SAT II Math I : Solving Quadratic Functions

Study concepts, example questions & explanations for SAT II Math I

varsity tutors app store varsity tutors android store

Example Questions

← Previous 1

Example Question #1 : Solving Quadratic Functions

A baseball is thrown straight up with an initial speed of 50 feet per second by a man standing on the roof of a 120-foot high building. The height of the baseball in feet, as a function of time in seconds , is modeled by the function

To the nearest tenth of a second, how long does it take for the baseball to hit the ground?

Possible Answers:

Correct answer:

Explanation:

When the baseball hits the ground, the height is 0, so we set . and solve for .

This can be done using the quadratic formula:

Set :

One possible solution:

 

We throw this out, since time must be positive.

The other:

This solution, we keep. The baseball hits the ground in about 4.7 seconds.

Example Question #2 : Solving Quadratic Functions

Define  and  .

Find 

Possible Answers:

Correct answer:

Explanation:

By definition, , so

Example Question #1 : Solving Quadratic Equations

Factor the above function to find the roots of the quadratic equation.

Possible Answers:

Correct answer:

Explanation:

Factoring a quadratic equation means doing FOIL backwards. Recall that when you use FOIL, you start with two binomials and end with a trinomial:

Now, we're trying to go the other direction -- starting with a trinomial, and going back to two factors.

Here, -3 is equal to , and -2 is equal to . We can use this information to find out what and  are, separately. In other words, we have to find two factors of -3 that add up to -2.

Factors of -3:

  • 3*-1 (sum = 2)
  • -3*1 (sum = -2)

Thus our factored equation should look like this:

The roots of the quadratic equation are the values of x for which y is 0.

We know that anything times zero is zero. So the entire expression equals zero when at least one of the factors equals zero.

Example Question #334 : Intermediate Single Variable Algebra

Find the roots of the function:

Possible Answers:

Correct answer:

Explanation:

Factor:

Double check by factoring:

Add together:

Therefore:

Example Question #1 : Finding Zeros Of A Polynomial

Solve for x.

Possible Answers:

x = –4, –3

x = 5

x = 5, 2

x = 4, 3

x = –5, –2

Correct answer:

x = 5, 2

Explanation:

1) Split up the middle term so that factoring by grouping is possible.

Factors of 10 include:

1 * 10= 10    1 + 10 = 11

2 * 5 =10      2 + 5 = 7

–2 * –5 = 10    –2 + –5 = –7 Good!

2) Now factor by grouping, pulling "x" out of the first pair and "-5" out of the second.

3) Now pull out the common factor, the "(x-2)," from both terms.

4) Set both terms equal to zero to find the possible roots and solve using inverse operations.

x – 5 = 0,  x = 5

x – 2 = 0, x = 2

Example Question #2 : Finding Roots

Solve for x.

Possible Answers:

x = 5, 2

x = –5, –2

x = –4

x = 2

x = –4, 4

Correct answer:

x = –4

Explanation:

1) First step of solving any equation: combine like terms. With quadratics, the easiest step to take is to set the expression equal to zero.

2) There are two ways to do this problem. The first and most intuitive method is standard factoring.

16 + 1 = 17

8 + 2 = 10

4 + 4 = 8

3) Then follow the usual steps, pulling out the common factor from both pairs, "x" from the first and "4" from the second.

4) Pull out the "(x+4)" to wind up with:

5) Set each term equal to zero.

x + 4 = 0, x = –4

But there's a shortcut! Assuming the terms are arranged by descending degree (i.e., ), and the third term is both a perfect square whose square root is equal to half of the middle term, mathematicians use a little trick. In this case, the square root of 16 is 4. 4 * 2=8, so the trick will work. Take the square root of the first and last term, then stick a plus sign in between them and square the parentheses. 

And x, once again, is equal to –4.

Example Question #1 : Solving Quadratic Functions

Solve for :

Possible Answers:

Correct answer:

Explanation:

To solve for , you need to isolate it to one side of the equation. You can subtract the  from the right to the left. Then you can add the 6 from the right to the left:

Next, you can factor out this quadratic equation to solve for . You need to determine which factors of 8 add up to negative 6:

Finally, you set each binomial equal to 0 and solve for :

Example Question #3 : Use And Evaluate Square Roots And Cube Roots: Ccss.Math.Content.8.Ee.A.2

Solve the equation:

Possible Answers:

Correct answer:

Explanation:

Add 8 to both sides to set the equation equal to 0:

To factor, find two integers that multiply to 24 and add to 10.  4 and 6 satisfy both conditions. Thus, we can rewrite the quadratic of three terms as a quadratic of four terms, using the the two integers we just found to split the middle coefficient:

Then factor by grouping:

Set each factor equal to 0 and solve:

and

Example Question #2 : Finding Zeros Of A Polynomial

Find the roots of the following quadratic expression:

Possible Answers:

Correct answer:

Explanation:

First, we have to know that "finding the roots" means "finding the values of x which make the expression =0." So basically we are going to set the original expression = 0 and factor.

This quadratic looks messy to factor by sight, so we'll use factoring by composition. We multiply a and c together, and look for factors that add to b.

So we can use 8 and -3. We will re-write 5x using these numbers as 8x - 3x, and then factor by grouping.

Note the extra + sign we inserted to make sure the meaning is not lost when parentheses are added. Now we identify common factors to be "pulled" out.

Now we factor out the (3x + 4).

Setting each factor = 0 we can find the solutions.

So the solutions are x = 1/2 and x = -4/3, or {-4/3, 1/2}.

 

 

Example Question #3 : Solving Quadratic Functions

Find the roots of the following quadratic expression.

Possible Answers:

Correct answer:

Explanation:

First we remember that "find the roots" means "find the values of x for which this expression equals 0." So we set the expression = 0 and approach solving as normal.

Since solving this by sight is difficult, we'll use composition, multiplying a by c and finding factors which add to b.

So -9 and 5 will work; we will use them to rewrite -4x as -9x + 5x and then factor by grouping.

We identify common factors to "pull" out of each group.

And now we factor out x-3.

Setting each factor equal to 0 lets us solve for x. 

So our solutions are x = -5/3 and x = 3, which we write as x = {-5/3, 3}.

← Previous 1
Learning Tools by Varsity Tutors

Incompatible Browser

Please upgrade or download one of the following browsers to use Instant Tutoring: