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SAT Math : Squaring / Square Roots / Radicals

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Example Questions

Example Question #2 : How To Multiply Complex Numbers

$x^{4}-1$ has 4 roots, including the complex numbers. Take the product of $\left ( 5+2i \right )$ with each of these roots. Take the sum of these 4 results. Which of the following is equal to this sum?

Possible Answers:

4-10i

The correct answer is not listed.

10i

3+7i

Correct answer:

Explanation:

$x^{4}-1=\left ( x^{2} -1\right )\left ( x^{2}+1 \right )=\left ( x+1 \right )\left ( x-1 \right )\left ( x-i \right )\left ( x+i \right )$

This gives us roots of

$1,\ -1,\ i,\ -i$

The product of $\left ( 5+2i \right )$ with each of these gives us:

$(5+2i)\cdot i = 5i + 2i^2 = 5i +2(-1) = 5i-2 = -2 + 5i$

$(5+2i)\cdot (-i) = -5i -2i^2 = -5i -2(-1) = -5i +2 = 2-5i$

$(5+2i)\cdot 1 = 5+2i$

$(5+2i)\cdot (-1) = -5 -2i$

The sum of these 4 is:

[(-2+5i) + (2-5i)] + [(5+2i) + (-5-2i)] =

=[(-2)+2 +5i - 5i] + [5-5 + 2i - 2i] = [0] + [0] = 0

What we notice is that each of the roots has a negative. It thus makes sense that they will all cancel out. Rather than going through all the multiplication, we can instead look at the very beginning setup, which we can simplify using the distributive property:

$1(5+2i) + (-1)\cdot(5+2i)+ i(5+2i)+(-i) \cdot(5+2i) = (5+2i)\cdot[1+(-1)+i+(-i)]$

$= (5+2i)\cdot[0] = 0$

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Example Question #2386 : Sat Mathematics

Simplify:

$(3i)^{6}$

Possible Answers:

729 i

-729 i

None of the other responses gives the correct answer.

729

Correct answer:

Explanation:

Apply the Power of a Product Property:

$(3i)^{6} = 3 ^{6} \cdot i^{6} = 729 \cdot i^{6}$

A power of can be found by dividing the exponent by 4 and noting the remainder. 6 divided by 4 is equal to 1, with remainder 2, so

$i^{6} = i^{2} = -1$

Substituting,

$(3i)^{6} = 729 \cdot (-1) = -729$ .

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Example Question #3 : How To Multiply Complex Numbers

Multiply $3 + i \sqrt {3 }$ by its complex conjugate.

Possible Answers:

None of the other responses gives the correct answer.

Correct answer:

Explanation:

The complex conjugate of a complex number a+ bi is a - bi . The product of the two is the number

$a^{2}+b^{2}$ .

Therefore, the product of $3 + i \sqrt {3 }$ and its complex conjugate $3 - i \sqrt {3 }$ can be found by setting a = 3 and $b = \sqrt{3}$ in this pattern:

$a^{2}+b^{2} = 3^{2}+ (\sqrt{3})^{2} = 9 + 3 = 12$ ,

the correct response.

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Example Question #2 : How To Multiply Complex Numbers

Multiply $11 - i \sqrt{11}$ by its complex conjugate.

Possible Answers:

110i

132 i

110

132

Correct answer:

132

Explanation:

The complex conjugate of a complex number a - bi is a+ bi . The product of the two is the number

$a^{2}+b^{2}$ .

Therefore, the product of $11 - i \sqrt{11}$ and its complex conjugate $11 + i \sqrt{11}$ can be found by setting a = 11 and $b = \sqrt{11}$ in this pattern:

$a^{2}+b^{2} = 11^{2}+ (\sqrt{11})^{2} = 121+11 = 132$ ,

the correct response.

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Example Question #2 : How To Multiply Complex Numbers

What is the product of $\frac{3}{2}+ \frac{1}{4} i$ and its complex conjugate?

