Question 2134: 1) find two consecutive integers such that the sum of the first and the square of the second is 19
2)Find 3 consecutive integers such that the square of the first, decreased by 5 times the second is 9
Answer by sehar(13)   (Show Source):
You can put this solution on YOUR website! solution:
let the first no. be 'x', then the 2nd no. will be x+1.
now,according to question,
x+(x+1)^2=19
x+x^2+2x+1=19
x^2+3x-18=0
x^2+6x-3x-18=0
x(x+6)-3(x+6)=0
(x-3)(x+6)=0
x=3, x=-6
if x=3,then x+1=4
if x=-6 then x+1=-5
solution of 2:
let first no. 'x', the second 'x+1' & third 'x+2'
now,
x^2-5(x+1)=9
x^2-5x-5-9=0
x^2-5x-14=0
x^2-7x+2x-14=0
x(x-7)+2(x-7)=0
(x-7)(x+2)=0
so x=7, x=-2
so three consecutive nos. r 7,8,9 or -2,-1,0.
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