CA Unit 3 Multiplying Radicals

Front

What is the product of two radicals, \(\sqrt{a} \cdot \sqrt{b}\)?

Back

The product of two radicals is \(\sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b}\).

Front

How do you simplify the expression \(\sqrt{a} \cdot \sqrt{b} \cdot \sqrt{c}\)?

Back

You can simplify it as \(\sqrt{a} \cdot \sqrt{b} \cdot \sqrt{c} = \sqrt{a \cdot b \cdot c}\).

Front

What is the distributive property in the context of multiplying binomials?

Back

The distributive property states that \(a(b + c) = ab + ac\). This applies to multiplying binomials as well.

Tags

CCSS.HSA.APR.A.1

Front

How do you multiply two binomials that contain radicals, such as \((a + b)(c + d)\)?

Back

You apply the distributive property: \((a + b)(c + d) = ac + ad + bc + bd\).

Front

What is the result of multiplying \(\left(4\sqrt{2}-2\sqrt{3}\right)\left(5\sqrt{2}-\sqrt{3}\right)?

Back

The result is \(46 - 14\sqrt{6}\).

Tags

CCSS.HSN.RN.A.2

Front

How do you simplify the expression \(-2\sqrt{10} \left(\sqrt{3} - 4\sqrt{2}\right)?

Back

Distributing gives \(-2\sqrt{10} \cdot \sqrt{3} + 8\sqrt{20} = -2\sqrt{30} + 16\sqrt{5}\).

Tags

CCSS.HSN.RN.A.2

Front

What is the product of \(-2\sqrt{7}\) and \(3\sqrt{28}\)?

Back

The product is \(-84\).

Front

How do you simplify \(\sqrt{15} \left(3\sqrt{5} - 2\sqrt{3}\right)?

Back

Distributing gives \(3\sqrt{15 \cdot 5} - 2\sqrt{15 \cdot 3} = 15\sqrt{3} - 6\sqrt{5}\).

Tags

CCSS.HSN.RN.A.2

Front

What is the result of \(\sqrt{30} \cdot \sqrt{20}\)?

Back

The result is \(10\sqrt{6}\).

Front

What is the general formula for multiplying two radicals?

Back

The general formula is \(\sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b}\).