Quadratic equations - Passport to advanced math - Math

PSAT/NMSQT Prep 2019 - Princeton Review 2019

Quadratic equations
Passport to advanced math
Math

CHAPTER OBJECTIVES

By the end of this chapter, you will be able to:

1. Solve quadratic equations via algebra, graphing, or the quadratic formula

2. Sketch the graph of a given quadratic equation

3. Identify how various components of a quadratic equation are significant to its graph or a real-world scenario

SMARTPOINTS

Point Value

SmartPoint Category

40 Points

Quadratics

Prepare

INTRODUCTION TO QUADRATIC EQUATIONS

A quadratic equation or expression is simply one that contains a squared variable (x 2) as the highest-order term (also called highest-powered term). In standard form, a quadratic equation is written as ax 2 + bx + c = 0, where a, b, and c are constants. However, quadratics can be written in a variety of other forms as well, such as the following:

x 2 − 9 = 0

2r 2 − 8r +10 = 4

2(x − 3)2 = 8

(x − 2)(x + 3) = 6

Note

At first glance, the last equation might not look quadratic, but it is; it’s merely masquerading as a product of binomials. You’ll learn a strategy for unveiling its x 2 term shortly.

All quadratic equations have 0, 1, or 2 real solutions. When you are asked to find the solutions of a quadratic equation, all you need to do is equate the variable to a constant. Solutions might also be called roots, x-intercepts, or zeros.

Before you can solve, however, there is a step you must always complete: set the equation equal to 0. In other words, move everything to one side of the equation so that 0 is the only thing left on the other side. Once the quadratic equation is equal to 0, you can take one of three routes to determine how many solutions it has: algebra, graphing, or the quadratic formula. Read on for more information about these three techniques.

SOLVING QUADRATICS ALGEBRAICALLY

Using algebra is often necessary when working with quadratic equations, so getting comfortable with it is critical. We’ll start with a technique that is highly useful for manipulating quadratics: FOIL. FOIL is essential for putting a quadratic into standard form.

Expert Tip

If you get stuck on the algebra in a question about a quadratic equation, Picking Numbers can often help. Just remember that it might take more time than the algebraic route, so use good judgment if you’re in a bind—and remember that you can always skip the question and revisit it later.

FOIL

Whenever you see a pair of binomials on the PSAT, your default algebra strategy should be FOIL, which stands for First, Outer, Inner, Last. This acronym helps ensure that you don’t forget any terms when distributing. Multiply the first terms in each binomial together, then repeat with the outer, inner, and last terms. Then add the four products together, combining like terms as needed. Here is a generic scheme for the FOIL procedure:

(a + b)(c + d) = ac + ad + bc + bd

(Binomial 1)(Binomial 2) = First + Outer + Inner + Last

It is often tempting to FOIL in your head, but this is risky: it is very easy to lose a negative sign or switch a pair of coefficients (and arrive at a trap answer). Show all of your work when using FOIL.

Factoring

Factoring, also known as reverse-FOILing, allows you to go from a quadratic to a product of two binomials. This is a very powerful tool; once you have a binomial pair, you’re a few short algebraic steps away from finding the solution(s). The factoring process for a quadratic equation that is written in standard form (ax 2 + bx + c) is demonstrated in the following table:

Step

Scratchwork

Starting point: notice a, the coefficient in front of x 2, is equal to 1, a great condition for ­factoring.



x
 2 + 5x + 6 = 0 → (x ± ?)(x ± ?) = 0

1. What are the factors of c? Remember to include negatives.

factors of 6:

1 & 6, −1 & −6, 2 & 3, −2 & −3

2. Which factor pair, when added, equals b, the coefficient in front of x?

2 + 3 = 5

3. Write as a product of binomials.

(x + 2)(x + 3) = 0

4. Split the product of binomials into two ­equations set equal to 0.

x + 2 = 0, x + 3 = 0

5. Solve each equation.

x = −2 , x = −3

Factoring is easiest when a is 1, so whenever possible, try to simplify the expression so that is the case. In addition, if you see nice-looking numbers (integers, simple fractions) in the answer choices, this is a clue that factoring is possible. If you’re ever not sure that you’ve done your factoring correctly, go ahead and FOIL to check your work. You should get the expression you started with.

