Question

Find three consecutive negative integers such that 3 times the product of the first and third exceeds twice the product of the first and second by 32

Answer 1

Let the first negative integer be = -x
second negative integer = -x -1
third negative integer = -x -2

product of first and third = (-x) (-x -2) = x^2 + 2x
product of first and second = (-x) (-x -1) = x^2 + x

ATQ, 3(x^2 + 2x) - 2(x^2 + x) = 32
3x^2 + 6x -2x^2 - 2x = 32
x^2 + 4x - 32 = 0
x^2 + 8x - 4x - 32 = 0
x(x + 8) -4(x + 8) = 0
(x - 4)(x + 8) = 0
if x - 4 = 0, then x = 4
if x + 8 = 0, then x = -8


so if x = 4
then first negative no is = -x = -4
second negative no is = -x -1 = -5
third negative no is = -x -2 = -6


we cannot use/accept x = -8 because
if we take x = -8, then we get positive nos.
first no = -x = -(-8) = 8
second no = -x -1 = -(-8) - 1 = 8 - 1 = 7
third no = -x -2 = -(-8) - 2 = 8 -2 = 6

but we have to find negative numbers

so -4, -5 and -6 is the right answer