Question

Find two consecutive positive odd integers, sum of whose square is .

Answer 1

Hint: In this question many concepts are used like middle term split, factorization and identities. To solve these questions one should learn all the formulas of algebra.

Complete step-by-step answer:
Let first consecutive odd positive integer \[=x\]
Let second consecutive odd positive integer \[=x+2\]
Now, according to the question
\[\left( {{x}^{2}} \right)+{{\left( x+2 \right)}^{2}}=290\]
[using identity \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\]]
     ⇒\[{{x}^{2}}+{{\left( x \right)}^{2}}+{{\left( 2 \right)}^{2}}+2\times 2\times x=290\]
     ⇒ \[{{x}^{2}}+{{x}^{2}}+4+4x=290\]
     ⇒ \[2{{x}^{2}}+4x=290-4\]
     ⇒ \[2{{x}^{2}}+4x-286=0\]
Dividing by
     \[{{x}^{2}}+2x-143=0\]
Applying middle term split in above equation
     \[{{x}^{2}}+13x-11x-143=0\]
     \[x\left( x+13 \right)-11\left( x+13 \right)=0\]
     \[\left( x-11 \right)\left( x+13 \right)=0\]
     \[x=11,\,-13\]
Ans. \[=11,\,-13\]

Note: In this question, a lot of concepts are used like factorization, middle term split and odd positive integer. So go through the chapter factorization and then solve it.