Note

Sometimes, the two binomials factors will be identical. In this case, the quadratic equation will have only one real solution (because the two solutions are identical).

Completing the Square

For more difficult quadratics, you’ll need to turn to a more advanced strategy: completing the square. In this process, you’ll create a perfect square trinomial, which has the form (x + h)2 = k, where h and k are constants. This route takes some practice to master but will pay dividends when you sail through the most challenging quadratic equation questions on Test Day. The following table illustrates the procedure along with a corresponding example (even though the equation could have been factored).

Step

Scratchwork

Starting point:

x 2 + 6x − 7 = 0

1. Move the constant to the opposite side.

x 2 + 6x = 7

2. Divide b by 2, then square the quotient.

3. Add the number from the previous step to both sides of the equation, then factor.

x 2 + 6x + 9 = 7 + 9 → (x + 3)(x + 3) = 16 → (x +3)2 = 16

4. Take the square root of both sides.

5. Split the product into two equations and solve each one.

x + 3 = 4, x + 3 = − 4 → x = 1, x = −7

A note about completing the square: a needs to be 1 to use this process. You can divide the first term by a to convert the coefficient to 1, but if you start getting strange-looking fractions, it may be easier to use the quadratic formula instead.

Grouping

Although less commonly seen than other strategies, grouping is useful with more challenging quadratics, especially when the leading coefficient (the value of a) is not 1. You’ll need two x terms to use this route. The goal of grouping is to identify the greatest common factor (GCF) of the first two terms, repeat for the second two terms, then finally combine the two GCFs into a separate binomial. Check out the following example:

Step

Scratchwork

Starting point:

2x 2 − 7x − 15 = 0

1. You need to split the x term in two; the sum of the new terms’ coefficients must equal b, and their product must equal ac.

a × c = 2 × (−15) = −30, b = −7

new x term coefficients: 3 and −10

2x 2 − 10x + 3x − 15 = 0

2. What’s the GCF of the first pair of terms? How about the second pair of terms?

GCF of 2x 2 and −10x is 2x

GCF of 3x and —15 is 3



3. Factor out the GCFs for each pair of terms.

2x 2 − 10x + 3x − 15 = 0

2x(x − 5) + 3(x − 5) = 0

4. Factor out the newly formed binomial and combine the GCFs into another factor.

2x(x − 5) + 3(x − 5) = 0

(2x + 3)(x − 5) = 0

5. Split into two equations and solve as usual.

Straightforward Math

Sometimes you can get away with not having to FOIL or factor extensively, but you need to be able to spot patterns or trends. Don’t resort to complex techniques when some easy simplification will get the job done. Equations similar to the following examples are highly likely to appear on the PSAT.

No Middle Term

No Last Term

Squared Binomial

x 2 − 9 = 0

x 2 − 9x = 0

(x − 3)2 = 9

x 2 = 9

x(x − 9) = 0

(x − 3) =

x =

x = 0, x — 9 = 0

(x − 3) = ±3

x = ±3

x = 0, x = 9

x — 3 = 3 → x = 6

x — 3 = —3 → x = 0

Expert Tip

You can also factor x 2 − 9 to get (x + 3)(x − 3); this is called a difference of squares. Note that this only works when the terms are being subtracted.

Quadratic Formula

The quadratic formula can be used to solve any quadratic equation. However, because the math can often get complicated, use this as a last resort or when you need to find exact (i.e., not rounded, fractions, and/or radicals) solutions. If you see square roots in the answer choices, this is a clue to use the quadratic formula.

The quadratic formula that follows yields solutions to a quadratic equation that is written in standard form, ax 2 + bx + c = 0:

The ± sign that follows —b indicates that you will have two solutions, so remember to find both.

