Question

the product of two consecutive multiplies of 6 is 432. which are the numbers? ​

Answer 1

let the two numbers be x and x+6

let the two numbers be x and x+6x(x+6)=432

let the two numbers be x and x+6x(x+6)=432x^2+6x=432

let the two numbers be x and x+6x(x+6)=432x^2+6x=432(x+3)^2=432+9

let the two numbers be x and x+6x(x+6)=432x^2+6x=432(x+3)^2=432+9x+3^2=441

let the two numbers be x and x+6x(x+6)=432x^2+6x=432(x+3)^2=432+9x+3^2=441x+3=+or - 21

let the two numbers be x and x+6x(x+6)=432x^2+6x=432(x+3)^2=432+9x+3^2=441x+3=+or - 21x=21-3=18

let the two numbers be x and x+6x(x+6)=432x^2+6x=432(x+3)^2=432+9x+3^2=441x+3=+or - 21x=21-3=18x=-21-13=24

let the two numbers be x and x+6x(x+6)=432x^2+6x=432(x+3)^2=432+9x+3^2=441x+3=+or - 21x=21-3=18x=-21-13=24the value of x are 18 and 24

ʜᴏᴩᴇ ɪᴛ ʜᴇʟᴩꜱ :)

Answer 2

$\huge\star{\underline{\mathtt{\red{A}\pink{N}\green{S}\blue{W}\purple{E}\orange{R}\star}}}$

Let the two numbers be x and x+6

$= > x(x + 6) = 432$

$= > {x}^{2} + 6x = 432$

$= > ({x + 3})^{2} = 432 + 9$

$= > ( {x + 3})^{2} = 441$

$= > x + 3 = 21$

$= > x = 21 - 3$

$= > x = 18$

$x + 6 = > 18 + 6$

$x = 18 \: and \: \: 24$