Algebra II : Intermediate Single-Variable Algebra

Study concepts, example questions & explanations for Algebra II

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Example Questions

Example Question #41 : Sequences

The product of two consective positive odd integers is 143. Find both integers.

Possible Answers:

Correct answer:

Explanation:

If  is one odd number, then the next odd number is . If their product is 143, then the following equation is true.

Distribute into the parenthesis.

Subtract 143 from both sides.

This can be solved by factoring, or by the quadratic equation. We will use the latter.

We are told that both integers are positive, so .

The other integer is .

Example Question #1 : How To Use The Quadratic Function

Solve for :

Possible Answers:

The solution is undefined.

Correct answer:

Explanation:

To factor this equation, first find two numbers that multiply to 35 and sum to 12.  These numbers are 5 and 7.  Split up 12x using these two coefficients:

 

Example Question #2 : Quadratic Formula

A baseball that is thrown in the air follows a trajectory of , where  is the height of the ball in feet and  is the time elapsed in seconds. How long does the ball stay in the air before it hits the ground?

Possible Answers:

Between 3.5 and 4 seconds

Between 2.5 and 3 seconds

Between 2 and 2.5 seconds

Between 3 and 3.5 seconds

 Between 4 and 4.5 seconds 

Correct answer:

Between 3 and 3.5 seconds

Explanation:

To solve this, we look at the equation .

Setting the equation equal to 0 we get .

Once in this form, we can use the Quadratic Formula to solve for .

The quadratic formula says that if , then 

.

Plugging in our values:

 

Therefore or  and since we are looking only for positive values (because we can't have negative time), 3.4375 seconds is our answer.

Example Question #2 : Quadratic Formula

Use the quadratic formula to solve the following equation.

Possible Answers:

Correct answer:

Explanation:

First we want to put the equation into standard form; we do this by making sure the equation is = 0, so let's subtract 4 from both sides.

We could go straight to the quadratic formula from here, but quadratics are always easier to solve if we eliminate Greatest Common Factors first. In this case the GCF is 4 so let's divide both sides by 4.

Now we can compare against the standard form to find a, b, and c.

By pattern matching, we see that a = 1, b = -3, and c = 4. Now we substitute into the quadratic formula.

We can evaluate the square root of a negative number by factoring out the square root of -1 and calling it the imaginary number i. This gives our answer.

Example Question #3 : Quadratic Formula

Use the quadratic formula to solve the following equation:

Possible Answers:

Correct answer:

Explanation:

Since the equation is already in standard form, we can compare it to the general version of standard form to find a, b, and c.

By pattern matching it is clear that a = 2, b = 4, and c = -3. Now we can substitute into the quadratic formula.

Recall that when subtracting a negative number [as in - 4(2)(-3)], the result is addition:

Combine terms and simplify the radical:

Now eliminate common factors to reduce to lowest terms.

Example Question #2 : Quadratic Formula

Use the quadratic formula to find the solutions of the following equation:

Possible Answers:

Correct answer:

Explanation:

We begin by designating values of a, b and c by comparing the equation to the standard form.

By pattern matching it is clear that a = 2, b = -3, and c = -5. We can now substitute into the quadratic formula:

Take note that in the two cases where negatives/minus signs are multiplied together [-(-3) and -4(2)(-5)], they bceome positive:

Now we simplify and evaluate.

Note that in the previous step we listed the subtraction instance first, as that instance yields the smaller number and it is usually convenient to start sets with smaller numbers first. 

Example Question #6 : Quadratic Formula

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 

No solution

 and 

 and 

Correct answer:

No solution

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:

Since you have to take the square root of a negative number, this means that there are no real solutions.

Example Question #3 : Quadratic Formula

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 

 and 

No solution

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:

Use a calculator to determine the final values.

 

Example Question #8 : Quadratic Formula

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:

Use a calculator to determine the final values.

Example Question #11 : Quadratic Formula

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

Possible Answers:

 and 

No solution

 and 

 and 

 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.

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