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Algebra II : Algebra II

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Example Questions

Example Question #473 : Intermediate Single Variable Algebra

Use the quadratic equation to solve x^2-2x+1

Possible Answers:

x=1

x=2

x=-2

x=-1

Correct answer:

x=1

Explanation:

We use the quadratic equation to solve for x. The quadratic equation is:

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

In our case a=1, b=-2, c=1

Substituting these values into the quadratic equation we get:

$x=\frac{2+\sqrt{(-2)^2-4(1)(1)}}{2(1)}, x=\frac{2-\sqrt{(-2)^2-4(1)(1)}}{2(1)}$

$x=\frac{2+\sqrt{(4-4}}{2}, x=\frac{2-\sqrt{4-4}}{2}$

$x=\frac{2+0}{2}, x=\frac{2-0}{2}$

x=1, x=1

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Example Question #15 : Quadratic Formula

The height of a kicked soccer ball can be modeled with the equation

H(x)=-4.9x^2 +15x +0.5 ,

where the height is given in meters and is the time in seconds. At what time(s) will the ball be 2 meters off the ground?

Possible Answers:

x=0.412 seconds

x=1.244 seconds

x=0.104 seconds

x=2.958 seconds

x=1.244 seconds

x=0.104 seconds

Correct answer:

x=0.104 seconds

x=2.958 seconds

Explanation:

Set up the equation to solve for the time when the height is at 2 meters:

2=-4.9x^2+15x-0.5

Now put the equation into quadratic form ax^2+bx+c=0 so that we can solve it using the quadratic formula

-4.9x^2+15x-1.5=0 .

The quadratic equation is

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

where a=-4.9 , b=15 , and c=-1.5 .

Solving for gives us two possible values,

x=0.104 seconds

x=2.958 seconds.

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Example Question #471 : Intermediate Single Variable Algebra

Solve for .

x^2+4x+5=0

Possible Answers:

x=-2

$\textup{Answer is not present}$

x=2, -2

x=0, -4

Correct answer:

$\textup{Answer is not present}$

Explanation:

When applying the quadratic formula, the discriminant (portion under the square root) is negative and so there are no real roots of the equation shown.

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Example Question #475 : Intermediate Single Variable Algebra

Solve for x

(x+2)^2+2x+2=0

Possible Answers:

x=-9, -3

$x=3\pm\sqrt{3}$

x=-1, -5

$x=-3\pm{\sqrt{3}}$

Correct answer:

$x=-3\pm{\sqrt{3}}$

Explanation:

Once the square is multiplied out and the equation simplified, it yields x^2+6x+6=0 , a good time for the quadratic formula, $x=\frac{-b\pm{\sqrt{b^2-4ac}}}{2a}$ where a, b, c are the coefficients of the polynominal in descending order. Plug in a=1, b=6, c=6, and it yields $-3\pm{\frac{\sqrt{12}}{2}}$ , multiply out the square root and it yields $-3\pm{\sqrt{3}}$ .

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Example Question #476 : Intermediate Single Variable Algebra

Using the quadratic equation, find the roots of the following expression.

2x^2+5x - 12

Possible Answers:

x=-4

x=-1.5

x = -4

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

x =-4

x=1.5

x=4

x = 1.5

No real solutions

Correct answer:

x =-4

x=1.5

Explanation:

To find the roots of the quadratic expression, we must use the quadratic equation

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

Plugging in our values for , , and (, , and , respectively) we get the equation:

$x = \frac{-5\pm\sqrt{25-4(2)(-12)}}{2\cdot2}$

First, let's simplify the radical:

$x=\frac{-5\pm\sqrt{25+96}}{4}$

which becomes

$x=\frac{-5\pm\sqrt{121}}{4}$

$x=\frac{-5\pm11}{4}$

Now that we've simplified the radical, we need to solve for both solutions:

$x_1=\frac{-5+11}{4}=\frac{6}{4}=\frac{3}{2}=1.5$

and

$x_2=\frac{-5-11}{4}=\frac{-16}{4}=-4$

Therefore, the roots of this quadratic expression are x=1.5 and x=-4 .

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Example Question #311 : Quadratic Equations And Inequalities

Find the roots of the following equation using the quadratic formula:

f(x) = 3x^2 + 8x - 3

Express in simplest form.

