Question

three consecutive integers are such that the sum of the square of the smallest number and the product of the other two is 277. find the three integers

Answer 1

Let the numbers be x, x+1 and x+2.
We have -
${x}^{2} + (x + 1)(x + 2) = 277$
${x}^{2} + ( {x}^{2} + 2x + x + 2) = 277$
$2 {x}^{2} + 3x + 2 = 277$
2(x^2)+3x-275=0
2(x^2)-22x+25x-275=0
2x(x-11)+25(x-11)=0
(x-11)(2x+25)=0
x=11,x=-12.5
On solving, we get, x=-12.5 (np) or x=11
So, the 3 numbers are 11, 12, 13.