Question

production of two consecutive positive integers is 156. represent it in the form of quadratic expression​

Answer 1

Answer:

let x and x+1 be the integers

let x and x+1 be the integersx(x+1)=156

let x and x+1 be the integersx(x+1)=156x^2+x=156

let x and x+1 be the integersx(x+1)=156x^2+x=156x^2+x-156=0

let x and x+1 be the integersx(x+1)=156x^2+x=156x^2+x-156=0156=2*2*3*13

let x and x+1 be the integersx(x+1)=156x^2+x=156x^2+x-156=0156=2*2*3*13(2*2*3)*13=12*13 and 12*13=156

let x and x+1 be the integersx(x+1)=156x^2+x=156x^2+x-156=0156=2*2*3*13(2*2*3)*13=12*13 and 12*13=156factor

let x and x+1 be the integersx(x+1)=156x^2+x=156x^2+x-156=0156=2*2*3*13(2*2*3)*13=12*13 and 12*13=156factor(x+13)(x-12)=156

let x and x+1 be the integersx(x+1)=156x^2+x=156x^2+x-156=0156=2*2*3*13(2*2*3)*13=12*13 and 12*13=156factor(x+13)(x-12)=156x-12=0

let x and x+1 be the integersx(x+1)=156x^2+x=156x^2+x-156=0156=2*2*3*13(2*2*3)*13=12*13 and 12*13=156factor(x+13)(x-12)=156x-12=0x=12 and x+1=13

let x and x+1 be the integersx(x+1)=156x^2+x=156x^2+x-156=0156=2*2*3*13(2*2*3)*13=12*13 and 12*13=156factor(x+13)(x-12)=156x-12=0x=12 and x+1=1312 and 13 are the integers

let x and x+1 be the integersx(x+1)=156x^2+x=156x^2+x-156=0156=2*2*3*13(2*2*3)*13=12*13 and 12*13=156factor(x+13)(x-12)=156x-12=0x=12 and x+1=1312 and 13 are the integersx+13=0

let x and x+1 be the integersx(x+1)=156x^2+x=156x^2+x-156=0156=2*2*3*13(2*2*3)*13=12*13 and 12*13=156factor(x+13)(x-12)=156x-12=0x=12 and x+1=1312 and 13 are the integersx+13=0x=-13

let x and x+1 be the integersx(x+1)=156x^2+x=156x^2+x-156=0156=2*2*3*13(2*2*3)*13=12*13 and 12*13=156factor(x+13)(x-12)=156x-12=0x=12 and x+1=1312 and 13 are the integersx+13=0x=-13x+1=-12

let x and x+1 be the integersx(x+1)=156x^2+x=156x^2+x-156=0156=2*2*3*13(2*2*3)*13=12*13 and 12*13=156factor(x+13)(x-12)=156x-12=0x=12 and x+1=1312 and 13 are the integersx+13=0x=-13x+1=-12-12 and -13 is also a solution

let x and x+1 be the integersx(x+1)=156x^2+x=156x^2+x-156=0156=2*2*3*13(2*2*3)*13=12*13 and 12*13=156factor(x+13)(x-12)=156x-12=0x=12 and x+1=1312 and 13 are the integersx+13=0x=-13x+1=-12-12 and -13 is also a solutionusing arithmetic...

let x and x+1 be the integersx(x+1)=156x^2+x=156x^2+x-156=0156=2*2*3*13(2*2*3)*13=12*13 and 12*13=156factor(x+13)(x-12)=156x-12=0x=12 and x+1=1312 and 13 are the integersx+13=0x=-13x+1=-12-12 and -13 is also a solutionusing arithmetic...√156=12.48999

let x and x+1 be the integersx(x+1)=156x^2+x=156x^2+x-156=0156=2*2*3*13(2*2*3)*13=12*13 and 12*13=156factor(x+13)(x-12)=156x-12=0x=12 and x+1=1312 and 13 are the integersx+13=0x=-13x+1=-12-12 and -13 is also a solutionusing arithmetic...√156=12.48999truncate the number to get 12 and then 13 and then -12 and -13

Step-by-step explanation:

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