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SAT II Math I : Number Theory

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Example Question #41 : Sat Subject Test In Math I

Which of the following choices gives a sixth root of sixty-four?

Possible Answers:

$1 - i \sqrt{3 }$

All of these

$-1 - i \sqrt{3 }$

$-1 + i \sqrt{3 }$

$1 + i \sqrt{3 }$

Correct answer:

All of these

Explanation:

Let be a sixth root of 64. The question is to find a solution of the equation

$x^{6} = 64$ .

Subtracting 64 from both sides, this equation becomes

$x^{6} - 64= 0$

64 is a perfect square (of 8) The binomial at left can be factored first as the difference of two squares:

$(x^{3} ) ^{2}- 8 ^{2}= 0$

$( x^{3} + 8 )( x^{3} - 8 )= 0$

8 is a perfect cube (of 2), so the two binomials can be factored as the sum and difference, respectively, of two cubes:

$x^{3} + 8 = x^{3}+ 2 ^{3} = (x+2) (x^{2}- x \cdot 2 + 2 ^{2}) = (x+2) (x^{2}-2x+4)$

$x^{3} - 8 = x^{3}+ 2 ^{3} = (x-2) (x^{2}+x \cdot 2 + 2 ^{2}) = (x-2) (x^{2}+2x+4)$

The equation therefore becomes

$(x+2) (x^{2}-2x+4) (x-2) (x^{2}+2x+4) = 0$ .

By the Zero Product Principle, one of these factors must be equal to 0.

If x+ 2 = 0 , then x = -2 ; if x-2 = 0 , then x = 2 . Therefore, and 2 are sixth roots of 64. However, these are not choices, so we examine the other polynomials for their zeroes.

If $x^{2}-2x+4 = 0$ , then, setting a= 1, b= -2, c= 4 in the following quadratic formula:

$x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

$= \frac{- (-2) \pm \sqrt{ (-2)^{2}-4(1)(4)}}{2(1)}$

$= \frac{2 \pm \sqrt{ 4 - 16 }}{2}$

$= \frac{2 \pm \sqrt{ -12 }}{2}$

$= \frac{2 \pm\sqrt{4 } \cdot \sqrt{ -1} \cdot \sqrt{3 }}{2}$

$= \frac{2 \pm 2i \sqrt{3 }}{2}$

$=1 \pm i \sqrt{3 }$

If $x^{2}+2x+4= 0$ , then, setting a= 1, b=2, c= 4 in the quadratic formula:

$x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

$= \frac{- 2 \pm \sqrt{ 2^{2}-4(1)(4)}}{2(1)}$

$= \frac{-2 \pm \sqrt{ 4 - 16 }}{2}$

$= \frac{-2 \pm \sqrt{ -12 }}{2}$

$= \frac{-2 \pm\sqrt{4 } \cdot \sqrt{ -1} \cdot \sqrt{3 }}{2}$

$= \frac{-2 \pm 2i \sqrt{3 }}{2}$

$=-1 \pm i \sqrt{3 }$

Therefore, the set of sixth roots of 64 is

$\left \{ -4, 4,-1 - i \sqrt{3 }, -1 + i \sqrt{3 } ,1 - i \sqrt{3 }, 1 + i \sqrt{3 } \right \}$ .

All four choices appear in this set.

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Example Question #11 : Real And Complex Numbers

Let and be complex numbers. $\overline{x}$ and $\overline{y}$ denote their complex conjugates.

xy = 13+ 7i

x= 4 - 7 i

Evaluate $\overline{x} \cdot \overline{y}$ .

Possible Answers:

7 + 13i

None of these

-13 - 7i

7 - 13i

-13 + 7i

Correct answer:

None of these

Explanation:

Knowing the actual values of and is not necessary to solve this problem. The product of the complex conjugates of two numbers is equal to the complex conjugate of the product of the numbers; that is,

$\overline{x} \cdot \overline{y} = \overline{xy}$

xy = 13+ 7i , so $\overline{xy }= 13- 7i$ , and

$\overline{x} \cdot \overline{y} =13- 7i$ ,

which is not among the choices.

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Example Question #41 : Sat Subject Test In Math I

$\overline{x}$ denotes the complex conjugate of .

If x = 5 + 2i , then evaluate $x^{3} \cdot \overline{x} ^{3}$ .

Possible Answers:

1,000

9,261

24, 389

None of these

Correct answer:

24, 389

Explanation:

Applying the Power of a Product Rule:

$x^{3} \cdot \overline{x} ^{3}=(x \overline{x})^{3}$

The complex conjugate of an imaginary number x = a+ bi is $\overline{x} = a - bi$ ; the product of the two is

$x \overline{x} = a^{2} + b^{2}$

x = 5 + 2i , so, setting a = 5 , b = 2 in the above pattern:

$x \overline{x} = 5^{2} + 2^{2} = 25 + 4 = 29$

Consequently,

$x^{3} \cdot \overline{x} ^{3}=(x \overline{x})^{3} = 29 ^{3}= 24, 389$

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SAT II Math I : Number Theory

Study concepts, example questions & explanations for SAT II Math I

All SAT II Math I Resources

Example Questions

Example Question #41 : Sat Subject Test In Math I

Example Question #11 : Real And Complex Numbers

Example Question #41 : Sat Subject Test In Math I

All SAT II Math I Resources

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