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SAT Math : Squaring / Square Roots / Radicals

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Example Questions

Example Question #592 : Algebra

Evaluate:

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

Possible Answers:

-2i

Correct answer:

Explanation:

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:

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

= -1+ ( - i )+ 1 +i

= -1 + 1 - i + i

= 0

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Example Question #1 : How To Divide Complex Numbers

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

Possible Answers:

$\frac{1}{16}$

$\frac{1}{4}$

Correct answer:

Explanation:

${x^{-\frac{1}{2}}}$ 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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Example Question #2 : How To Divide Complex Numbers

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

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

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

$(2+i) \ast i$

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

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

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

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

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

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

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

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

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Example Question #3 : How To Divide Complex Numbers

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

$\Phi a = a + 2ai$

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

Possible Answers:

$\frac{ 2}{5} - \frac{1}{5} i$

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

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

$\frac{ 1}{5} - \frac{2}{5} i$

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

Correct answer:

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

Explanation:

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

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

$\Phi N = i$ , so

N (1+ 2i) = i , and

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

Rationalize the denominator by multiplying both numerator and denominator by the complex conjugate of the latter, which is 1 -2i :

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

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

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

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

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

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

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

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Example Question #4 : How To Divide Complex Numbers

Define an operation $\blacklozenge$ as follows:

For any two complex numbers and ,

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

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

Possible Answers:

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

$\frac{ 1 }{2} -\frac{3}{2} i$

$\frac{ 3 }{2} - \frac{1}{2} i$

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

Correct answer:

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

Explanation:

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

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

We can simplify each expression separately by rationalizing the denominators.

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

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

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

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

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

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

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

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

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

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

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

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

= 1 + i

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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Example Question #5 : How To Divide Complex Numbers

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

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

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

Possible Answers:

$\frac{19}{13} + \frac{ 16}{13}i$

$\frac{11 }{13} + \frac{ 16}{13}i$

$\frac{19}{13} - \frac{ 16}{13}i$

$\frac{11 }{13} - \frac{ 4}{13}i$

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

Correct answer:

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

Explanation:

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

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

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

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

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

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

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

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

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

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Example Question #6 : How To Divide Complex Numbers

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

$\Theta a = 2a i + i$

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

Possible Answers:

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

$- \frac{ 1 }{2}+ 500i$

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

$\frac{ 1 }{2}+ 500i$

$500 + \frac{1}{2}i$

Correct answer:

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

Explanation:

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:

2N i + i = 1,000

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.

2N i= 1,000 - i

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 i^2=-1 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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Example Question #191 : Exponents

Let 4x^2 = y^3 . What is the following equivalent to, in terms of :

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

Possible Answers:

$\pm\frac{1}{8}y^{\frac{3}{2}}$

$\frac{1}{2}\sqrt{y}$

$\pm\frac{1}{2}y$

$\frac{1}{4}y^2$

Correct answer:

$\pm\frac{1}{2}y$

Explanation:

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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Example Question #192 : Exponents

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

Possible Answers:

$\frac{18}{25} + \frac{ 6}{25}i$

$\frac{3}{4} + \frac{ 1}{4}i$

$\frac{17}{24} + \frac{ 1}{4}i$

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

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

Correct answer:

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

Explanation:

Multiply both numerator and denominator by the complex conjugate of the denominator, which is 7 + i :

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

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

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

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

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

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

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

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Example Question #193 : Exponents

Find the product of (3 + 4i)(4 - 3i) given that i is the square root of negative one.

Possible Answers:

7 + i

12 - 12i

24 + 7i

Correct answer: 24 + 7i

Explanation:

Distribute (3 + 4i)(4 - 3i)

3(4) + 3(-3i) + 4i(4) + 4i(-3i)

12 - 9i + 16i -12i²

12 + 7i - 12(-1)

12 + 7i + 12

24 + 7i

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SAT Math : Squaring / Square Roots / Radicals

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #592 : Algebra

Example Question #1 : How To Divide Complex Numbers

Example Question #2 : How To Divide Complex Numbers

Example Question #3 : How To Divide Complex Numbers

Example Question #4 : How To Divide Complex Numbers

Example Question #5 : How To Divide Complex Numbers

Example Question #6 : How To Divide Complex Numbers

Example Question #191 : Exponents

Example Question #192 : Exponents

Example Question #193 : Exponents

All SAT Math Resources

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