Algebra II : Intermediate Single-Variable Algebra

Study concepts, example questions & explanations for Algebra II

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Example Questions

Example Question #471 : Intermediate Single Variable Algebra

Solve for . Use the quadratic formula to find your solution. Use a calculator to estimate the value to the closest hundredth.

Possible Answers:

 and 

 and 

 and 

No solution

 and 

Correct answer:

 and 

Explanation:

Recall that the quadratic formula is defined as:

For this question, the variables are as follows:

Substituting these values into the equation, you get:

Separate this expression into two fractions and simplify to determine the final values.

Example Question #471 : Intermediate Single Variable Algebra

Solve the quadratic equation with the quadratic formula.

Possible Answers:

Correct answer:

Explanation:

Based on the quadratic equation:

,

, and 

Given the quadratic formula:

We have:

Simplfying,

 

Example Question #473 : Intermediate Single Variable Algebra

Use the quadratic equation to solve 

Possible Answers:

Correct answer:

Explanation:

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

In our case 

Substituting these values into the quadratic equation we get:

Example Question #15 : Quadratic Formula

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

,

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:

 seconds

 seconds

or

 seconds

 seconds 

or

 seconds

 seconds

 seconds

Correct answer:

 seconds 

or

 seconds

Explanation:

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

Now put the equation into quadratic form  so that we can solve it using the quadratic formula

.

The quadratic equation is

,

where ,  , and .

Solving for  gives us two possible values,

 seconds

or

 seconds.

 

 

 

Example Question #471 : Intermediate Single Variable Algebra

Solve for .

Possible Answers:

Correct answer:

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.

Example Question #475 : Intermediate Single Variable Algebra

Solve for x

Possible Answers:

Correct answer:

Explanation:

Once the square is multiplied out and the equation simplified, it yields , a good time for the quadratic formula,  where a, b, c are the coefficients of the polynominal in descending order. Plug in a=1, b=6, c=6, and it yields , multiply out the square root and it yields .

Example Question #476 : Intermediate Single Variable Algebra

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

Possible Answers:

No real solutions

Correct answer:

Explanation:

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

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

First, let's simplify the radical:

which becomes

or

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

and

Therefore, the roots of this quadratic expression are  and .

Example Question #311 : Quadratic Equations And Inequalities

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

Express in simplest form.

Possible Answers:

Correct answer:

Explanation:

Remember the quadratic equation. For any quadratic polynomial, , the roots of the function are given by:

In this situation, we have , so .

Substituting into the formula, we get the roots at:

Simplifying gives us:

.

Example Question #11 : Quadratic Formula

Which of the following is the correct solution when    is solved using the quadratic equation?

Possible Answers:

Correct answer:

Explanation:

Example Question #311 : Quadratic Equations And Inequalities

Solve the equation using the quadratic formula.

Possible Answers:

Correct answer:

Explanation:

The quadratic formula is

.

Setting , ,  yields,

 

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