Algebra II - Square Roots

Evaluate the radical: $\sqrt{8}\times \sqrt{40} \times \sqrt{24}$

$16\sqrt{30}$

$12\sqrt{30}$

$14\sqrt{30}$

$18\sqrt{30}$

$11\sqrt{21}$

Explanation

In order to get the most simplified answer, do not multiply all the numbers together and combine as one radical.

Rewrite each radicals in their most simplified form.

$\sqrt8 = \sqrt4 \cdot \sqrt2 = 2\sqrt2$

$\sqrt{40} = \sqrt{4} \cdot \sqrt{10} = 2\sqrt{10}$

$\sqrt{24}= \sqrt{4}\cdot \sqrt 6 = 2\sqrt6$

Multiply the terms.

$2\sqrt{2} \cdot 2\sqrt{10} \cdot2\sqrt{6} = (2 \cdot\sqrt2)(2 \cdot\sqrt2 \cdot \sqrt5)(2 \cdot\sqrt2 \cdot \sqrt3)$

Notice that multiplying a square root of a number by itself will leave only the integer and will eliminate the radical.

$\sqrt2 \cdot \sqrt2 =2$

The terms become:

$=(2 \cdot\sqrt2)(2 \cdot\sqrt2 \cdot \sqrt5)(2 \cdot\sqrt2 \cdot \sqrt3)$

$= 2\cdot2\cdot2\cdot2 (\sqrt2 \cdot \sqrt5\cdot \sqrt3)$

$=16\sqrt{30}$

The answer is: $16\sqrt{30}$

Simplify:

$\sqrt{256x^5y^6z^3}+\sqrt{169x^3y^2}$

$16x^2y^3z\sqrt{xz}+13xy\sqrt{x}$

$16x^2y^3z\sqrt{x^2z}+13xy\sqrt{x}$

$16xy^3z\sqrt{xz}+13xy\sqrt{x}$

$14x^2y^3z\sqrt{xz}+13xy\sqrt{x}$

$(16x^2y^3z+13xy)\sqrt{2xz}$

Explanation

To simplify, we must find the squares that are underneath each radical (this can be easier to see after some factoring):

$\sqrt{256\cdot x^4\cdot x\cdot y^6\cdot z^2\cdot z} + \sqrt{169\cdot x^2\cdot x\cdot y^2}$

The squares are easier to identify now! We can pull them out of the radical after taking their square root, leaving behind everything that is not a square:

$16x^2y^3z\sqrt{xz}+13xy\sqrt{x}$

Evaluate: $\sqrt{0}-\sqrt1-\sqrt9-\sqrt{16}$

$\textup{Undefined.}$

Explanation

Evaluate each square root. The square root of a number evaluates into a number which multiplies by itself to achieve the number in the square root.

$\sqrt0 =0$

$\sqrt{1} = 1$

$\sqrt{9} =3$

$\sqrt{16}=4$

Substitute the terms back into the expression.

$\sqrt{0}-\sqrt1-\sqrt9-\sqrt{16} = 0-1-3-4 = -8$

The answer is:

Evaluate: $-\sqrt{121}+\sqrt{64}-\sqrt{16}$

Explanation

Evaluate each square root. The square root of the number is equal to a number multiplied by itself.

$-\sqrt{121}+\sqrt{64}-\sqrt{16} =-11+8-4 =-7$

The answer is:

Simplify: $\sqrt{100}-\sqrt{64}-\sqrt{169}$

Explanation

Evaluate by solving each square root first. The square root of a number is a number that multiplies by itself to achieve the number inside the square root.

$\sqrt{100} =10$

$\sqrt{64}=8$

$\sqrt{169} = 13$

Rewrite the expression.

$\sqrt{100}-\sqrt{64}-\sqrt{169} = 10-8-13$

The answer is:

Solve: $\sqrt{16}+\sqrt{196}-\sqrt{36}$

Explanation

Solve each radical. The square root determines a number that multiples by itself to equal the number inside the square root.

$\sqrt{16}=4$

$\sqrt{196}=14$

$\sqrt{36}=6$

Rewrite the expression.

$\sqrt{16}+\sqrt{196}-\sqrt{36} = 4+14-6$

The answer is:

Simplify by rationalizing the denominator:

$\frac{4}{\sqrt{11} - \sqrt{3}}$

$\frac{ \sqrt{11}+ \sqrt{3} }{2 }$

$\frac{ \sqrt{11}- \sqrt{3} }{2 }$

$\frac{ 2 \sqrt{11}+ 2\sqrt{3} }{7 }$

$\frac{ 2 \sqrt{11}- 2\sqrt{3} }{7 }$

$\frac{7}{4}$

Explanation

Multiply the numerator and the denominator by the conjugate of the denominator, which is $\sqrt{11}+ \sqrt{3}$ . Then take advantage of the distributive properties and the difference of squares pattern: