Question

determine two consecutive negative even integers whose product is 24.
(completing the square method)​

Answer 1

Solution :

$\bf{\red{\underline{\bf{Given\::}}}}$

Integers whose product is 24.

$\bf{\red{\underline{\bf{To\:find\::}}}}$

The two consecutive negative even integers.

$\bf{\red{\underline{\bf{Explanation\::}}}}$

Let the two consecutive number be r and (r-2).

A/q

$\longrightarrow\sf{r(r-2)=24}\\\\\longrightarrow\sf{r^{2} -2r-24=0}\\\\\longrightarrow\sf{r^{2} -6r+4r-24=0}\\\\\longrightarrow\sf{r(r-6)+4(r-6)=0}\\\\\longrightarrow\sf{(r-6)(r+4)=0}\\\\\longrightarrow\sf{r-6=0\:\:Or\:\:r+4=0}\\\\\longrightarrow\sf{\pink{r\neq 6\:\:\:Or\:\:\:r=-4}}$

Thus;

$\star\underline{\sf{The\:1st\:consecutive\:negative\:even\:Integers\:is\:r=\boxed{\bf{-4}}}}}\\\star\underline{\sf{The\:2nd\:consecutive\:negative\:even\:Integers\:is\:(r-2)=(-4-2)=\boxed{\bf{-6}}}}$

Answer 2

The numbers are -6 and -4

Step-by-step explanation:  

let the two consecutive negative even integers  be -2n and -2 (n+1)  

Then Atq,

The product of two consecutive negative even integers is 24.  

i.e  

( -2n) × (-2(n+1)) = 24  

4×n×(n+1)  = 24  

n(n+1) = 6  

n² +n =6  

n² +n -6 =0  

n² +3n -2n -6=0  

n(n+3) -2(n+3) =0  

(n-2) (n+3) =0  

n = -2  

n= 3  

taking n = 2 because n must be a positive integer

hence for n = 2  

-2 (n+1) = -6  

hence , The numbers are -6 and -4