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SAT Math : Algebra

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Example Questions

Example Question #41 : Squaring / Square Roots / Radicals

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

Possible Answers:

$\displaystyle - 540,000 i$

None of the other choices gives the correct response.

$\displaystyle -540,000$

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

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

Correct answer:

None of the other choices gives the correct response.

Explanation:

Apply the Power of a Product Rule:

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

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

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

Applying the Product of Powers Rule:

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

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

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

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

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

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

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

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

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

$\displaystyle =540,000$

This is not among the given choices.

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Example Question #44 : Squaring / Square Roots / Radicals

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

Evaluate

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

Possible Answers:

$\displaystyle 1$

$\displaystyle 80$

$\displaystyle -196$

196 $\displaystyle 196$

$\displaystyle -80$

Correct answer:

196 $\displaystyle 196$

Explanation:

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

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

The easiest way to solve this problem is to add $\displaystyle x$ and $\displaystyle y$, then square the sum.

The complex conjugate of a complex number a+bi $\displaystyle a+bi$ is a-bi $\displaystyle a-bi$.

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

so $\displaystyle y$ is the complex conjugate of this;

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

and

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

Substitute 14 for x+y $\displaystyle x+y$:

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

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Example Question #45 : Squaring / Square Roots / Radicals

$\displaystyle x = 4+ 7i$; $\displaystyle y$ is the complex conjugate of $\displaystyle x$.

Evaluate

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

Possible Answers:

$\displaystyle -132-224i$

196 $\displaystyle 196$

$\displaystyle 64$

$\displaystyle -132+224i$

$\displaystyle -196$

Correct answer:

$\displaystyle 64$

Explanation:

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

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

The easiest way to solve this problem is to add $\displaystyle x$ and $\displaystyle y$, then square the sum.

The complex conjugate of a complex number a+bi $\displaystyle a+bi$ is a-bi $\displaystyle a-bi$.

$\displaystyle x = 4+ 7i$,

so $\displaystyle y$ is the complex conjugate of this;

$\displaystyle y = 4 - 7i$,

and

$\displaystyle x+y = (4+7i) + (4-7i) = 4+4 +7i - 7i = 8$

Substitute 8 for x+y $\displaystyle x+y$:

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

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Example Question #46 : Squaring / Square Roots / Radicals

$\displaystyle x = 5+ 8i$; $\displaystyle y$ is the complex conjugate of $\displaystyle x$.

Evaluate

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

Possible Answers:

256 $\displaystyle 256$

100 $\displaystyle 100$

$\displaystyle -156-320 i$

$\displaystyle -256$

$\displaystyle -156+320 i$

Correct answer:

$\displaystyle -256$

Explanation:

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

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

The easiest way to solve this problem is to subtract $\displaystyle x$ and $\displaystyle y$, then square the difference.

The complex conjugate of a complex number a+bi $\displaystyle a+bi$ is a-bi $\displaystyle a-bi$.

$\displaystyle x = 5+ 8i$,

so $\displaystyle y$ is the complex conjugate of this;

$\displaystyle y = 5 - 8i$

$\displaystyle x-y = (5+8i) - (5-8i) = 5- 5 + 8i + 8i = 16i$

Substitute 16i $\displaystyle 16i$ for x-y $\displaystyle x-y$:

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

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

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

the correct choice.

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Example Question #2401 : Sat Mathematics

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

Simplify: $\displaystyle (2+3i)(1-i)$

Possible Answers:

-1+5i $\displaystyle -1+5i$

-1+i $\displaystyle -1+i$

-1-i $\displaystyle -1-i$

5-i $\displaystyle 5-i$

5+i $\displaystyle 5+i$

Correct answer:

5+i $\displaystyle 5+i$

Explanation:

Use FOIL to multiply complex numbers as follows:

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

$\displaystyle 2-2i+3i-3i^2$

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

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

$\displaystyle 2-2i+3i+3$

Combining like terms gives:

5+i $\displaystyle 5+i$

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Example Question #621 : Algebra

Simplify: $\displaystyle (3+2i)(3-2i)$

Possible Answers:

13-12i $\displaystyle 13-12i$

$\displaystyle 5$

5-12i $\displaystyle 5-12i$

$\displaystyle 13$

Correct answer:

$\displaystyle 13$

Explanation:

Use FOIL:

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

Combine like terms:

=9-4i^2 $\displaystyle =9-4i^2$

But since i^2=-1 $\displaystyle i^2=-1$, we know

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

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Example Question #49 : Squaring / Square Roots / Radicals

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

Evaluate

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

Possible Answers:

100 $\displaystyle 100$

$\displaystyle 28$

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

$\displaystyle -100$

$\displaystyle -28$

Correct answer:

$\displaystyle -28$

Explanation:

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

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

The easiest way to solve this problem is to subtract $\displaystyle x$ and $\displaystyle y$, then square the difference.

The complex conjugate of a complex number a+bi $\displaystyle a+bi$ is a-bi $\displaystyle a-bi$.

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

so $\displaystyle y$ is the complex conjugate of this;

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

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

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

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

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

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

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

= -28 $\displaystyle = -28$

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Example Question #211 : Exponents

Raise 7+2i $\displaystyle 7+2i$ to the fourth power.

Possible Answers:

$\displaystyle 2,809 +2,520 i$

None of these

$\displaystyle 2,025+2,968i$

$\displaystyle 3,593+2,968i$

$\displaystyle 1,241 +2,520 i$

Correct answer:

$\displaystyle 1,241 +2,520 i$

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:

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

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

Using the binomial square pattern to square 7+2i $\displaystyle 7+2i$:

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

Applying the Power of a Product Property:

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

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

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

$\displaystyle = 49 + 28i -4$

$\displaystyle = 45 + 28i$

Square this using the same steps:

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

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

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

$\displaystyle = 2,025 +2,520 i -784$

$\displaystyle = 1,241 +2,520 i$

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Example Question #51 : Squaring / Square Roots / Radicals

Raise 8 - 3i $\displaystyle 8 - 3i$ to the fourth power.

