## find two consecutive odd positive integers ,sum of whose squares is 290​

Question

find two consecutive odd positive integers ,sum of whose squares is 290​

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2 months 2021-07-27T22:45:37+00:00 2 Answers 2 views 0

11 and 13

Step-by-step explanation:

let the two consecutive odd prime integers be = x and x+2

the sum of both the numbers are 290

so , x^2+(x+2)^2=290

value of x => (x+2)^2=x^2+2x+4

=> x^2+x^2+2x+4=290

=> 2x^2+4x-286=0

=> x^2+2x-143=0

We need to find two numbers whose sum is 2 and product is -143

there are two numbers +13 and -11 whose sum is 2 and product is -143

therefore , 2x=+13x11x

by putting this in the quadratic equation ,

=> x^2+2x-143=0

=> x^2+13x-11x-143=0

=> x(x+13)-11(x-13)=0

=> (x+13)(x-11)=0

first number is : x+13=0

x=> 13

-13 is a negative odd number and we are asked to find two consecutive positive odd numbers therefore , -13 Isn’t our answer

second number is : x11=0

x=> 11

we have taken the second odd number as x+2

x+2=11+2=13

So , the correct answers are 11 and 13 .

let the 2 numbers be x&(x+2)

Step-by-step explanation:

the sum of the squares of numbers=290

x^2+(x+2)^2=290

x^2+x^2+2^2+2×2×x=290

2x^2+4+4x=290

x^2+2x+2=290/2

x^2+2x=145-2

x^2+2x-143=0

x^2-11x+13x-143=0

x(x-11) +13(x-11)=0

x-11=0 x+13=0

x=11 x=-13

the two consecutive numbers are 11 &13.