Possible Answers:

$\frac{35}{16} i$

$\frac{35}{16}$

$-\frac{35}{16} i$

$-\frac{35}{16}$

The correct response is not among the other choices.

Correct answer:

The correct response is not among the other choices.

Explanation:

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

The product of a+bi and a - bi is equal to $a^{2}+ b^{2}$ , so set $a = \frac{3}{2} , b= \frac{1}{4}$ in this expression, and evaluate:

$a^{2}+ b^{2} = \left ( \frac{3}{2} \right )^{2} + \left (\frac{1}{4} \right )^{2} = \frac{9}{4} + \frac{1}{16} = \frac{37}{16}$ .

This is not among the given responses.

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Example Question #1 : How To Multiply Complex Numbers

Multiply and simplify:

$\sqrt{-35} \cdot \sqrt{-5}$

Possible Answers:

$5i \sqrt{7}$

$7i \sqrt{5}$

$-7i \sqrt{5}$

$-5i \sqrt{7}$

None of the other choices gives the correct response.

Correct answer:

None of the other choices gives the correct response.

Explanation:

The two factors are both square roots of negative numbers, and are therefore imaginary. Write both in terms of before multiplying:

$\sqrt{-35} = i \sqrt{35}$

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

Therefore, using the Product of Radicals rule:

$\sqrt{-35} \cdot \sqrt{-5}$

$= i \sqrt{35} \cdot i \sqrt{5}$

$= i \cdot i \cdot \sqrt{35} \cdot \sqrt{5}$

$= i ^{2}\cdot \sqrt{35 \cdot 5 }$

$=-1 \cdot \sqrt{175 }$

$=-1 \cdot \sqrt{25 }\cdot \sqrt{7}$

$=-1 \cdot 5\cdot \sqrt{7}$

$=-5\sqrt{7}$

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Example Question #7 : How To Multiply Complex Numbers

x = 6+ 3i

y= 6 - 3i

Evaluate $x^{3} + 3x^{2} y + 3 xy^{2} + y^{3}$

Possible Answers:

-216 i

1,728

- 1,728

1,728 -216 i

216 i

Correct answer:

1,728

Explanation:

$x^{3} + 3x^{2} y + 3 xy^{2} + y^{3}$ is recognizable as the cube of the binomial x+y . That is,

$x^{3} + 3x^{2} y + 3 xy^{2} + y^{3} = (x+y) ^{3}$

Therefore, setting x = 6+ 3i and y= 6 - 3i and evaluating:

$x^{3} + 3x^{2} y + 3 xy^{2} + y^{3} =[ (6+3i) + (6-3i) ]^{3}$

$= (6+6+3i -3i) ^{3}$

$=12 ^{3}$

= 1,728 .

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Example Question #613 : Algebra

x = 7+ 4i

y= 7 - 4i

Evaluate $x^{3} - 3x^{2} y + 3 xy^{2} - y^{3}$

Possible Answers:

None of the other choices gives the correct response.

343

512i

-512i

Correct answer:

-512i

Explanation:

$x^{3} - 3x^{2} y + 3 xy^{2} - y^{3}$ is recognizable as the cube of the binomial x-y . That is,

$x^{3} - 3x^{2} y + 3 xy^{2} - y^{3}= (x-y) ^{3}$

Therefore, setting x = 7+ 4i and y= 7 - 4i and evaluating:

$x^{3} - 3x^{2} y + 3 xy^{2} - y^{3}=[ (7+4i) - (7-4 i) ]^{3}$

$= (7-7 +4i+4i ) ^{3}$

$=(8i)^{3}$

Applying the Power of a Product Rule and the fact that $i^{3} = -i$ :

$(8i)^{3} = 8 ^{3} i ^{3} = 512 (-i) = -512i$ ,

the correct value.

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Example Question #8 : How To Multiply Complex Numbers

Raise $6- i\sqrt{7}$ to the power of 3.

