Question

(1) Find three successive even natural numbers, the sum of whose squares is 308.
(2) Find three consecutive odd integers, the sum of whose squares is 83.​

Answer 1

1)

let the first number be 'x'

then the other two numbers will become (x+2) and (x+4).

the sum of the squares of these numbers will be equal to 308.

x^2+(x+2)^2+(x+4)^2=308 (a+b)^2= a^2+b^2+2ab

x^2+x^2+4x+4+x^2+8x+16=308

3x^2+12x+20=308

3x^2+12x=288

3x(x+4)=288

now we will use hid and trial method to find the value of x

first we will put value of x as 1 and then solve; the answer will not be equal to 288 then we will take value of x as 2 and then solve and then also the answer will not be equal to 288. if we keep doing like this then will we reach 8 and when we put value of x as 8

3×8(8+4)

3×8×12

=288

lhs=rhs

hence the value of x will be 8 and so the three consecutive even numbers will become 8,10,12.

2)

let the integer be 'x'

then the other two integers will become

(x+2) and (x+4)

then the sum of squares of these number will become 83.

x^2+(x+2)^2+(x+4)^2 (a+b)^2= a^2+b^2+2ab

x^2+x^2+4x+4+x^2+8x+16=83

3x^2+12x+20=83

3x(x+4)=63

now we will use the same method as we used in question first and will put values of x as 1 then 2 then 3 and so on until we get the desired answer which us 63 in this question.

SO, in this question when we will put the value of x as 3 then we will get

3×3(3+4)

3×3×7

=63

lhs=rhs

so the value of x is 3 and the other two numbers will be 5 and 7 respectively.

Answer 2

Step-by-step explanation:

1)

let the first number be 'x'

then the other two numbers will become (x+2) and (x+4).

the sum of the squares of these numbers will be equal to 308.

x^2+(x+2)^2+(x+4)^2=308 (a+b)^2= a^2+b^2+2ab

x^2+x^2+4x+4+x^2+8x+16=308

3x^2+12x+20=308

3x^2+12x=288

3x(x+4)=288

now we will use hid and trial method to find the value of x

first we will put value of x as 1 and then solve; the answer will not be equal to 288 then we will take value of x as 2 and then solve and then also the answer will not be equal to 288. if we keep doing like this then will we reach 8 and when we put value of x as 8

3×8(8+4)

3×8×12

=288

lhs=rhs

hence the value of x will be 8 and so the three consecutive even numbers will become 8,10,12.

2)

let the integer be 'x'

then the other two integers will become

(x+2) and (x+4)

then the sum of squares of these number will become 83.

x^2+(x+2)^2+(x+4)^2 (a+b)^2= a^2+b^2+2ab

x^2+x^2+4x+4+x^2+8x+16=83

3x^2+12x+20=83

3x(x+4)=63

now we will use the same method as we used in question first and will put values of x as 1 then 2 then 3 and so on until we get the desired answer which us 63 in this question.

SO, in this question when we will put the value of x as 3 then we will get

3×3(3+4)

3×3×7

=63

lhs=rhs

so the value of x is 3 and the other two numbers will be 5 and 7 respectively.