Certainly! Let's solve each part of the question step by step.
### Part (a): Calculate [tex]\(19 \times 21\)[/tex]
Step 1: Write down the numbers to be multiplied:
[tex]\[ 19 \times 21 \][/tex]
Step 2: Perform the multiplication:
[tex]\[ 19 \times 21 = 399 \][/tex]
Therefore, the product of [tex]\(19\)[/tex] and [tex]\(21\)[/tex] is:
[tex]\[ 399 \][/tex]
### Part (b): If [tex]\( x + y = 5 \)[/tex] and [tex]\( xy = 3 \)[/tex], find the value of [tex]\( x^2 + y^2 \)[/tex]
Step 1: Recall the given expressions:
[tex]\[ x + y = 5 \][/tex]
[tex]\[ xy = 3 \][/tex]
Step 2: Use the identity for the square of a binomial:
[tex]\[ (x + y)^2 = x^2 + y^2 + 2xy \][/tex]
Step 3: Plug in the given values into the identity:
[tex]\[ (x + y)^2 = 5^2 = 25 \][/tex]
[tex]\[ xy = 3 \][/tex]
Step 4: Rearrange the identity to solve for [tex]\( x^2 + y^2 \)[/tex]:
[tex]\[ x^2 + y^2 = (x + y)^2 - 2xy \][/tex]
[tex]\[ x^2 + y^2 = 25 - 2 \cdot 3 \][/tex]
[tex]\[ x^2 + y^2 = 25 - 6 \][/tex]
[tex]\[ x^2 + y^2 = 19 \][/tex]
Therefore, the value of [tex]\( x^2 + y^2 \)[/tex] is:
[tex]\[ 19 \][/tex]
### Summary:
- The product of [tex]\( 19 \times 21 \)[/tex] is [tex]\( 399 \)[/tex].
- Given [tex]\( x + y = 5 \)[/tex] and [tex]\( xy = 3 \)[/tex], the value of [tex]\( x^2 + y^2 \)[/tex] is [tex]\( 19 \)[/tex].
