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High School Math : Algebra II

Study concepts, example questions & explanations for High School Math

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8 Diagnostic Tests 613 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #11 : Understanding Quadratic Equations

Write an equation with the given roots:

$\frac{3}{4}, \frac{1}{3}$

Possible Answers:

12x^2-13x+3=0

13x^2-13x+3=0

14x^2-13x+3=0

13x^2-12x+2=0

13x^2-12x+3=0

Correct answer:

12x^2-13x+3=0

Explanation:

To write an equation, find the sum and product of the roots. The sum is the negative coefficient of , and the product is the integer.

Sum: $(\frac{3}{4})+(\frac{1}{3})=\frac{13}{12}$

Product: $(\frac{3}{4})\cdot (\frac{1}{3})=\frac{1}{4}$

Subtract the sum and add the product.

The equation is:

$x^2 - \frac{13}{12}x+\frac{1}{4}=0$

Multiply the equation by :

12x^2+-13x+3=0

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Example Question #52 : Intermediate Single Variable Algebra

Write an equation with the given roots:

$2 \pm \sqrt{3}$

Possible Answers:

x^2-4x+1=0

x^2-3x+1=0

x^2-3x+2=0

x^2-2x+2=0

x^2-4x+2=0

Correct answer:

x^2-4x+1=0

Explanation:

To write an equation, find the sum and product of the roots. The sum is the negative coefficient of , and the product is the integer.

Sum: $(2 + \sqrt{3})+(2 - \sqrt{3})=4$

Product: $(2 + \sqrt{3})\cdot (2 - \sqrt{3})=4-3=1$

Subtract the sum and add the product.

The equation is:

x^2 -4x+1=0

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Example Question #53 : Intermediate Single Variable Algebra

Write an equation with the given roots:

$5 \pm 3i$

Possible Answers:

x^2-10x+14=0

x^2-10x+34=0

x^2-8x+24=0

x^2-8x+34=0

x^2-10x+24=0

Correct answer:

x^2-10x+34=0

Explanation:

To write an equation, find the sum and product of the roots. The sum is the negative coefficient of , and the product is the integer.

Sum: (5 + 3i)+(5-3i)=10

Product: $(5+3i)\cdot (5-3i)=25-9i^2=34$

Subtract the sum and add the product.

The equation is:

x^2-10x+34=0

Report an Error

Example Question #51 : Intermediate Single Variable Algebra

Write an equation with the given roots:

$-\frac{3}{5}, \frac{1}{2}$

Possible Answers:

11x^2+x-3=0

10x^2+x-3=0

7x^2+x-3=0

8x^2+x-3=0

9x^2+x-3=0

Correct answer:

10x^2+x-3=0

Explanation:

To write an equation, find the sum and product of the roots. The sum is the negative coefficient of , and the product is the integer.

Sum: $(\frac{-3}{5})+(\frac{1}{2})=\frac{-1}{10}$

Product: $(\frac{-3}{5})\cdot (\frac{1}{2})=\frac{-3}{10}$

Subtract the sum and add the product.

The equation is:

$x^2 + \frac{1}{10}x-\frac{3}{10}=0$

Multiply the equation by :

10x^2+x-3=0

Report an Error

Example Question #1 : Understanding Quadratic Roots

Write an equation with the given roots:

$4 \pm \sqrt{3}$

Possible Answers:

x^2-8x+12=0

2x^2-9x+13=0

x^2-7x+13=0

x^2-9x+13=0

x^2-8x+13=0

Correct answer:

x^2-8x+13=0

Explanation:

To write an equation, find the sum and product of the roots. The sum is the negative coefficient of , and the product is the integer.

Sum: $(4 + \sqrt{3})+(4 - \sqrt{3})=8$

Product: $(4 + \sqrt{3})\cdot (4 - \sqrt{3})=16-3=13$

Subtract the sum and add the product.

The equation is:

x^2 -8x+13=0

Report an Error

Example Question #51 : Intermediate Single Variable Algebra

Find the zeros.

$f(t)=t^{2}-4t-21$

Possible Answers:

x=7,3

x=-7,3

x=7,-7

x=-3,7

x=3,-3

Correct answer:

x=-3,7

Explanation:

Factor the equation to (t-7)(t+3) . Set both equal to zero and you get and . Remember, the zeros of an equation are wherever the function crosses the -axis.

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Example Question #52 : Intermediate Single Variable Algebra

Find the zeros.

$f(x)=x^{2}-10x$

Possible Answers:

x=0,10

x=-10

x=10

x=0

x=1,-1

Correct answer:

x=0,10

Explanation:

Factor out an from the equation so that you have x(x-10) . Set and x-10 equal to . Your roots are and .

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Example Question #53 : Intermediate Single Variable Algebra

Find the zeros.

$f(x)=\frac{1}{5}(x+9)^{2}(x^{2}-4)$

Possible Answers:

x=-9,-2,2

x=-9

x=-3,3,-9

x=-2,2

x=2,-9

Correct answer:

x=-9,-2,2

Explanation:

Set (x+9)^2 equal to zero and you get . Set (x^2-4) equal to zero as well and you get and because when you take a square root, your answer will be positive and negative.

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Example Question #1 : Solving Quadratic Equations

Find the zeros.

$f(x)=27x^{2}-45x-18$

Possible Answers:

$x=\frac{1}{3},2$

$x=\frac{1}{3}$

$x=9,-\frac{1}{3},2$

$x=-\frac{1}{3},2$

x=-2,-9

Correct answer:

$x=-\frac{1}{3},2$

Explanation:

Factor out a from the entire equation. After that, you get 9(3x^2-5x-2) . Factor the expression to (3x+1)(x-2) . Set both of those equal to zero and your answers are $-\frac{1}{3}$ and .

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Example Question #2 : Solving Quadratic Equations

Find the zeros.

$f(x)=25x^{2}-81$

Possible Answers:

$x=-\frac{9}{5}$

x=9,-9

x=5

$x=\frac{9}{5}$

$x=\frac{9}{5},-\frac{9}{5}$

Correct answer:

$x=\frac{9}{5},-\frac{9}{5}$

Explanation:

This expression is the difference of perfect squares. Therefore, it factors to (5x-9)(5x+9) . Set both of those equal to zero and your answers are $\frac{9}{5}$ and $-\frac{9}{5}$ .

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High School Math : Algebra II

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #11 : Understanding Quadratic Equations

Example Question #52 : Intermediate Single Variable Algebra

Example Question #53 : Intermediate Single Variable Algebra

Example Question #51 : Intermediate Single Variable Algebra

Example Question #1 : Understanding Quadratic Roots

Example Question #51 : Intermediate Single Variable Algebra

Example Question #52 : Intermediate Single Variable Algebra

Example Question #53 : Intermediate Single Variable Algebra

Example Question #1 : Solving Quadratic Equations

Example Question #2 : Solving Quadratic Equations

All High School Math Resources

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