Question

the sum of the squares of two consecutive odd positive integers is 290 find them.

Answer 1

we can let the 2 consecutive odd number be x and x+2

Acc,
⇒x²+(x+2)²=290
⇒x²+x²+4x+4=290
⇒2x²+4x+4=290
⇒2x²+4x=290-4
⇒2x²+4x=286
⇒x²+2x=143[Taking 2 in common]
⇒x²+2x-143=0
⇒x²+13x-11x-143=0
⇒x(x+13)-11(x+13)=0
⇒(x+13)(x-11)
=========================================
x+13=0
x=-13
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x-11=0
x=11
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As 11 is a posite number we are taking the value of x to be 11
⇒(x+2)=11+2=13
=========================================
Hence the two consecutive odd positive integers are 11 and 13.

Answer 2

Hey there!

Let the first number be x,
then the consecutive odd number is x+2.

Given

➫x²+(x+2)²=290

➫x²+x²+4+4x=290

➫2x²+4x+4=290

➫2(x²+2x+2)=290

➫x²+2x+2=145

➫x²+2x+2-145=0

➫ x²+2x-143=0

➫x²+13x-11x-143=0

➫x(x+13)-11(x+13)=0

➫(x+13)(x-11)=0

➫x≠-13,x=11

So, x+2=13

11,13 are the numbers whose sum of squares is 290

11²+13²=121+169=290

hope helped!