Possible Answers:

$\displaystyle 7,633 - 7,008i$

None of these

$\displaystyle 721-5,280 i$

$\displaystyle 3,025 - 7,008i$

$\displaystyle 5,329-5,280 i$

Correct answer:

$\displaystyle 721-5,280 i$

Explanation:

The easiest way to find $(8 - 3i)^{4}$ $\displaystyle (8 - 3i)^{4}$ is to note that

$(8 - 3i)^{4} =[ (8 - 3i)^{2} ]^{2}$ $\displaystyle (8 - 3i)^{4} =[ (8 - 3i)^{2} ]^{2}$.

Therefore, we can find the fourth power of 8 - 3i $\displaystyle 8 - 3i$ by squaring 8 - 3i $\displaystyle 8 - 3i$, then squaring the result.

Using the binomial square pattern to square 8 - 3i $\displaystyle 8 - 3i$:

$(8 - 3i)^{2} = 8^{2} - 2 \cdot 8 \cdot 3i + (3i)^{2}$ $\displaystyle (8 - 3i)^{2} = 8^{2} - 2 \cdot 8 \cdot 3i + (3i)^{2}$

Applying the Power of a Product Property:

$(8 - 3i)^{2} = 8^{2} - 2 \cdot 8 \cdot 3i + 3^{2} \cdot i^{2}$ $\displaystyle (8 - 3i)^{2} = 8^{2} - 2 \cdot 8 \cdot 3i + 3^{2} \cdot i^{2}$

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

$(8 - 3i)^{2} = 64 - 48 i + 9 (-1)$ $\displaystyle (8 - 3i)^{2} = 64 - 48 i + 9 (-1)$

$\displaystyle = 64 - 48 i -9$

$\displaystyle = 55 - 48 i$

Square this using the same steps:

$( 55 - 48 i )^{2} = 55^{2} - 2 \cdot 55 \cdot 48 i + (48i)^{2}$ $\displaystyle ( 55 - 48 i )^{2} = 55^{2} - 2 \cdot 55 \cdot 48 i + (48i)^{2}$

$= 3,025-5,280 i + 48^{2} i^{2}$ $\displaystyle = 3,025-5,280 i + 48^{2} i^{2}$

$\displaystyle = 3,025-5,280 i + 2,304(-1)$

$\displaystyle = 3,025-5,280 i - 2,304$

$\displaystyle = 721-5,280 i$

Therefore, $(8 - 3i)^{4} = 721-5,280 i$ $\displaystyle (8 - 3i)^{4} = 721-5,280 i$

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Example Question #211 : Exponents

Raise 5+4i $\displaystyle 5+4i$ to the third power.

Possible Answers:

$\displaystyle 365 + 236i$

$\displaystyle -115 + 236i$

$\displaystyle -115 + 364i$

$\displaystyle 365 + 364i$

None of these

Correct answer:

$\displaystyle -115 + 236i$

Explanation:

To raise any expression A+B $\displaystyle A+B$ to the third power, use the pattern

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

Setting $\displaystyle A = 5, B = 4i$:

$(5+4i)^{3} = 5^{3} + 3\cdot 5^{2} \cdot 4i + 3 \cdot 5 \cdot (4i)^{2} + (4i)^{3}$ $\displaystyle (5+4i)^{3} = 5^{3} + 3\cdot 5^{2} \cdot 4i + 3 \cdot 5 \cdot (4i)^{2} + (4i)^{3}$

Taking advantage of the Power of a Product Rule:

$(5+4i)^{3} = 5^{3} + 3\cdot 5^{2} \cdot 4i + 3 \cdot 5 \cdot 4^{2} \cdot i^{2} +4 ^{3} \cdot i ^{3}$ $\displaystyle (5+4i)^{3} = 5^{3} + 3\cdot 5^{2} \cdot 4i + 3 \cdot 5 \cdot 4^{2} \cdot i^{2} +4 ^{3} \cdot i ^{3}$

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

$(5+4i)^{3} = 125 + 3\cdot 25 \cdot 4 i + 3 \cdot 5 \cdot 16 \cdot (-1) +64 \cdot (- i )$ $\displaystyle (5+4i)^{3} = 125 + 3\cdot 25 \cdot 4 i + 3 \cdot 5 \cdot 16 \cdot (-1) +64 \cdot (- i )$

$(5+4i)^{3} = 125+ 300i -240 -64i$ $\displaystyle (5+4i)^{3} = 125+ 300i -240 -64i$

Collecting real and imaginary terms:

$(5+4i)^{3} = 125 -240 + 300i -64i$ $\displaystyle (5+4i)^{3} = 125 -240 + 300i -64i$

$(5+4i)^{3} = -115 + 236i$ $\displaystyle (5+4i)^{3} = -115 + 236i$

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SAT Math : Algebra

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #41 : Squaring / Square Roots / Radicals

Example Question #44 : Squaring / Square Roots / Radicals

Example Question #45 : Squaring / Square Roots / Radicals

Example Question #46 : Squaring / Square Roots / Radicals

Example Question #2401 : Sat Mathematics

Example Question #621 : Algebra

Example Question #49 : Squaring / Square Roots / Radicals

Example Question #211 : Exponents

Example Question #51 : Squaring / Square Roots / Radicals

Example Question #211 : Exponents

All SAT Math Resources

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