Algebra Honors (Period 6 & 7)

Multiplication of Binomials Containing Radicals 11-9

Chapter 5 taught us how to multiply binomials-- we can use those methods when multiplying binomials that contain square root radicals.
(6 + √11)(6 - √11)
The pattern is
(a +b)(a -b) = a² - b²
so using that we get
6² - (√11)²
36 - 11 = 25

Simplify (3 + √5)²
The pattern here is
(a + b)² = a² + 2ab + b²
so ( 3 + √5)² =
3² + 2[(3)(√5)] + (√5)² =
9 + 6√5 + 5 =
14 + 6√5

Simplify (2 √3 - 5√7)²
The pattern here is (a - b)² = a² - 2ab + b²

(2 √3 - 5√7)² =
(2 √3)² -2[(2)(5)(√3)(√7)] +(5√7)² =
4(3) -20√21 +25(7) =
12 -20√21+ 175 =
187 -20√21

If both b and d are nonnegative, then the binomials
a√b + c√d AND a√b - c√d are called conjugates of one another. COnjugates differ ONLY in the sign of one term

if a, b, c, and d are all integers then the product (a√b + c√d)(a√b - c√d) will be an integer... see the first example!!

Conjugates can be used to rationalize binomial denominators that contain radicals.. getting rid of the radicals in the denominator

Rationalize
3/(5- 2√7)

3/(5- 2√7) = [ 3/(5- 2√7)] × [((5+ 2√7)/(5+2√7)]
This doesn't show well here hopefully you can remember what was done in class...
= 3(5 +2√7)/25-(2√7)² =
(15+6√7)/25-28 =
(15+6√7)/-3 =
15/-3 +6√7/-3 =
-5 -2√7



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