Question

If \[x + y = \dfrac{7}{2}\] and \[xy = \dfrac{5}{2}\], find the value of:
(a) \[x - y\]
(b) \[{x^2} - {y^2}\]

Answer 1

Hint: Here, we have to find the value of \[x - y\] and \[{x^2} - {y^2}\]. We will use the algebraic identity for the sum of two numbers to find the sum of squares of \[x\] and \[y\]. Then, using the algebraic identity for the square of difference of two numbers, we will find the value of \[x - y\]. Finally, using the algebraic identity for the product of the sum and difference of two numbers, we will find the value of \[{x^2} - {y^2}\].

Formula Used: The square of the sum of two numbers \[a\] and \[b\] is given by the algebraic identity \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\].
The square of the difference of two numbers \[a\] and \[b\] is given by the algebraic identity \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\].
The product of the sum and difference of two numbers \[a\] and \[b\] is given by the algebraic identity \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\].

Complete step-by-step answer:
(a)
We know that the square of the sum of two numbers \[a\] and \[b\] is given by the algebraic identity \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\].
Using the algebraic identity, we get the square of the sum of the numbers \[x\] and \[y\] as
\[{\left( {x + y} \right)^2} = {x^2} + {y^2} + 2xy\]
Substituting \[x + y = \dfrac{7}{2}\] and \[xy = \dfrac{5}{2}\] in the equation, we get
\[ \Rightarrow {\left( {\dfrac{7}{2}} \right)^2} = {x^2} + {y^2} + 2\left( {\dfrac{5}{2}} \right)\]
Simplifying the expression, we get
\[ \Rightarrow \dfrac{{49}}{4} = {x^2} + {y^2} + 5\]
Subtracting 5 from both sides, we get
\[\begin{array}{l} \Rightarrow \dfrac{{49}}{4} - 5 = {x^2} + {y^2} + 5 - 5\\ \Rightarrow \dfrac{{49 - 20}}{4} = {x^2} + {y^2}\\ \Rightarrow \dfrac{{29}}{4} = {x^2} + {y^2}\end{array}\]
Using the algebraic identity, we get the square of the difference of the numbers \[x\] and \[y\] as
\[{\left( {x - y} \right)^2} = {x^2} + {y^2} - 2xy\]
Substituting \[{x^2} + {y^2} = \dfrac{{29}}{4}\] and \[xy = \dfrac{5}{2}\] in the equation, we get
\[ \Rightarrow {\left( {x - y} \right)^2} = \dfrac{{29}}{4} - 2\left( {\dfrac{5}{2}} \right)\]
Simplifying the equation, we get
\[ \Rightarrow {\left( {x - y} \right)^2} = \dfrac{{29}}{4} - 5\]
Subtracting the terms of the expression, we get
\[\begin{array}{l} \Rightarrow {\left( {x - y} \right)^2} = \dfrac{{29 - 20}}{4}\\ \Rightarrow {\left( {x - y} \right)^2} = \dfrac{9}{4}\end{array}\]
Applying the square root of both sides, we get
\[ \Rightarrow x - y = \sqrt {\dfrac{9}{4}} = \dfrac{3}{2}\]
Therefore, the value of \[x - y\] is \[\dfrac{3}{2}\].
(b)
We will use the algebraic identity for the product of the sum and difference of two numbers to find the value of \[{x^2} - {y^2}\].
We know that the product of the sum and difference of two numbers \[a\] and \[b\] is given by the algebraic identity \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\].
Using the algebraic identity, we get the product of the sum and difference of the numbers \[x\] and \[y\] as
\[{x^2} - {y^2} = \left( {x + y} \right)\left( {x - y} \right)\]
Substituting \[x + y = \dfrac{7}{2}\] and \[x - y = \dfrac{3}{2}\] in the equation, we get
\[ \Rightarrow {x^2} - {y^2} = \left( {\dfrac{7}{2}} \right)\left( {\dfrac{3}{2}} \right)\]
Multiplying the terms of the expression, we get
\[ \Rightarrow {x^2} - {y^2} = \dfrac{{21}}{4}\]
Therefore, the value of \[{x^2} - {y^2}\] is \[\dfrac{{21}}{4}\].

Note: We have assumed that \[x > y\]. The answer will be different if \[x < y\].
Taking the square root of both sides \[{\left( {x - y} \right)^2} = \dfrac{9}{4}\], we get
\[\begin{array}{l} \Rightarrow x - y = - \sqrt {\dfrac{9}{4}} \\ \Rightarrow x - y = - \dfrac{3}{2}\end{array}\]
The negative sign is taken because it is assumed that \[x < y\].
Substituting \[x + y = \dfrac{7}{2}\] and \[x - y = - \dfrac{3}{2}\] in the equation \[{x^2} - {y^2} = \left( {x + y} \right)\left( {x - y} \right)\], we get
\[ \Rightarrow {x^2} - {y^2} = \left( {\dfrac{7}{2}} \right)\left( { - \dfrac{3}{2}} \right)\]
Multiplying the terms of the expression, we get
\[ \Rightarrow {x^2} - {y^2} = - \dfrac{{21}}{4}\]
Therefore, if \[x < y\], then the values of \[x - y\] and \[{x^2} - {y^2}\] will be \[ - \dfrac{3}{2}\] and \[ - \dfrac{{21}}{4}\] respectively.