Radical Conjugates
Use radical conjugates in Grade 10 algebra to rationalize denominators. Multiply by conjugate pairs to eliminate radicals from denominators using the difference of squares identity.
Key Concepts
If $a$, $b$, $c$, and $d$ are rational numbers, then $a\sqrt{b} + c\sqrt{d}$ and $a\sqrt{b} c\sqrt{d}$ are conjugates of each other, and their product is a rational number.
Example 1: The conjugate of $ 4 + \sqrt{3}$ is $ 4 \sqrt{3}$. Example 2: The conjugate of $\sqrt{5} + \sqrt{2}$ is $\sqrt{5} \sqrt{2}$. Example 3: $\frac{5}{\sqrt{6} + \sqrt{3}} = \frac{5(\sqrt{6} \sqrt{3})}{(\sqrt{6})^2 (\sqrt{3})^2} = \frac{5(\sqrt{6} \sqrt{3})}{3}$.
Radical conjugates are like secret agent twins! They look almost identical but have one opposite sign. When they multiply together, they use their special power—the difference of squares formula, $(a+b)(a b)=a^2 b^2$—to eliminate the radicals completely. This trick is your go to move for cleaning up binomials in the denominator, leaving behind a nice, rational number.
Common Questions
What are radical conjugates?
Radical conjugates are expressions of the form a√b + c√d and a√b - c√d. Their product a²b - c²d contains no radical, making them useful for rationalizing denominators.
How do you use conjugates to rationalize a denominator like 1/(√3 + √2)?
Multiply numerator and denominator by the conjugate (√3 - √2): (√3 - √2)/((√3)² - (√2)²) = (√3 - √2)/(3 - 2) = √3 - √2. The denominator becomes rational.
Why does multiplying by a conjugate eliminate the radical?
The product (a + b)(a - b) = a² - b². For radical conjugates, (√3 + √2)(√3 - √2) = 3 - 2 = 1. The cross terms cancel, leaving only rational numbers.