Question

If a = 2, b = =2, find the value of ${{a}^{2}}+{{b}^{2}}$ .

Answer 1

Hint: The given problem is related to the square of a number. The square of a number is defined as the product of a number with itself.

Complete step by step answer:
Before proceeding with the solution, we must understand the concept of the square of a number. The square of a number is defined as the product of a number with itself. The square of a number is always positive. Except for 1 and 0, the square of a number can never be equal to the number itself. In case of 1 and 0, the square is equal to the number, i.e. ${{1}^{2}}=1$ and ${{0}^{2}}=0$ . Now, in the given question, we are given the values of a and b as 2 and -2 respectively. We are asked to find the value of ${{a}^{2}}+{{b}^{2}}$ . First, we will find
 the value of ${{a}^{2}}$ . We know, the square of a number is the product of the number with itself. So, ${{a}^{2}}=a\times a$ . We are given a = 2. So, ${{a}^{2}}=2\times 2=4$ . Now, we will find the value of ${{b}^{2}}$ . By the definition of square of a number, we can say that the value of ${{b}^{2}}$ is equal to $b\times b$ . So, ${{b}^{2}}=\left( -2 \right)\times \left( -2 \right)=4$ . Now, we will calculate the value of ${{a}^{2}}+{{b}^{2}}$. The value of ${{a}^{2}}+{{b}^{2}}$ is equal to 4 + 4 = 8. So, ${{a}^{2}}+{{b}^{2}}$ = 8.

Note: Alternate Solution: We can solve the given problem using algebraic identities. We know, ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ . So, we can say ${{a}^{2}}+{{b}^{2}}={{\left( a-b \right)}^{2}}+2ab$ . On substituting a= 2 and b = -2 in the right-hand side of the equation, we get ${{a}^{2}}+{{b}^{2}}={{\left( 2-\left( -2 \right) \right)}^{2}}+\left( 2\times 2\times \left( -2 \right) \right)$ .
$\Rightarrow {{a}^{2}}+{{b}^{2}}={{\left( 4 \right)}^{2}}-8$
$\Rightarrow {{a}^{2}}+{{b}^{2}}=16-8=8$
Hence, the value of ${{a}^{2}}+{{b}^{2}}$ , when a = 2, and b = -2, is equal to 8.