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PSAT Math : Algebra

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Example Questions

Example Question #1 : Squaring / Square Roots / Radicals

According to Heron's Formula, the area of a triangle with side lengths of a, b, and c is given by the following:

where s is one-half of the triangle's perimeter.

What is the area of a triangle with side lengths of 6, 10, and 12 units?

Possible Answers:

48√77

4√14

8√14

14√2

12√5

Correct answer:

8√14

Explanation:

We can use Heron's formula to find the area of the triangle. We can let a = 6, b = 10, and c = 12.

In order to find s, we need to find one half of the perimeter. The perimeter is the sum of the lengths of the sides of the triangle.

Perimeter = a + b + c = 6 + 10 + 12 = 28

In order to find s, we must multiply the perimeter by one-half, which would give us (1/2)(28), or 14.

Now that we have a, b, c, and s, we can calculate the area using Heron's formula.

Hero2

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Example Question #1 : How To Factor A Common Factor Out Of Squares

Simplify the radical expression.

$\sqrt[3]{-64x^{6}y^{12}}$

Possible Answers:

$-4x^{2}y^{4}$

4ix^2y^4

-2x^3y^4

2ix^3y^4

Correct answer:

$-4x^{2}y^{4}$

Explanation:

$\sqrt[3]{-64x^{6}y^{12}}$

Look for perfect cubes within each term. This will allow us to factor out of the radical.

$\sqrt[3]{-64}*\sqrt[3]{x^6}*\sqrt[3]{y^{12}}$

$\sqrt[3]{-4*-4*-4}*\sqrt[3]{x^2*x^2*x^2}*\sqrt[3]{y^4*y^4*y^4}$

Simplify.

$-4x^{2}y^4$

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Example Question #2 : Squaring / Square Roots / Radicals

Simplify the expression.

$\sqrt{5}(2\sqrt{3}+\sqrt{12})$

Possible Answers:

$3\sqrt{2}$

$4\sqrt{30}$

$4\sqrt{15}$

$10\sqrt{3}$

$10\sqrt{15}$

Correct answer:

$4\sqrt{15}$

Explanation:

Use the distributive property for radicals.

$\sqrt{5}(2\sqrt{3}+\sqrt{12})$

Multiply all terms by $\sqrt{5}$ .

$(\sqrt{5}*2\sqrt{3})+(\sqrt{5}*\sqrt{12})$

Combine terms under radicals.

$2\sqrt{3*5}+\sqrt{12*5}$

$2\sqrt{15}+\sqrt{60}$

Look for perfect square factors under each radical. has a perfect square of . The can be factored out.

$2\sqrt{15}+\sqrt{4*15}$

$2\sqrt{15}+2\sqrt{15}$

Since both radicals are the same, we can add them.

$4\sqrt{15}$

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Example Question #1 : How To Divide Complex Numbers

Simplify by rationalizing the denominator:

$\frac{20 + 6i}{8i}$

Possible Answers:

$-\frac{ 3 }{4} + \frac{ 5}{2}i$

$\frac{ 5}{2} - \frac{ 3 }{4}i$

$\frac{ 3 }{4} - \frac{ 5}{2}i$

$\frac{ 3 }{4} + \frac{ 5}{2}i$

$- \frac{ 5}{2} + \frac{ 3 }{4}i$

Correct answer:

$\frac{ 3 }{4} - \frac{ 5}{2}i$

Explanation:

Multiply both numerator and denominator by :

$\frac{20 + 6i}{8i}$

$=\frac{\left (20 + 6i \right ) \cdot i}{8i \cdot i}$

$=\frac{ 20i + 6i^{2} }{8i ^{2}}$

$=\frac{ 20i + 6(-1) }{8 (-1)}$

$=\frac{ 20i -6 }{-8}$

$=\frac{ -6+ 20i }{-8}$

$=\frac{ -6 }{-8} + \frac{ 20i }{-8}$

$=\frac{ 3 }{4} - \frac{ 5}{2}i$

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Example Question #1 : How To Divide Complex Numbers

Divide:

$-100 \div 5i$

Possible Answers:

