Squaring / Square Roots / Radicals - SAT Math

Card 0 of 424

Question

Simplify:

Answer

If you don't already have the pattern memorized, use FOIL. It's best to write out the parentheses twice (as below) to avoid mistakes:

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Question

Simplify the radical.

$\sqrt{3283}$

Answer

We can break the square root down into 2 roots of 67 and 49. 49 is a perfect square and reduces to 7.

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Question

Simplify:

Answer

If you don't already have the pattern memorized, use FOIL. It's best to write out the parentheses twice (as below) to avoid mistakes:

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Question

x2 = 36

Quantity A: x

Quantity B: 6

Answer

x2 = 36 -> it is important to remember that this leads to two answers.

x = 6 or x = -6.

If x = 6: A = B.

If x = -6: A < B.

Thus the relationship cannot be determined from the information given.

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Question

According to Heron's Formula, the area of a triangle with side lengths of a, b, and c is given by the following:

where s is one-half of the triangle's perimeter.

What is the area of a triangle with side lengths of 6, 10, and 12 units?

Answer

We can use Heron's formula to find the area of the triangle. We can let a = 6, b = 10, and c = 12.

In order to find s, we need to find one half of the perimeter. The perimeter is the sum of the lengths of the sides of the triangle.

Perimeter = a + b + c = 6 + 10 + 12 = 28

In order to find s, we must multiply the perimeter by one-half, which would give us (1/2)(28), or 14.

Now that we have a, b, c, and s, we can calculate the area using Heron's formula.

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Question

Simplify the expression.

$\sqrt{5}(2\sqrt{3}+\sqrt{12})$

Answer

Use the distributive property for radicals.

$\sqrt{5}(2\sqrt{3}+\sqrt{12})$

Multiply all terms by $\sqrt{5}$ .

$(\sqrt{5}2\sqrt{3})+(\sqrt{5}\sqrt{12})$

Combine terms under radicals.

$2\sqrt{35}+\sqrt{125}$

$2\sqrt{15}+\sqrt{60}$

Look for perfect square factors under each radical. has a perfect square of . The can be factored out.

$2\sqrt{15}+\sqrt{4*15}$

$2\sqrt{15}+2\sqrt{15}$

Since both radicals are the same, we can add them.

$4\sqrt{15}$

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Question

Simplify the radical expression.

$\sqrt[3]{-64x^{6}y^{12}}$

Answer

$\sqrt[3]{-64x^{6}y^{12}}$

Look for perfect cubes within each term. This will allow us to factor out of the radical.

$\sqrt[3]{-64}\sqrt[3]{x^6}\sqrt[3]{y^{12}}$

$\sqrt[3]{-4-4-4}\sqrt[3]{x^2x^2x^2}\sqrt[3]{y^4y^4y^4}$

Simplify.

$-4x^{2}y^4$

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Question

Which of the following expressions is equal to the following expression?

$\sqrt{(27)(45)(125)}$

Answer

$\sqrt{(27)(45)(125)}$

First, break down the component parts of the square root:

$\sqrt{(3^{3})(5\times 3^{2})(5^{3})}$

Combine like terms in a way that will let you pull some of them out from underneath the square root symbol:

$\sqrt{(5^{4})(3^4)(3)}$

Pull out the terms with even exponents and simplify:

$(5^{2})(3^{2})\sqrt{3}=225\sqrt{3}$

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Question

Which of the following is equal to the following expression?

$\sqrt{(16)(8)+(32)(20)}$

Answer

$\sqrt{(16)(8)+(32)(20)}$

First, break down the components of the square root:

$\sqrt{(2^{4})(2^{3})+(2^{5})(2^{2})\times 5}$

Combine like terms. Remember, when multiplying exponents, add them together:

$\sqrt{(2^{7})+(2^{7})\times5}$

Factor out the common factor of :

$\sqrt{(2^{7})(1+5)}$

$\sqrt{(2^7)\times6}$

Factor the :

$\sqrt{(2^7)\times2\times3}$

Combine the factored with the :

$\sqrt{(2^{8})\times3}$

Now, you can pull $\sqrt{2^8}$ out from underneath the square root sign as :

$2^4\sqrt{3}$

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Question

Which of the following expression is equal to

$\sqrt{(45)(9)+(27)(36)}$

Answer

$\sqrt{(45)(9)+(27)(36)}$

When simplifying a square root, consider the factors of each of its component parts:

$\sqrt{(3^2)\times5\times(3^{2})+(3^{3})(2^2)(3^2)}$

Combine like terms:

$\sqrt{5(3^4)+(2^2)(3^5)}$

Remove the common factor, :

$\sqrt{(3^4)\times(5+3\times(2^2))}$

Pull the $\sqrt{3^4}$ outside of the equation as :

$(3^2)\sqrt{5+12}=9\sqrt{17}$

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Question

Simplify: $\sqrt{-3}+\sqrt{-9}+\sqrt{-16}$

Answer

Rewrite $\sqrt{-3}+\sqrt{-9}+\sqrt{-16}$ in their imaginary terms.

