Squaring / Square Roots / Radicals - SAT Math

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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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Question

Define an operation $\ast$ so that for any two complex numbers and :

$a \ast b = \frac{a + b}{a}$

Evaluate $(2+i) \ast i$ .

Answer

$a \ast b = \frac{a + b}{a}$ , so

$(2+i) \ast i = \frac{(2+ i )+ i}{2+i} = \frac{2 + 2i}{2+i}$

Rationalize the denominator by multiplying both numerator and denominator by the complex conjugate of the latter, which is :

$(2+i) \ast i$

$= \frac{2 + 2i}{2+i}$

$= \frac{\left (2 + 2i \right )\left (2-i \right )}{\left (2+i \right )\left (2-i \right )}$

$= \frac{ 2 \cdot 2 - 2 \cdot i + 2i \cdot 2 - 2i \cdot i}{2 ^{2}+1^{2}}$

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

$= \frac{ 4 - 2 i + 4i - 2 \cdot (-1)}{5}$

$= \frac{ 4 - 2 i + 4i +2}{5}$

$= \frac{ 6 + 2 i }{5}$

$= \frac{ 6 }{5} + \frac{ 2 }{5}i$

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Question

Simplify the expression by rationalizing the denominator, and write the result in standard form: $\frac{5+i}{7-i}$

$\left (\textup{Note: } i=\sqrt{-1}\right\ )$

Answer

Multiply both numerator and denominator by the complex conjugate of the denominator, which is :

$\frac{5+i}{7-i}$

$= \frac{(5+i)(7+i)}{(7-i)(7+i)}$

$= \frac{5 \cdot 7 + 5 \cdot i + 7 \cdot i + i \cdot i }{(7^{2}+1 ^{2}) }$

$= \frac{35 +5i + 7 i +(-1) }{(49+1) }$

$= \frac{34 +12i }{50 }$

$= \frac{34 }{50 } + \frac{ 12i }{50 }$

$= \frac{17}{25} + \frac{ 6}{25}i$

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Question

Define an operation $\ast$ so that for any two complex numbers and :

$a \ast b = \frac{a + b}{a}$

Evaluate $(3+2i) \ast 2$

Answer

$a \ast b = \frac{a + b}{a}$ , so

$(3+2i) \ast 2 = \frac{(3+2i) + 2}{3+2i } = \frac{5+2i}{3+2i }$

Rationalize the denominator by multiplying both numerator and denominator by the complex conjugate of the latter, which is :

$(3+2i) \ast 2 = \frac{5+2i}{3+2i }$

$= \frac{\left (5+2i \right )\left (3-2i \right )}{\left (3+2i \right )\left (3-2i \right )}$

$= \frac{5 \cdot 3 - 5 \cdot 2i + 2i \cdot 3 - 2i \cdot 2i }{3^{2}+2^{2}}$

$= \frac{5 \cdot 3 - 5 \cdot 2i + 2i \cdot 3 - 2 \cdot 2 \cdot i^{2} }{9+4}$

$= \frac{15 - 10i + 6i -4 \cdot (-1) }{9+4}$

$= \frac{15 - 10i + 6i +4}{13}$

$= \frac{19 -4i}{13}$

$= \frac{19 }{13} - \frac{ 4}{13}i$

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Question

Define an operation $\Phi$ such that, for any complex number ,

$\Phi a = a + 2ai$

If $\Phi N = i$ , then evaluate .

Answer

$\Phi a = a + 2ai$ , so

$\Phi N = N + 2Ni = N (1+ 2i)$

$\Phi N = i$ , so

, and

$N=\frac{ i}{ 1+ 2i}$

Rationalize the denominator by multiplying both numerator and denominator by the complex conjugate of the latter, which is :

$N=\frac{ i}{ 1+ 2i}$

$=\frac{ i( 1- 2i)}{( 1+ 2i)( 1- 2i)}$

$=\frac{ i \cdot 1- i \cdot 2i}{1 ^{2}+2^{2}}$

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

$=\frac{ i - 2(-1)}{5}$

$=\frac{ i +2}{5}$

$=\frac{ 2}{5} + \frac{1}{5} i$

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Question

Define an operation $\Theta$ such that for any complex number ,

$\Theta a = 2a i + i$

If $\Theta N = 1,000$ , evaluate .

Answer

First substitute our variable N in where ever there is an a.

Thus, $\Theta a = 2a i + i$ , becomes $\Theta N = 2N i + i$ .

Since $\Theta N = 1,000$ , substitute:

In order to solve for the variable we will need to isolate the variable on one side with all other constants on the other side. To do this, apply the oppisite operation to the function.

First subtract i from both sides.

Now divide by 2i on both sides.

$\frac{2N i}{2i}= \frac{1,000 - i}{2i}$

From here multiply the numerator and denominator by i to further solve.

$N= \frac{1,000 - i}{2i}$

$= \frac{(1,000 - i) \cdot i }{2i \cdot i}$

$= \frac{1,000\cdot i - i\cdot i }{2 \cdot i^{2}}$

Recall that by definition. Therefore,

$= \frac{1,000 i - (-1) }{2 \cdot (-1)}$

$= \frac{1,000 i +1 }{-2}$

$= \frac{ 1 }{-2}+ \frac{1,000 i }{-2}$

$=- \frac{ 1 }{2}- 500i$ .

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Question

Define an operation $\blacklozenge$ as follows:

For any two complex numbers and ,

$a \blacklozenge b = \frac{a}{2i} + \frac{2i}{b}$

Evaluate $(1+ i) \blacklozenge (1+ i)$ .

Answer

$a \blacklozenge b = \frac{a}{2i} + \frac{2i}{b}$ , so

$(1+ i) \blacklozenge (1+ i) = \frac{1+ i}{2i} + \frac{2i}{1+ i}$

We can simplify each expression separately by rationalizing the denominators.

$\frac{1+ i}{2i}$

$= \frac{\left (1+ i \right )\cdot i}{2i \cdot i}$

$= \frac{1 \cdot i+ i \cdot i}{2 \cdot i ^{2}}$

$= \frac{ i+(-1)}{2 (-1)}$

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

$= \frac{ 1 }{2} - \frac{1}{2} i$

$\frac{2i}{1+ i}$

$= \frac{2i(1- i)}{(1+ i)(1- i)}$

$= \frac{2i \cdot 1- 2i \cdot i}{1^{2}+1^{2}}$

$= \frac{2i - 2i ^{2}}{1+1}$

$= \frac{2i - 2(-1)}{2}$

$= \frac{2i +2}{2}$

Therefore,

$(1+ i) \blacklozenge (1+ i) = \frac{1+ i}{2i} + \frac{2i}{1+ i}$

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

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

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

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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

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

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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