20i

20 - 20i

-20i

20 + 20 i

Correct answer:

20i

Explanation:

Rewrite in fraction form, and multiply both numerator and denominator by :

$-100 \div 5i$

$= \frac{-100}{5i}$

$= \frac{-100 \cdot i }{5i \cdot i }$

$= \frac{-100 i }{5i ^{2}}$

$= \frac{-100 i }{5 (-1)}$

$= \frac{-100 i }{-5}$

$= \frac{-100 }{-5}\cdot i$

= 20i

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Example Question #2 : Complex Numbers

Simplify by rationalizing the denominator:

$\frac{ 25- 12i}{ - 10i}$

Possible Answers:

$\frac{ 5 }{ 2} + \frac{6}{ 5} i$

$\frac{6}{ 5} - \frac{ 5 }{ 2}i$

$-\frac{ 5 }{ 2} + \frac{6}{ 5} i$

$-\frac{6}{ 5} + \frac{ 5 }{ 2}i$

$\frac{6}{ 5} + \frac{ 5 }{ 2}i$

Correct answer:

$\frac{6}{ 5} + \frac{ 5 }{ 2}i$

Explanation:

Multiply both numerator and denominator by :

$\frac{ 25- 12i}{ - 10i}$

$=\frac{\left ( 25- 12i \right )\cdot i}{ - 10i \cdot i }$

$=\frac{ 25i- 12i ^{2} }{ - 10i ^{2} }$

$=\frac{ 25i- 12 (-1)}{ - 10 (-1) }$

$=\frac{ 25i+ 12}{ 10}$

$=\frac{12+ 25i }{ 10}$

$=\frac{12 }{ 10} + \frac{ 25i }{ 10}$

$=\frac{6}{ 5} + \frac{ 5 }{ 2}i$

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Example Question #3 : How To Divide Complex Numbers

Simplify by rationalizing the denominator:

$\frac{ 21i}{3+ i\sqrt{5}}$

Possible Answers:

$-\frac{ 21 \sqrt{5} }{ 4}+\frac{63 }{ 4} i$

$\frac{ 3\sqrt{5} }{ 2} - \frac{ 9 }{ 2} i$

$\frac{ 3\sqrt{5} }{ 2} + \frac{ 9 }{ 2} i$

$\frac{ 21 \sqrt{5} }{ 4}+ \frac{63 }{ 4} i$

$-\frac{ 3\sqrt{5} }{ 2} + \frac{ 9 }{ 2} i$

Correct answer:

$\frac{ 3\sqrt{5} }{ 2} + \frac{ 9 }{ 2} i$

Explanation:

Multiply the numerator and the denominator by the complex conjugate of the denominator, which is $3- i\sqrt{5}$ :

$\frac{ 21i}{3+ i\sqrt{5}}$

$= \frac{ 21i\left (3- i\sqrt{5} \right )}{\left (3+ i\sqrt{5} \right )\left (3- i\sqrt{5} \right )}$

$= \frac{21i \cdot 3- 21i \cdot i\sqrt{5} }{ 3^{2}+ \left ( \sqrt{5} \right )^{2}}$

$= \frac{ 63i - 21i^{2} \sqrt{5}}{ 9+5}$

$= \frac{ 63i - 21 (-1)\sqrt{5}}{ 14}$

$= \frac{ 21 \sqrt{5}+ 63i }{ 14}$

$= \frac{ 21 \sqrt{5} }{ 14}+ \frac{63 i}{ 14}$

$= \frac{ 3\sqrt{5} }{ 2} + \frac{ 9 }{ 2} i$

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Example Question #2 : How To Divide Complex Numbers

Simplify by rationalizing the denominator:

$\frac{2+3i}{4+ 3i}$

Possible Answers:

$-\frac{1 }{25} + \frac{ 18 }{25} i$

$\frac{17 }{25} + \frac{ 6 }{25} i$

$\frac{17}{7} + \frac{ 6 }{7} i$

-1 + 18i

$-\frac{1 }{7} + \frac{ 18 }{7} i$

Correct answer:

