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SAT Math : SAT Mathematics

Study concepts, example questions & explanations for SAT Math

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Example Questions

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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Example Question #31 : Complex Numbers

Evaluate $\left ( 4i\right )^{3} \cdot (3i)^{4}$ .

Possible Answers:

None of the other choices gives the correct response.

5,184i

-5,184

5,184

-5,184i

Correct answer:

-5,184i

Explanation:

Apply the Power of a Product Rule:

$\left ( 4i\right )^{3} \cdot (3i)^{4}$

$= 4^{3}\cdot i^{3} \cdot 3^{4} \cdot i^{4}$

$i^{3} = i^{2} \cdot i = (-1 ) \cdot i = -i$ ,

and

$i^{4} = (i^{2})^{2} = (-1)^{2} = 1$ ,

so, substituting and evaluating:

$4^{3}\cdot i^{3} \cdot 3^{4} \cdot i^{4}$

$= 64 \cdot (-i) \cdot 81 \cdot 1$

$= - 64 \cdot 81 \cdot 1 \cdot i$

=-5,184i

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Example Question #32 : Complex Numbers

Raise $8 + i \sqrt{5}$ to the power of 4.

Possible Answers:

$4,761+2,208\sqrt{5}$

$3,481+2,208\sqrt{5}$

$4,761+1,888 i \sqrt{5}$

$3,481+2,208\sqrt{5}$

$2,201+1,888 i \sqrt{5}$

Correct answer:

$2,201+1,888 i \sqrt{5}$

Explanation:

The easiest way to find $\left (8 + i \sqrt{5} \right )^{4}$ is to note that

$\left (8 + i \sqrt{5} \right )^{4} =\left [ \left (8 + i \sqrt{5} \right )^{2} \right ]^{2}$ .

Therefore, we can find the fourth power of $8 + i \sqrt{5}$ by squaring $8 + i \sqrt{5}$ , then squaring the result.

Using the binomial square pattern to square $8 + i \sqrt{5}$ :

$(8 + i \sqrt{5} )^{2} = 8^{2} + 2 \cdot 8 \cdot i \sqrt{5} + (i \sqrt{5} ) ^{2}$

Applying the Power of a Product Property:

$(8 + i \sqrt{5} )^{2} = 8^{2} + 2 \cdot 8 \cdot i \sqrt{5} + i^{2} \cdot \left ( \sqrt{5} \right) ^{2}$

Since $i^{2} = -1$ by definition:

$(8 + i \sqrt{5} )^{2} = 8^{2} + 2 \cdot 8 \cdot i \sqrt{5} + (-1) \cdot 5$

$(8 + i \sqrt{5} )^{2} = 64 +16 i \sqrt{5} -5$

$= 59 +16 i \sqrt{5}$

Square this using the same steps:

$(59 +16 i \sqrt{5} )^{2} = 59^{2} + 2 \cdot 59 \cdot 16 i \sqrt{5} + (16i \sqrt{5} ) ^{2}$

$= 59^{2} + 2 \cdot 59 \cdot 16 i \sqrt{5} + 16^{2} \cdot i^{2} \cdot (\sqrt{5} ) ^{2}$

$=3,481+1,888 i \sqrt{5} + 256 \cdot (-1) \cdot 5$

$=3,481+1,888 i \sqrt{5} -1,280$

$=2,201+1,888 i \sqrt{5}$ ,

the correct response.

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Example Question #31 : Complex Numbers

Evaluate $\left ( 2 i \sqrt{5}\right )^{5} \cdot \left ( 3 i \sqrt{5} \right )^{3}$

Possible Answers:

None of the other choices gives the correct response.

- 540,000 i

$-108,000 \sqrt{5}$

-540,000

$-108,000 i \sqrt{5}$

Correct answer:

None of the other choices gives the correct response.

Explanation:

Apply the Power of a Product Rule:

$\left ( 2 i \sqrt{5}\right )^{5} \cdot \left ( 3 i \sqrt{5} \right )^{3}$

$= 2 ^{5} \cdot i^{5} \cdot \left ( \sqrt{5}\right )^{5} \cdot 3 ^{3} \cdot i ^{3} \cdot \left ( \sqrt{5} \right )^{3}$

$= 2 ^{5} \cdot 3 ^{3} \cdot i^{5}\cdot i ^{3} \cdot \left ( \sqrt{5}\right )^{5} \cdot \left ( \sqrt{5} \right )^{3}$

Applying the Product of Powers Rule:

