Question

Each pa
total are
3. Two consecutive positive numbers are such that
the sum of their squares is 113. Find the two numbers.
perimet​

Answer 1

Step-by-step explanation:

let ist no.= x

other no. = x+1

ATQ

${ x}^{2} + {(x + 1)}^{2} = 113 \\ {x}^{2} + {x}^{2} + 1 + 2x = 113 \\ 2 {x}^{2} + 2x - 112 = 0 \\ {x}^{2} + x - 56 = 0 \\ {x}^{2} - 7x + 8x - 56 = 0 \\ x(x - 7) + 8(x - 7) = 0 \\ (x + 8)(x - 7) = 0 \\ x = - 8 \: and 7$

since no.are positive

therefore x=-8 is rejected

ist no.=7

other no.=8

Answer 2

Given :-

Sum of two positive consecutive numbers = 113

To Find :-

The first number.

The second number.

Analysis :-

Consider the two numbers as any 2 consecutive variables.

Make an equation to get the value of the variable.

Substitute the value of the variable in the 2 numbers and solve it.

Solution :-

Let us consider the two consecutive numbers as 'x' and 'x + 1'

Given that, sum of two consecutive numbers = 113

Making an equation,

$\sf x^{2} + (x+1)^{2}=113$

Opening the brackets,

$\sf x^{2}+x^{2}+2x+1=113$

$\sf 2x^{2}+2x-112=0$

$\sf x^{2}+x-56=0$

$\sf (x+8)(x-7)=0$

$\sf x=-8 \ or \ 7$

The numbers cannot be negative,

Therefore, the two consecutive numbers are 7 and 8.