$i=\sqrt{-1}$

$\sqrt{-3}+\sqrt{-9}+\sqrt{-16}=i\sqrt3+3i+4i = i\sqrt3 +7i$

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Question

For $i=\sqrt{-1}$ , what is the sum of $\frac{3}{2} + \frac{1}{4} i$ and its complex conjugate?

Answer

The complex conjugate of a complex number is , so $\frac{3}{2} + \frac{1}{4} i$ has $\frac{3}{2} - \frac{1}{4} i$ as its complex conjugate. The sum of the two numbers is

$\left (\frac{3}{2} + \frac{1}{4} i \right ) + \left ( \frac{3}{2} - \frac{1}{4} i \right )$

$= \frac{3}{2} + \frac{1}{4} i + \frac{3}{2} - \frac{1}{4} i$

$= \frac{3}{2} + \frac{3}{2} + \frac{1}{4} i - \frac{1}{4} i$

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Question

Add $5 + i \sqrt {5 }$ and its complex conjugate.

Answer

The complex conjugate of a complex number is . Therefore, the complex conjugate of $5 + i \sqrt {5 }$ is $5 - i \sqrt {5 }$ ; add them by adding real parts and adding imaginary parts, as follows:

$(5 + i \sqrt {5 } )+( 5 - i \sqrt {5 })$

$= 5 + 5 + i \sqrt {5 } - i \sqrt {5 }$

,

the correct response.

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Question

Add $13 - i \sqrt{13}$ to its complex conjugate.

Answer

The complex conjugate of a complex number is . Therefore, the complex conjugate of $13 - i \sqrt{13}$ is $13 + i \sqrt{13}$ ; add them by adding real parts and adding imaginary parts, as follows:

$(13 - i \sqrt{13} )+( 13 + i \sqrt{13})$

$(13 - i \sqrt{13} )+( 13 + i \sqrt{13})$

$= 13+13 - i \sqrt {13 }+ i \sqrt {13}$

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Question

An arithmetic sequence begins as follows:

Give the next term of the sequence

Answer

The common difference of an arithmetic sequence can be found by subtracting the first term from the second:

$d = a_{2} - a_{1}$

Add this to the second term to obtain the desired third term:

$a_{3} = a_{2} +d = 3i + ( -3 + 3i) = -3 + 3i + 3i = -3 + 6i$ .

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Question

Evaluate:

$i^{6} + i^{7}+ i^{8}+ i^{9}$

Answer

A power of can be evaluated by dividing the exponent by 4 and noting the remainder. The power is determined according to the following table:

$\begin{matrix} \textrm{\underline{Rem}}& \textrm{\underline{Power}} \ 0& 1\ 1& i\ 2& -1\ 3& -i \end{matrix}$

$6 \div 4 = 1 \textrm{ R }2$ , so $i^{6} = -1$

$7 \div 4 = 1 \textrm{ R }3$ , so $i^{7} = -i$

$8 \div 4 = 2 \textrm{ R }0$ , so $i^{8} = 1$

$9 \div 4 = 2 \textrm{ R }1$ , so $i^{9} = i$

Substituting:

$i^{6} + i^{7}+ i^{8}+ i^{9}$

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Question

Evaluate:

$9i^{9}+ 10 i^{10}+ 11 i^{11}+ 12 i^{12}$

Answer

A power of can be evaluated by dividing the exponent by 4 and noting the remainder. The power is determined according to the following table:

$\begin{matrix} \textrm{\underline{Rem}}& \textrm{\underline{Power}} \ 0& 1\ 1& i\ 2& -1\ 3& -i \end{matrix}$

$9 \div 4 = 2 \textrm{ R }1$ , so $i^{9} = i$

$10 \div 4 = 2 \textrm{ R }2$ , so $i^{10} = -1$

$11 \div 4 = 2 \textrm{ R }3$ , so $i^{11} = - i$

$12 \div 4 = 3 \textrm{ R }0$ , so $i^{12} = 1$

Substituting:

$9i^{9}+ 10 i^{10}+ 11 i^{11}+ 12 i^{12}$

Collect real and imaginary terms:

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Question

Simplify:

Answer

It can be easier to line real and imaginary parts vertically to keep things organized, but in essence, combine like terms (where 'like' here means real or imaginary):

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Question

Let . What is the following equivalent to, in terms of :

$\frac{x}{\sqrt{y}}$

Answer

Solve for x first in terms of y, and plug back into the equation.

$\4x^2 = y^3 \ \x^2 = \frac{1}{4} y^3 \ \x = \pm \frac{1}{2} y\sqrt{y}$

Then go back to the equation you are solving for:

$\frac{x}{\sqrt{y}}$ substitute in $x=\pm\frac{1}{2}y\sqrt{y}$

$\ \frac{x}{\sqrt{y}}=\frac{\pm\frac{1}{2}y\sqrt{y}}{\sqrt{y}} \ \\frac{x}{\sqrt{y}}=\pm\frac{1}{2}y$

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Question

For which of the following values of is the value of ${x^{-\frac{1}{2}}}$ least?

Answer

${x^{-\frac{1}{2}}}$ is the same as $\frac{1}{\sqrt{x}}$ , which means that the bigger the answer to $\sqrt{x}$ is, the smaller the fraction will be.

Therefore, is the correct answer because

$\frac{1}{\sqrt{16}} = \frac{1}{4}$ .

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