 QUESTION

# If $x+y=17$ and ${{x}^{2}}+{{y}^{2}}=167$, what is the value of ‘xy’?(a) $17+4\sqrt{114}$ (b) 61(c) $48+\sqrt{167}$ (d) 122

Hint: Square the first equation and simplify it using the algebraic identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$. Substitute the values given in the second equation and rearrange the terms to calculate the value of ‘xy’.

We know that $x+y=17$ and ${{x}^{2}}+{{y}^{2}}=167$. We have to calculate the value of ‘xy’.
We will solve this question using algebraic identities. We will square the equation $x+y=17$ on both sides.
Thus, we have ${{\left( x+y \right)}^{2}}={{\left( 17 \right)}^{2}}=289.....\left( 1 \right)$.
We know the algebraic identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$.
Thus, we can rewrite equation (1) as ${{\left( x+y \right)}^{2}}={{x}^{2}}+{{y}^{2}}+2xy=289.....\left( 2 \right)$.
We know that ${{x}^{2}}+{{y}^{2}}=167$. Substituting this equation in equation (2), we have $167+2xy=289$.
Simplifying the above equation by rearranging the terms, we have $2xy=289-167=122$.
Dividing the above equation by 2 on both sides, we have $xy=\dfrac{122}{2}=61$.
Note: We can’t solve this question without using the algebraic identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ as it’s difficult to calculate the exact value of variables ‘x’ and ‘y’. We can verify an algebraic identity using the substitution method. In this method, we substitute the values for the variables and perform an arithmetic operation.