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Chapter 10: Problem 24

What happens to the radical terms whenever we multiply \((a+b)(a-b)\) where the binomials contain square roots?

Short Answer

Expert verified
When multiplying binomials \((a+b)\) and \((a-b)\) containing square roots, the radical terms cancel out, resulting in a difference of squares expression: \(a^2 - b^2\).

Step by step solution

01

Recognize the expressions involved

We have two binomials, \((a+b)\) and \((a-b)\). Notice that \(a\) and \(b\) can contain square roots (radical terms).
02

Use distributive property to expand the expression

Apply the distributive property to multiply the two expressions: \((a+b)(a-b) = a(a-b) + b(a-b)\).
03

Further distribute

Now, distribute \(a\) and \(b\) in each term from the previous step: \begin{align*} a(a-b) + b(a-b) &= a\cdot a - a\cdot b + b\cdot a - b\cdot b \\ \end{align*}
04

Use the associative property of addition

Rearrange the terms to group the square root terms together using the associative property of addition: \begin{align*} a\cdot a - a\cdot b + b\cdot a - b\cdot b &= a\cdot a - b\cdot b + (b\cdot a - a\cdot b) \end{align*}
05

Observe the resulting expression

We can observe that the sum of the square root terms ended up in the expression \((b\cdot a - a\cdot b)\). Since addition is commutative, this is equivalent to \(0\), and we are left with: \begin{align*} a\cdot a - b\cdot b &= a^2 - b^2 \end{align*} So when multiplying binomials \((a+b)\) and \((a-b)\) containing square roots, the resulting expression is a difference of squares with no remaining square root terms.

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