Study the example below for the product of conjugate pairs.
[tex]\begin{array}{l}
(3+\sqrt{7})(3-\sqrt{7}) \\
=9-3 \sqrt{7}+3 \sqrt{7}-7 \\
=9-7 \\
=2
\end{array}[/tex]
Now, find the product [tex]$(2+\sqrt{5})(2-\sqrt{5})$[/tex].
The product is $\square$.
Answer (2)
The product of the conjugate pair ( 2 + s q r t 5 ) ( 2 − s q r t 5 ) is − 1 . We multiplied the terms using the distributive property, simplified, and combined like terms to reach this result.
;
Multiply the conjugate pair ( 2 + 5 ) ( 2 − 5 ) using the distributive property.
Simplify the expression: 4 − 2 5 + 2 5 − 5 .
Combine like terms, canceling the square root terms: 4 − 5 .
Calculate the final result: − 1 .
Explanation
Understanding the Problem We are asked to find the product of the conjugate pair ( 2 + 5 ) ( 2 − 5 ) . Let's recall what happens when we multiply conjugate pairs.
Multiplying the Terms To find the product, we use the distributive property (also known as the FOIL method): ( 2 + 5 ) ( 2 − 5 ) = 2 ( 2 ) + 2 ( − 5 ) + 5 ( 2 ) + 5 ( − 5 )
Simplifying the Expression Now, let's simplify each term:
2 ( 2 ) = 4 2 ( − 5 ) = − 2 5 5 ( 2 ) = 2 5 5 ( − 5 ) = − 5
So, we have: 4 − 2 5 + 2 5 − 5
Combining Like Terms Notice that the terms − 2 5 and 2 5 cancel each other out: 4 − 2 5 + 2 5 − 5 = 4 − 5
Final Calculation Finally, subtract the numbers: 4 − 5 = − 1 Thus, the product of ( 2 + 5 ) ( 2 − 5 ) is − 1 .
Examples
Conjugate pairs are useful in various areas of mathematics, such as simplifying expressions, rationalizing denominators, and solving equations. For instance, in electrical engineering, when dealing with complex impedances, conjugate pairs are used to find the power factor and perform impedance matching to maximize power transfer. Imagine you're designing an audio amplifier and need to ensure the speaker receives the maximum power output from the amplifier. By using conjugate matching, you can minimize signal reflections and ensure efficient energy transfer, leading to better sound quality and performance.
