High School

The product of two consecutive negative integers is 600. What is the value of the lesser integer?

A. [tex] -60 [/tex]

B. [tex] -30 [/tex]

C. [tex] -25 [/tex]

D. [tex] -15 [/tex]

Answer :

Let the two consecutive negative integers be [tex]$n$[/tex] and [tex]$n+1$[/tex]. The product of these integers is given by

[tex]$$
n(n+1) = 600.
$$[/tex]

Expanding the left side, we have

[tex]$$
n^2 + n = 600.
$$[/tex]

Subtracting 600 from both sides yields the quadratic equation:

[tex]$$
n^2 + n - 600 = 0.
$$[/tex]

For a quadratic equation of the form [tex]$ax^2+bx+c=0$[/tex], the discriminant is given by

[tex]$$
D = b^2 - 4ac.
$$[/tex]

Here, [tex]$a=1$[/tex], [tex]$b=1$[/tex], and [tex]$c=-600$[/tex]. Plugging in these values:

[tex]$$
D = 1^2 - 4(1)(-600) = 1 + 2400 = 2401.
$$[/tex]

Taking the square root of the discriminant:

[tex]$$
\sqrt{D} = \sqrt{2401} = 49.
$$[/tex]

Now, applying the quadratic formula

[tex]$$
n = \frac{-b \pm \sqrt{D}}{2a},
$$[/tex]

we find the two possible solutions:

[tex]$$
n = \frac{-1 + 49}{2} = \frac{48}{2} = 24
$$[/tex]

and

[tex]$$
n = \frac{-1 - 49}{2} = \frac{-50}{2} = -25.
$$[/tex]

Since the problem specifies consecutive negative integers, we select the negative solution. Therefore, the lesser integer is

[tex]$$
\boxed{-25}.
$$[/tex]

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