RBQE - Simplify Radical Expressions (Lesson)

Simplify Radical Expressions

Let's think of some common perfect squares:

x	1	2	3	4	5	6	7	8	9	10	11	12
x²	1	4	9	16	25	36	49	64	81	100	121	144

Examples

$LaTeX: \sqrt[2]{25}=5$ $\sqrt[2]{25}=5$
$LaTeX: \sqrt[2]{121}=11$ $\sqrt[2]{121}=11$
$LaTeX: \sqrt[2]{-16}=\:does\:not\:exist$ $\sqrt[2]{-16}=\:does\:not\:exist$

$LaTeX: \sqrt[2]{-16}$ $\sqrt[2]{-16}$ does not have an answer because 4² = 16 and (-4)² = 16, there is no number that you can multiply by itself to get -16.

Product Property is square root of a times b equals the square root of a times the square root of b. The quotient property is the square root of a divided by b which equals the square root of a divided by the square root of b.

Examples

$LaTeX: \sqrt[2]{\frac{64}{100}}=\frac{\sqrt[2]{64}}{\sqrt[2]{100}}=\frac{8}{10}=\frac{4}{5}$ $\sqrt[2]{\frac{64}{100}}=\frac{\sqrt[2]{64}}{\sqrt[2]{100}}=\frac{8}{10}=\frac{4}{5}$
$LaTeX: \sqrt[2]{18}\cdot\sqrt[2]{2}=\sqrt[2]{36}=6$ $\sqrt[2]{18}\cdot\sqrt[2]{2}=\sqrt[2]{36}=6$
$LaTeX: \sqrt[2]{\frac{100}{25}}=\sqrt[2]{4}=2$ $\sqrt[2]{\frac{100}{25}}=\sqrt[2]{4}=2$

Let's look at some examples that are not perfect squares:

Example

The diagonal of a square is 20 in, what are the side lengths?

Step One: Draw a picture & label everything you know!

a square with a diagonal of 20 in.

Step Two: Use the Pythagorean theorem to set up an equation & solve:

$LaTeX: a^2+b^2=c^2\\x^2+x^2=\left(20\right)^2\\2x^2=400\\x^2=200\\\:x=\pm\sqrt[2]{200}\\x=\pm\sqrt[2]{100}\cdot\sqrt[2]{2}\\x=\pm10\sqrt[2]{2}$ $a^2+b^2=c^2\\x^2+x^2=\left(20\right)^2\\2x^2=400\\x^2=200\\\:x=\pm\sqrt[2]{200}\\x=\pm\sqrt[2]{100}\cdot\sqrt[2]{2}\\x=\pm10\sqrt[2]{2}$

The answer must be positive because you can not have a negative length!

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