*25=38+12
50=50

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