Question

three consecutive positive integer are such that the sum of square of the first and product of other two is 46 find the integer

Answer 1

Let the numbers are x,x+1,x+2.
As per the question,
x^2 +(x+1)(x+2) = 46.
i.e. 2x^2 +3x-44 = 0.
Now, By applying quadratic formula, you will get
x= -22/4, 4.
But x can't be negative.
So, Numbers will be 4,5 and 6.

Answer 2

The number 4, 5 and 6 are three consecutive positive integer.

Solution:

Let the first number be x, then the other numbers will be (x+1),(x+2)

According to question:-

$\begin{array}{l}{x^{2}+(x+1)(x+2)=46} \\ \\{x^{2}+x^{2}+2 x+x+2=46}\end{array}$

On simplifying above equation we get, $2 x^{2}+3 x-44 = 0$

$2 x^{2}+11 x-8 x-44=0$

x(2x+11)-4 (2x+11)=0

(x-4)(2x+11)

On solving we will get $x=-\frac{11}{2}, 4$

Since x can’t be negative.

Now substituting value of x in x+1 and x+2 we get

x = 4 and x=6

So numbers will be 4, 5 and 6.