
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
Answer
615.6k+ views
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’.
Complete step-by-step answer:
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 $xy=\dfrac{122}{2}=61$.
Hence, the value of ‘xy’ is 61, which is option (b).
An algebraic identity is an equality that holds for all possible values of its variables. We can prove each of the identities by performing basic algebraic operations such as addition, multiplication, subtraction, and division. They are used for the factorization of the polynomials. That’s why they are useful in the computation of algebraic expressions. An algebraic expression differs from an algebraic identity in the way that the value of algebraic expression changes with the change in variables. However, an algebraic identity is equality which holds for all possible values of variables.
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.
Complete step-by-step answer:
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 $xy=\dfrac{122}{2}=61$.
Hence, the value of ‘xy’ is 61, which is option (b).
An algebraic identity is an equality that holds for all possible values of its variables. We can prove each of the identities by performing basic algebraic operations such as addition, multiplication, subtraction, and division. They are used for the factorization of the polynomials. That’s why they are useful in the computation of algebraic expressions. An algebraic expression differs from an algebraic identity in the way that the value of algebraic expression changes with the change in variables. However, an algebraic identity is equality which holds for all possible values of variables.
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.
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