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Question Number 217321 by ArshadS last updated on 10/Mar/25

Find three consecutive integers such that the product of the first two is 16 more than twice the third integer. Provide all possible solutions.

Findthreeconsecutiveintegerssuchthattheproductofthefirsttwois16morethantwicethethirdinteger.Provideallpossiblesolutions.

Answered by Rasheed.Sindhi last updated on 10/Mar/25

let x−1,x,x+1 are required numbers According to given: x(x−1)=2(x+1)+16 x^2 −x=2x+2+16 x^2 −3x−18=0 x^2 −6x+3x−18=0 x(x−6)+3(x−6)=0 (x−6)(x+3)=0 x=6,−3 x=6 : 6−1^(x−1) ,6^(x) ,6+1^(x+1) 5,6,7 x=−3 : −3−1^(x−1) ,−3^(x) ,−3+1^(x+1) −4,−3,−2

letx−1,x,x+1arerequirednumbersAccordingtogiven:x(x−1)=2(x+1)+16x2−x=2x+2+16x2−3x−18=0x2−6x+3x−18=0x(x−6)+3(x−6)=0(x−6)(x+3)=0x=6,−3x=6:6−1x−1,6x,6+1x+15,6,7x=−3:−3−1x−1,−3x,−3+1x+1−4,−3,−2

Commented by ArshadS last updated on 10/Mar/25

Right!

Right!

Answered by mehdee7396 last updated on 10/Mar/25

a(a+1)=2a+4+16 a^2 −a−20=0 (a−5)(a+4)=0⇒a=5 or a=−4 1)5,6,7 & 2)−4,−3,−2

a(a+1)=2a+4+16a2−a−20=0(a−5)(a+4)=0⇒a=5ora=−41)5,6,7&2)−4,−3,−2

Commented by ArshadS last updated on 10/Mar/25

Ok, thanks sir!

Ok,thankssir!

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