Question

Quadratic Equation



Find two consecutive even integers whose product is 168​

Answer 1

Step-by-step explanation:

Let the two consecutive even integers be (2x) , (2x+2).

According to question,

=> (2x)(2x+2) = 168

=> 4x(x+1) = 168

=> x(x+1) = 168/4

=> x^2 + x = 42

=> x^2 + x - 42 = 0

=> x^2 + 7x - 6x - 42 = 0

=> (x+7)(x-6) = 0

=> x = -7 , 6

So, when x = -7, then

integers are 2x and 2x+2 i.e 2(-7) and 2(-7)+2

i.e -14 , -12.

and when x= 6 then,

integers are 2x and 2x+2 i.e 2(6) and 2(6)+2

i.e 12 and 14.

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Answer 2

Answer:

let the integers 2n and 2n+2

product = 168

2n(2n+2)= 168

4n²+4n -168= 0

4(n²+n-42)=0

n²+n-42=0÷4

n²+7n-6n-42=0

n(n+7)-6(n+7)=0

(n+7)(n-6)=0

n+7=0 or n-6=0

n=-7 or n= 6

the integers are 2*-7 , 2*-7+2 = -14,-16

or

2 *6, 2*6+2 = 12,14

Step-by-step explanation:

2*2*2*3*7

2*84

3*56

4*42

6*28

7*24

8*21

12*14