## 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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Mathematics
2 months
2021-07-27T22:45:37+00:00
2021-07-27T22:45:37+00:00 2 Answers
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## Answers ( )

Answer: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=+13x–11xby 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=0x=>–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 :

x–11=0x=>11we have taken the second odd number as

x+2x+2=11+2=13So,thecorrectanswersare11and13.Answer: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.