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High School Math : High School Math

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Example Questions

Example Question #3 : Absolute Value

$\textup{What are all possible values of }b\textup{ when }\left | 3b-4\right |=11\textup{ ?}$

Possible Answers:

$\frac{19}{3}, -5$

$5, -\frac{7}{3}$

$5, -\frac{3}{7}$

$5, -\frac{11}{3}$

$\frac{7}{3}, -5$

Correct answer:

$5, -\frac{7}{3}$

Explanation:

$\textup{Set up two equations and solve for both solutions:}$

$3b-4=11\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;3b-4 = -11$

$3b = 15\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \; \; \; \; \; \; 3b=-7$

$b = 5\;\;\:\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;b=-\frac{7}{3}$

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Example Question #14 : Algebra Ii

Solve:

$|x-5|\leqslant 0$

Possible Answers:

x=5

All real numbers

$x=5 \; or \; x=-5$

No solution

$-5\leqslant x\leqslant 5$

Correct answer:

x=5

Explanation:

The absolute value can never be negative, so the equation is ONLY valid at zero.

The equation to solve becomes x-5=0 .

x=5

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Example Question #11 : Algebra Ii

Solve for :

$\left | 2x-5\right |=3x$

Possible Answers:

x=-5,0

x=-1,5

x=-5

x=1

x=5

Correct answer:

x=1

Explanation:

To solve absolute value equations, we must understand that the absoute value function makes a value positive. So when we are solving these problems, we must consider two scenarios, one where the value is positive and one where the value is negative.

$\left | 2x-5\right |=3x$

2x-5=3x

and

2x-5=-(3x)

This gives us:

-5=x and 5x=5

$x=-5\ and\ 1$

However, this question has an outside of the absolute value expression, in this case . Thus, any negative value of will make the right side of the equation equal to a negative number, which cannot be true for an absolute value expression. Thus, x=-5 is an extraneous solution, as $\left | 2x-5\right |$ cannot equal a negative number.

Our final solution is then

x=1

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Example Question #12 : Algebra Ii

Solve for .

$3\left | x+2 \right |=18$

Possible Answers:

x=-8,-4

x=4

x=8,4

x=8,-4

x=-8,4

Correct answer:

x=-8,4

Explanation:

$3\left | x+2 \right |=18$

Divide both sides by 3.

$\left | x+2 \right |=6$

Consider both the negative and positive values for the absolute value term.

$x+2=6\ or\ x+2=-6$

Subtract 2 from both sides to solve both scenarios for .

$x=-8\ or\ x=4$

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Example Question #13 : Algebra Ii

Find the -intercepts for the graph given by the equation:

$y=\left | 2x-6\right |-8$

Possible Answers:

$x=1\ and\ 7$

$x=-7\ and\ 1$

$x=-1\ and\ 7$

$x=-1\ and\ 1$

$x=-7\ and\ 1$

Correct answer:

$x=-1\ and\ 7$

Explanation:

To find the -intercepts, we must set y=0 .

$y=\left | 2x-6\right |-8$

$0=\left | 2x-6\right |-8$

$\left | 2x-6\right |=8$

Now we must set up our two scenarios:

2x-6=8 and 2x-6=-8

2x=14 and 2x=-2

x=7 and x=-1

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Example Question #1 : Understanding Imaginary And Complex Numbers

Simplify the radical.

$\sqrt{-250}$

Possible Answers:

15i

$25i\sqrt{10}$

No solution

$5i\sqrt{10}$

Correct answer:

$5i\sqrt{10}$

Explanation:

$\sqrt{-250}$

First, factor the term in the radical.

-250=(-1)(25)(10)

$\sqrt{-250}=(\sqrt{-1})(\sqrt{25})(\sqrt{10})$

Now, we can simplify.

$\sqrt{-1}=i\ \text{and}\ \sqrt{25}=5$

$(\sqrt{-1})(\sqrt{25})(\sqrt{10})=(i)(5)(\sqrt{10})$

$5i\sqrt{10}$

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Example Question #2 : Understanding Imaginary And Complex Numbers

Multiply: (3 + 2i)(4 - 3 i)

Possible Answers:

6+ i

6-i

6 - 17i

18-17i

18-i

Correct answer:

18-i

Explanation:

FOIL:

(3 + 2i)(4 - 3 i)

$= 3 \cdot 4 -3 \cdot 3i + 2i \cdot4 - 2i \cdot3i$

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

= 12 -9i + 8i - 6 (-1)

= 12 -9i + 8i +6

= 18-i

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Example Question #1 : Imaginary Numbers

Multiply: (5+4i) (5-4i)

Possible Answers:

25+16i

25-16i

Correct answer:

Explanation:

Since 5+4i and 5-4i are conmplex conjugates, they can be multiplied according to the following pattern:

$(5+4i) (5-4i) = 5 ^{2} + 4^{2} = 25 + 16 = 41$

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Example Question #1 : Imaginary Numbers

Multiply:

$\left (7 + i \sqrt{2 } \right )\left (7 - i \sqrt{2 } \right )$

Possible Answers:

$51 + 2i\sqrt{2}$

49-2i

$51 - 2i\sqrt{2}$

Correct answer:

Explanation:

Since $7 + i \sqrt{2 }$ and $7 - i \sqrt{2 }$ are conmplex conjugates, they can be multiplied according to the following pattern:

$\left (7 + i \sqrt{2 } \right )\left (7 - i \sqrt{2 } \right ) = 7^{2} +\left ( \sqrt{2} \right )^{2} = 49 + 2 = 51$

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Example Question #5 : Understanding Imaginary And Complex Numbers

Evaluate: $i ^{45}$

Possible Answers:

45i

Correct answer:

Explanation:

$i ^{N}$ can be evaluated by dividing by 4 and noting the remainder. Since $45 \div 4 = 11 \textrm{ R } 1$ - that is, since dividing 45 by 4 yields remainder 1:

$i ^{45} = i ^{1} = i$

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All High School Math Resources

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High School Math : High School Math

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #3 : Absolute Value

Example Question #14 : Algebra Ii

Example Question #11 : Algebra Ii

Example Question #12 : Algebra Ii

Example Question #13 : Algebra Ii

Example Question #1 : Understanding Imaginary And Complex Numbers

Example Question #2 : Understanding Imaginary And Complex Numbers

Example Question #1 : Imaginary Numbers

Example Question #1 : Imaginary Numbers

Example Question #5 : Understanding Imaginary And Complex Numbers

All High School Math Resources

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