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SAT Math : Squaring / Square Roots / Radicals

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Example Questions

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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Example Question #211 : Exponents

Remember that $i= \sqrt{-1}$ .

Simplify: (2+3i)(1-i)

Possible Answers:

-1+i

5+i

5-i

-1+5i

-1-i

Correct answer:

5+i

Explanation:

Use FOIL to multiply complex numbers as follows:

(2+3i)(1-i)

2-2i+3i-3i^2

Since $i= \sqrt{-1}$ , it follows that i^2 = -1 , so then:

$2-2i+3i-3 \cdot (-1)$

2-2i+3i+3

Combining like terms gives:

5+i

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Example Question #2401 : Sat Mathematics

Simplify: (3+2i)(3-2i)

Possible Answers:

5-12i

13-12i

Correct answer:

Explanation:

Use FOIL:

(3+2i)(3-2i)=9-6i+6i-4i^2

Combine like terms:

=9-4i^2

But since i^2=-1 , we know

$9-4i^2=9-4 \cdot (-1)=9+4=13$

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Example Question #49 : Squaring / Square Roots / Radicals

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

Evaluate

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

Possible Answers:

100

$\textup{None of the other choices gives the correct response}$

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 .

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

so is the complex conjugate of this;

$x = 5- i \sqrt{7}$

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

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

$=( 2i \sqrt{7} )^{2}$

Taking advantage of the Power of a Product Rule and the fact that $i^{2} = -1$ :

$= 2^{2} \cdot i ^{2}\cdot (\sqrt{7}) ^{2}$

$=4 \cdot (-1 )\cdot 7$

= -28

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Example Question #21 : How To Multiply Complex Numbers

Raise 7+2i to the fourth power.

Possible Answers:

1,241 +2,520 i

3,593+2,968i

None of these

2,025+2,968i

2,809 +2,520 i

Correct answer:

1,241 +2,520 i

Explanation:

By the Power of a Power Rule, the fourth power of any number is equal to the square of the square of that number:

$x^{4} = x ^{2 \cdot 2} =( x^{2})^{2}$

Therefore, one way to raise 7+2i to the fourth power is to square it, then to square the result.

Using the binomial square pattern to square 7+2i :

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

Applying the Power of a Product Property:

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

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

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

= 49 + 28i -4

= 45 + 28i

Square this using the same steps:

$(45 + 28i )^{2} = 45 ^{2} + 2 \cdot 45 \cdot 28 i + (28 i) ^{2}$

$= 45 ^{2} + 2 \cdot 45 \cdot 28 i + 28^{2} i ^{2}$

$= 45 ^{2} + 2 \cdot 45 \cdot 28 i + 28^{2} (-1)$

= 2,025 +2,520 i -784

= 1,241 +2,520 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 #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

Example Question #211 : Exponents

Example Question #2401 : Sat Mathematics

Example Question #49 : Squaring / Square Roots / Radicals

Example Question #21 : How To Multiply Complex Numbers

All SAT Math Resources

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