$2 ^{5} \cdot 3 ^{3} \cdot i^{5}\cdot i ^{3} \cdot \left ( \sqrt{5}\right )^{5} \cdot \left ( \sqrt{5} \right )^{3}$

$= 2 ^{5} \cdot 3 ^{3} \cdot i^{5+3} \cdot \left ( \sqrt{5}\right )^{5+3}$

$= 2 ^{5} \cdot 3 ^{3}\cdot \left ( \sqrt{5}\right )^{8} \cdot i^{8}$

raised to any multiple of 4 is equal to 1, and $\sqrt{5} = 5 ^{\frac{1}{2}}$ , so, substituting and evaluating:

$2 ^{5} \cdot 3 ^{3}\cdot \left ( \sqrt{5}\right )^{8} \cdot i^{8}$

$= 2 ^{5} \cdot 3 ^{3}\cdot \left ( 5^{\frac{1}{2}} \right )^{8} \cdot 1$

$= 2 ^{5} \cdot 3 ^{3}\cdot 5^{ \frac{1}{2} \cdot 8} \cdot 1$

$= 2 ^{5} \cdot 3 ^{3}\cdot 5^{4} \cdot 1$

$= 32 \cdot 27 \cdot 625 \cdot 1$

=540,000

This is not among the given choices.

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Example Question #44 : Squaring / Square Roots / Radicals

$x = 7 + 2 i \sqrt{5}$ ; is the complex conjugate of .

Evaluate

$x^{2} + 2xy + y^{2}$ .

Possible Answers:

196

Correct answer:

196

Explanation:

$x^{2} + 2xy + y^{2}$ conforms to the perfect square trinomial pattern

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

The easiest way to solve this problem is to add and , then square the sum.

The complex conjugate of a complex number a+bi is a-bi .

$x = 7 + 2 i \sqrt{5}$ ,

so is the complex conjugate of this;

$y = 7 - 2 i \sqrt{5}$ ,

and

$x+y = ( 7 + 2 i \sqrt{5} ) + ( 7 - 2 i \sqrt{5}) = 7+ 7+ 2 i \sqrt{5} - 2 i \sqrt{5} = 14$

Substitute 14 for x+y :

$x^{2} + 2xy + y^{2} = (x+y) ^{2} = 14^{2} = 196$ .

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Example Question #45 : Squaring / Square Roots / Radicals

x = 4+ 7i ; is the complex conjugate of .

Evaluate

$x^{2} + 2xy + y^{2}$ .

Possible Answers:

-132-224i

196

-132+224i

Correct answer:

Explanation:

$x^{2} + 2xy + y^{2}$ conforms to the perfect square trinomial pattern

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

The easiest way to solve this problem is to add and , then square the sum.

The complex conjugate of a complex number a+bi is a-bi .

x = 4+ 7i ,

so is the complex conjugate of this;

y = 4 - 7i ,

and

x+y = (4+7i) + (4-7i) = 4+4 +7i - 7i = 8

Substitute 8 for x+y :

$x^{2} + 2xy + y^{2} = (x+y) ^{2} = 8^{2} = 64$ .

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Example Question #46 : Squaring / Square Roots / Radicals

x = 5+ 8i ; is the complex conjugate of .

Evaluate

$x^{2} - 2xy + y^{2}$ .

Possible Answers:

256

100

-156-320 i

-156+320 i

Correct answer:

Explanation:

$x^{2} - 2xy + y^{2}$ conforms to the perfect square trinomial pattern

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

The easiest way to solve this problem is to subtract and , then square the difference.

The complex conjugate of a complex number a+bi is a-bi .

x = 5+ 8i ,

so is the complex conjugate of this;

y = 5 - 8i

x-y = (5+8i) - (5-8i) = 5- 5 + 8i + 8i = 16i

Substitute 16i for x-y :

$x^{2} - 2xy + y^{2} = (x-y) ^{2} =( 16 i)^{2} = 16 ^{2} \cdot i^{2}$

By definition, $i^{2} = -1$ , so, substituting,

$16 ^{2} \cdot i^{2} = 256 (-1 ) = -256$ ,

the correct choice.

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SAT Math : SAT Mathematics

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

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

Example Question #31 : Complex Numbers

Example Question #32 : Complex Numbers

Example Question #31 : Complex Numbers

Example Question #44 : Squaring / Square Roots / Radicals

Example Question #45 : Squaring / Square Roots / Radicals

Example Question #46 : Squaring / Square Roots / Radicals

All SAT Math Resources

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