Possible Answers:

$342- 101i \sqrt{7}$

$90- 101i \sqrt{7}$

$216 - 7i\sqrt{7}$

$90- 115i\sqrt{7}$

$342- 115i\sqrt{7}$

Correct answer:

$90- 101i \sqrt{7}$

Explanation:

To raise any expression A-B to the third power, use the pattern

$\left (A+B \right )^{3} = A^{3} - 3A^{2} B + 3 AB^{2} - B^{3}$

Setting $A = 6, B = i \sqrt{7}$ :

$(6- i \sqrt{7} )^{3} = 6^{3} - 3\cdot 6^{2} \cdot i \sqrt{7} + 3 \cdot 6 \cdot ( i \sqrt{7})^{2} - ( i \sqrt{7})^{3}$

Taking advantage of the Power of a Product Rule:

$(6- i \sqrt{7} )^{3} = 6^{3} - 3\cdot 6^{2} \cdot i \sqrt{7} + 3 \cdot 6 \cdot i^{2} \cdot ( \sqrt{7})^{2} - i ^{3}\cdot (\sqrt{7})^{3}$

Since $i^{2} = -1$ ,

and

$i^{3} = i^{2} \cdot i = -1 \cdot i = -i$ :

$(6- i \sqrt{7} )^{3}=216 - 3\cdot 36 \cdot i \sqrt{7} + 3 \cdot 6 \cdot 7 \cdot (-1) - (-i) \cdot 7\cdot \sqrt{7}$

$(6- i \sqrt{7} )^{3} =216 - 108i \sqrt{7} -126 + 7i\sqrt{7}$

Collecting real and imaginary terms:

$(6- i \sqrt{7} )^{3} =216 -126 - 108i \sqrt{7} + 7i\sqrt{7}$

$(6- i \sqrt{7} )^{3} =90- 101i \sqrt{7}$

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Example Question #24 : Complex Numbers

Raise 6-7i to the power of 3.

Possible Answers:

None of the other choices gives the correct response.

1,099 - 1,099 i

-666 - 1,099 i

-666 - 413 i

1,099 - 413 i

Correct answer:

-666 - 413 i

Explanation:

To raise any expression A-B to the third power, use the pattern

$\left (A+B \right )^{3} = A^{3} - 3A^{2} B + 3 AB^{2} - B^{3}$

Setting A = 6, B = 7i :

$(6-7i)^{3} = 6^{3} - 3\cdot 6^{2} \cdot 7i + 3 \cdot 6 \cdot (7i)^{2} - (7i)^{3}$

Taking advantage of the Power of a Product Rule:

$(6-7i)^{3} = 6^{3} - 3\cdot 6^{2} \cdot 7i + 3 \cdot 6 \cdot 7^{2} \cdot i^{2} - 7^{3} \cdot i^{3}$

Since $i^{2} = -1$ ,

and

$i^{3} = i^{2} \cdot i = -1 \cdot i = -i$ :

$(6-7i)^{3} =216 - 3\cdot 36 \cdot 7i + 3 \cdot 6 \cdot 49 \cdot (-1) - 343 \cdot (-i)$

$(6-7i)^{3} =216 - 756 i -882 + 343 i$

Collecting real and imaginary terms:

$(6-7i)^{3} =216 -882 - 756 i + 343 i$

$(6-7i)^{3} =-666 - 413 i$

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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 #2 : How To Multiply Complex Numbers

Example Question #2386 : Sat Mathematics

Example Question #3 : How To Multiply Complex Numbers

Example Question #2 : How To Multiply Complex Numbers

Example Question #2 : How To Multiply Complex Numbers

Example Question #1 : How To Multiply Complex Numbers

Example Question #7 : How To Multiply Complex Numbers

Example Question #613 : Algebra

Example Question #8 : How To Multiply Complex Numbers

Example Question #24 : Complex Numbers

All SAT Math Resources

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