Question

Three consecutive positive integers are taken such that the sum of the square of the first and the product of the other two is 154. Find the integers.​

Answer 1

x,  x+1  and x+2

According to question

$x^2+(x+1)(x+2)=154\\\\x^2+x(x+2)+1(x+2)=154\\\\x^2+x^2+2x+x+2=154\\\\2x^2+2x+x+2=154\\\\2x^2+3x+2=154\\\\2x^2+3x+2-154=0\\\\2x^2+3x-152=0\\\\2x^2+19x-16x-152=0\\\\x(2x+19)-8(2x+19)=0\\\\(2x+19)\:(x-8)\\\\\\2x+19=0\\\\2x=-19\\\\x=-\dfrac{19}{2}\:\bigg(x\neq\dfrac{-19}{2}\bigg)\\\\\\x-8=0\\\\x=8\:\bigg(x=8\bigg)$

Here we get 'x' = 8

Now we put the value of 'x' we get,

First consecutive = x = 8

Second consecutive = x + 1 = 8 + 1 = 9

Third consecutive = x + 2 = 8 + 2 = 10

Therefore three consecutive positive integers are 8, 9 and 10

Answer 2

$\bf{\large{\underline{\underline{Answer:-}}}}$

Required integers are 8, 9, 10

$\bf{\large{\underline{\underline{Explanation:-}}}}$

Given :- Three consecutive integers are such that sum of square first an product of other two numbers is 154

To find :- Required integers

Solution :-

Let the three consecutive integers be x, (x + 1), (x + 3)

Square of first integer = x²

Product of other two integers = (x + 1)(x + 2) = x² + (1 + 2)x + 1(2)

[Since (x + a)(x + b) = x² + (a + b)x + ab]

= x² + 3x + 2

Product of other two integers = x² + 3x + 2

Sum of square of first integer and product of other two integers = 154

According to the question :-

Equation formed :-

${x}^{2} + ( {x}^{2} + 3x + 2) = 154$

${x}^{2} + {x}^{2} + 3x + 2 = 154$

$2{x}^{2} + 3x + 2 = 154$

$2 {x}^{2} + 3x = 154 - 2$

$2 {x}^{2} + 3x = 152$

$2 {x}^{2} + 3x = 152$

$2 {x}^{2} + 3x - 152 = 0$

By splitting the middle term :-

$2 {x}^{2} + 19x - 16x - 152 = 0$

$x(2x + 19) - 8(2x + 19) = 0$

$(2x + 19)(x -8) = 0$

Now equate the products of the polynomial to 0

$2x + 19 = 0$

$2x = - 19$

$x = \dfrac{ - 19}{2}$

Given that numbers are positive integers. But here it is rational number. So, -19/2 is not possible.

$x - 8 = 0$

$x = 8$

Therefore first integer = x = 8

Second integer = (x + 1) = (8 + 1) = 9

Third integer = (x + 2) = (8 + 2) = 10

$\bf{\large{\underline{\underline{Verification:-}}}}$

${8}^{2} + 9(10) = 154$

$64 + 90 = 154$

$154 = 154$

So, required integers are 8, 9, 10