Simplifying Radical Expressions

Simplifying Radical Expressions exercise

• Determine the distance $c$.

  Hints

  You can use the Pythagorean theorem to determine the missing length.

  Don't forget to simplify your answer.

  You can simplify your answer using the product property of roots.

  Solution

  The distance the ships have traveled and the distance between them form a right angle. We can use the Pythagorean theorem to find the distance between the two ships. The distance between them is the hypotenuse. We have

  $a^2 + b^2 = c^2$.

  Substituting in known values,

  $(2x)^2 + (4x)^2 = c^2$,

  and simplifying gives

  $4x^2 + 16x^2 = c^2$

  $20x^2 = c^2$.

  Taking the root of both sides gives

  $\sqrt{20x^2} = c$.

  Now we have an expression for $c$, and we must simplify it as much as possible. We can do this using the product property of roots:

  $c = \sqrt{20x^2}$.

  We can start by separating the radicand (the term the root operator is applied to) into a product of two terms, one of which contains only perfect squares:

  $c = \sqrt{4x^2 \times 5}$.

  We know that the product property of roots states that

  $\sqrt{ab} = \sqrt{a} \times \sqrt{b}$.

  We can apply this to our solution, to arrive at

  $c = \sqrt{4x^2} \times \sqrt{5}$.

  We can then simplify this expression to a final answer:

  $c = 2x \times \sqrt{5}$.

• Explain how to multiply and divide radical expressions.

  Hints

  The name of the property includes either "product" or "quotient", which gives you a hint about what operation it involves.

  The properties don't use addition or subtraction.

  Solution

  The product property of roots is used to simplify terms that involve multiplication.

  The property states that the root of the product of two terms is equal to the product of the root of each term:

  $\sqrt{ab} = \sqrt{a} \times \sqrt{b}$.

  The quotient property of roots is used to simplify terms that involve division.

  The property states that the root of a rational number is equal to the root of the numerator divided by the root of the denominator:

  $\sqrt{\dfrac{a}{b}} = \dfrac{\sqrt{a}}{\sqrt{b}}$.

• Decide which radical expressions can be simplified.

  Hints

  Look for factor pairs of the radicand, where one pair is a perfect square.

  Keep in mind that $4$, $9$, $16$, $25$, and $x^2$ are all perfect squares.

  Solution

  All of these radical terms involve multiplication, not division. Therefore we should try to simplify each expression using the product property of roots. A radical term can be simplified using this property if its radicand has a factor that is also a perfect square.

  Let's look for factors of each radicand, keeping in mind that $4$, $9$, $16$, $25$, and $x^2$ are all perfect squares.

  For the first radical term, $\sqrt{32x^2}$, the radicand $32x^2$ has the factor $16x^2$. The square root of this factor is $4x$, so it is a perfect square. Therefore this radical term can be factored. We can prove this by factoring it:

  $~$

  $\sqrt{32x^2}$

  $= \sqrt{16x^2 \times 2}$

  $= \sqrt{16x^2} \times \sqrt{2}$

  $= 4x \times \sqrt{2}$ ✓

  $~$

  Similarly, for the other radical expressions...

  • $\sqrt{14}$ the radicand $14$ has no factors which are a perfect square. ✗

  • $\sqrt{18x}$ the radicand $18x$ has the factor $9$, which is a perfect square. ✓

  • $\sqrt{50}$ the radicand $50$ has the factor $25$, which is a perfect square. ✓

• Determine the missing length.

  Hints

  It may help to assign a variable to the side length of one of the squares.

  You can find the side length of the square using the area of the square, and the fact that the area is equal to the square of the side length.

  Don't forget to simplify radical terms.

  Solution

  Jack labels the side length of a square as $s$. He therefore knows that $x = 2s$.

  He can also say that the area of the square, $a$, is $a = s^2$.

  He knows that the area of the square is $32$ units, so he substitutes this in to get $32 = s^2$.

  He applies opposite operations to isolate $s$: $\sqrt{32} = s$.

  Now he must simplify the radical to find the value of $s$:

  $s = \sqrt{32}$

  $s = \sqrt{16 \times 2}$.

  Using the product property of roots, he gets

  $s = \sqrt{16} \times \sqrt{2}$.

  Simplifying gives

  $s = 4 \times \sqrt{2}$.

  Jack can now find the missing value $x$:

  $x = 2s$

  $x = 2 \times 4 \times \sqrt{2}$

  $x = 8 \sqrt{2}$.

• Label the parts of a radical expression.

  Hints

  Pay close attention to spelling.

  The number that indicates what root is being taken is called the index.

  Solution

  The small number is called the index. It indicates which root is being taken. It is equal to two in the case of a square root.

  The boxy shape that looks a bit like a check mark is called the radical.

  The large number, inside the radical, is called the radicand.

• Simplify the radical expressions.

  Hints

  The different expressions sometimes share common factors. It may help to start by simplifying each radical expression without looking at the answers.

  Each expression can be simplified using the product property of roots.

  Solution

  Each expression can be simplified using the product property of roots. Let's simplify each radical expression.

  For $\sqrt{32}$, $16$ is both a factor of the radicand, and is a perfect square. Let's factor this number out of the radicand: $\sqrt{16 \times 2}$. Now we can used the product property of roots to simplify the radical expression: $\sqrt{16} \times \sqrt{2}$. This simplifies to $4 \times \sqrt{2}$.

  Similarly, we can simplify $\sqrt{18}$ into $3 \times \sqrt{2}$.

  The radical expression $\sqrt{12}$ becomes $2 \times \sqrt{3}$.

  And the expression $\sqrt{48}$ simplifies to $4 \times \sqrt{3}$.