$\frac{17 }{25} + \frac{ 6 }{25} i$

Explanation:

Multiply the numerator and the denominator by the complex conjugate of the denominator, which is 4 - 3i :

$\frac{2+3i}{4+ 3i}$

$=\frac{\left (2+3i \right )\left (4-3i \right )}{\left (4+ 3i \right )\left (4-3i \right )}$

$=\frac{2 \cdot 4 - 2 \cdot 3i +3i \cdot 4 - 3i \cdot 3i }{4^{2}+3^{2}}$

$=\frac{8 - 6i +12i - 9 i ^{2}}{16+ 9}$

$=\frac{8 - 6i +12i - 9(-1)}{25}$

$=\frac{8 - 6i +12i +9}{25}$

$=\frac{17 + 6i }{25}$

$=\frac{17 }{25} + \frac{ 6 }{25} i$

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Example Question #72 : Exponents

Simplify by rationalizing the denominator:

$\frac{15i}{4- 3i}$

Possible Answers:

$\frac{45 }{7} + \frac{ 60}{7}i$

$-\frac{9}{5} + \frac{ 12}{5}i$

$-\frac{45 }{7} + \frac{ 60}{7}i$

$\frac{9}{5} + \frac{ 12}{5}i$

$\frac{9}{5} -\frac{ 12}{5}i$

Correct answer:

$-\frac{9}{5} + \frac{ 12}{5}i$

Explanation:

Multiply the numerator and the denominator by the complex conjugate of the denominator, which is 4 +3i :

$\frac{15i}{4- 3i}$

$= \frac{15i\left (4+ 3i \right )}{\left (4- 3i \right )\left (4+ 3i \right )}$

$= \frac{15i\cdot 4+ 15i\cdot 3i }{4^{2}+3^{2}}$

$= \frac{60 i + 45i^{2} }{16+9}$

$= \frac{60 i + 45 (-1)}{25}$

$= \frac{-45 + 60 i}{25}$

$= \frac{-45 }{25} + \frac{ 60 i}{25}$

$= - \frac{9}{5} + \frac{ 12}{5}i$

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Example Question #511 : Algebra

Define an operation $\ast$ as follows:

For all complex numbers a, b ,

$a \ast b = \frac{ai}{b}$ .

Evaluate $(2+ i) \ast (2-i)$ .

Possible Answers:

$- \frac{4}{5} +\frac {3 }{5}i$

$-\frac {3 }{5}i$

$\frac {3 }{5}i$

$\frac{4}{5} -\frac {3 }{5}i$

Correct answer:

$- \frac{4}{5} +\frac {3 }{5}i$

Explanation:

$a \ast b = \frac{ai}{b}$

$(2+ i) \ast (2-i)= \frac{(2+ i)i}{2-i}$

$= \frac{2i+ i^{2}}{2-i}$

$= \frac{2i+ (-1)}{2-i}$

$= \frac{-1+2i }{2-i}$

Multiply the numerator and the denominator by the complex conjugate of the denominator, which is 2 + i . The above becomes:

$\frac{\left (-1+2i \right )\left (2+i \right )}{\left (2-i \right )\left (2+i \right )}$

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

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

$= \frac{-2 -i+ 4i + 2(-1) }{5}$

$= \frac{-2 -i+ 4i -2 }{5}$

$= \frac{-4 +3i }{5}$

$=- \frac{4}{5} +\frac {3 }{5}i$

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PSAT Math : Algebra

Study concepts, example questions & explanations for PSAT Math

All PSAT Math Resources

Example Questions

Example Question #1 : Squaring / Square Roots / Radicals

Example Question #1 : How To Factor A Common Factor Out Of Squares

Example Question #2 : Squaring / Square Roots / Radicals

Example Question #1 : How To Divide Complex Numbers

Example Question #1 : How To Divide Complex Numbers

Example Question #2 : Complex Numbers

Example Question #3 : How To Divide Complex Numbers

Example Question #2 : How To Divide Complex Numbers

Example Question #72 : Exponents

Example Question #511 : Algebra

All PSAT Math Resources

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