Possible Answers:

$x = \frac{-8\pm10}{6}$

$x = \frac{-4\pm5}{3}$

$x = \frac{-8\pm2\sqrt{7}}{6}$

$x = \frac{-4\pm\sqrt{7}}{3}$

$x = \frac{-8\pm\sqrt{28}}{6}$

Correct answer:

$x = \frac{-4\pm5}{3}$

Explanation:

Remember the quadratic equation. For any quadratic polynomial, ax^2 + bx + c , the roots of the function are given by:

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

In this situation, we have 3x^2 + 8x -3 , so a = 3, b = 8, c = -3 .

Substituting into the formula, we get the roots at:

$x = \frac{-8\pm\sqrt{8^2-4(3\cdot-3)}}{2\cdot3}$

Simplifying gives us:

$x = \frac{-8\pm\sqrt{64+36}}{6} = \frac{-8\pm\sqrt{100}}{6} = \frac{-8\pm10}{6} = \frac{-4\pm5}{3}$ .

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Example Question #2 : How To Use The Quadratic Function

Which of the following is the correct solution when x^2-10x-11=0 is solved using the quadratic equation?

Possible Answers:

$\frac{-10\pm 12}{2}$

$10\pm 12$

$\frac{10\pm \sqrt{56}}{2}$

$\frac{10\pm 12}{2}$

Correct answer:

$\frac{10\pm 12}{2}$

Explanation:

$x=\frac{-(-10)\pm \sqrt{(-10)^2-4(1)(-11)}}{2(1)}$

$=\frac{10\pm \sqrt{100+44}}{2}$

$=\frac{10\pm \sqrt{144}}{2}$

$=\frac{10\pm 12}{2}$

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Example Question #311 : Quadratic Equations And Inequalities

Solve the equation using the quadratic formula.

$x^2+5x+\frac{9}{4}=0$

Possible Answers:

7,9

$\frac{1}{2}, \frac{9}{2}$

-5,2

$-\frac{1}{2}, -\frac{9}{2}$

Correct answer:

$-\frac{1}{2}, -\frac{9}{2}$

Explanation:

The quadratic formula is

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

Setting a=1 , b=5 , $c=\frac{9}{4}$ yields,

$\\ \frac{-5\pm\sqrt{5^2-4(1)(\frac{9}{4})}}{2(1)}\\ \\=\frac{-5\pm\sqrt{25-9}}{2} \\ \\=-\frac{5}{2}\pm2 \\ \\ = -\frac{1}{2}, -\frac{9}{2}$

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Example Question #481 : Intermediate Single Variable Algebra

Find the roots of: y=x^2-x+100

Possible Answers:

$\pm10$

Since the quadratic cannot be factored, there are no roots.

$\frac{1\pm \sqrt{401}}{2}$

$\frac{1\pm i\sqrt{399}}{2}$

$\frac{1}{2}\pm i\sqrt{399}$

Correct answer:

$\frac{1\pm i\sqrt{399}}{2}$

Explanation:

Identify the values of , , and in the standard form of the parabola.

y=ax^2+bx+c

a= 1

b=-1

c=100

Calculate the discriminant.

b^2-4ac=1-4(1)(100)=-399

Since the discriminant is less than zero, the quadratic is irreducible and there are no real roots. However, there are complex roots. Use the quadratic formula to determine the complex roots.

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

$x= \frac{1\pm \sqrt{-399}}{2(1)}$

$x= \frac{1\pm i\sqrt{399}}{2}$

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Example Question #482 : Intermediate Single Variable Algebra

Use the quadratic formula to find the roots of x^2+5x-14 .

Possible Answers:

x= 0

x=5 and x=-14

x=7 and x=-2

no solution

x=-7 and x=2

Correct answer:

x=-7 and x=2

Explanation:

The parent function of a quadratic is represented as ax^2+bx+c . The quadratic formula is $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ . In this case a=1 , b=5 , and c=-14 . Replacing these values into the quadratic forumula will give you the solutions to the quadratic.

$x=\frac{-5\pm \sqrt{5^2-4(1)(-14)}}{2*1}$

$x=\frac{-5\pm \sqrt{25-(-56)}}{2}$

$x=\frac{-5\pm \sqrt{81}}{2}$

$x=\frac{-5\pm 9}{2}$

x=-7 and x=2

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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #473 : Intermediate Single Variable Algebra

Example Question #15 : Quadratic Formula

Example Question #471 : Intermediate Single Variable Algebra

Example Question #475 : Intermediate Single Variable Algebra

Example Question #476 : Intermediate Single Variable Algebra

Example Question #311 : Quadratic Equations And Inequalities

Example Question #2 : How To Use The Quadratic Function

Example Question #311 : Quadratic Equations And Inequalities

Example Question #481 : Intermediate Single Variable Algebra

Example Question #482 : Intermediate Single Variable Algebra

All Algebra II Resources

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