The expression under the radical (b2 − 4ac) is called the discriminant, and its value determines the number of real solutions. If this quantity is positive, the equation has two distinct real solutions; if it is equal to 0, there is only one distinct real solution; and if it is negative, there are no real solutions.

Note

Being flexible and familiar with your strengths on Test Day is essential. By doing so, you can identify the path to the answer to a quadratics question that is the most efficient for you.

On the next few pages, you’ll get to try applying some of these strategies to test-like PSAT questions. Let’s start with a FOIL question:

1. Which of the following is an equivalent form of the expression (x — 4)(x + 2) ?

1.

2.

3.

4.

Work through the Kaplan Method for Math step-by-step to solve this question. The following table shows Kaplan’s strategic thinking on the left, along with suggested math scratchwork on the right.

Strategic Thinking

Math Scratchwork

Step 1: Read the question, identifying and organizing important information as you go

You’re asked to identify the quadratic expression equivalent to (x — 4)(x + 2).


Step 2: Choose the best strategy to answer the question

Are you presented with anything familiar in the question stem? How about in the answer choices? What’s the best route to the answer?

You have a product of two binomials in the stem and quadratic expressions written in standard form in the answer choices, so FOIL is the quickest route.

Follow the standard FOIL procedure, then simplify.

First + Outer + Inner + Last

Step 3: Check that you answered the right question

You correctly expanded the quadratic using FOIL and got an exact match for (B), the correct answer.





Expert Tip

Although you could pick numbers here, remember you have only a few seconds to solve questions like this on Test Day. FOIL is much faster and should be your preferred method.

Use the strategies you’ve learned in this section to simplify the rational expression that follows:

2. Which of the following is equivalent to ?

1.

2.

3.

4.

Expert tip

When you encounter a quadratic expression in the numerator and/or denominator of a rational expression, try to factor the quadratics. Chances are that one or more factors in the numerator will cancel with one or more factors in the denominator. Always factor the easier quadratic first, which may provide a hint as to how to factor the more difficult quadratic.

Work through the Kaplan Method for Math step-by-step to solve this question. The following table shows Kaplan’s strategic thinking on the left, along with suggested math scratchwork on the right.

Strategic Thinking

Math Scratchwork

Step 1: Read the question, identifying and organizing important information as you go

You need to find the simplified expression that is equivalent to the one given.


Step 2: Choose the best strategy to answer the question

What familiar pieces do you see?

There are a few x 2 terms, so you should be thinking about quadratics and factoring. Also, whenever you’re given a fraction, think about ways to cancel terms.

What should be done first?

Examine the numerator first. It’s an example of a perfect square trinomial, so it’s easy to factor. The denominator is a bit more involved. Factor out a 2 to get 1 for the x 2 coefficient, then factor the quadratic as usual. If the numerator and denominator have any factors in common, cancel those factors.

The answer doesn’t match any of the choices. Did I make a mistake?

Not necessarily. In this case, you’ll notice your choice is very close to (C); all you need to do is factor a —1 out of your answer’s numerator to get the whole expression to match.

Step 3: Check that you answered the right question

After factoring out a negative, the expression matches (C).





Expert Tip

Something seemingly trivial like a negative sign can separate your answer from the correct answer. If you’ve checked your math for errors but found none, look for ways to alter your answer’s appearance so that it matches an answer choice. When in doubt, pick numbers and plug a few test cases into your expression and the answer choices to find a match.

3. What value of x satisfies the equation ?

1.

2.

3. 3

4.

This question is full of radicals, but don’t panic. You can use the Kaplan Method for Math to efficiently tackle this kind of question on Test Day. The following table shows Kaplan’s strategic thinking on the left, along with suggested math scratchwork on the right.

Strategic Thinking

Math Scratchwork

Step 1: Read the question, identifying and organizing important information as you go

The question asks for a solution to the given equation.


Step 2: Choose the best strategy to answer the question

The equation is a quadratic, so think about factoring. Remember to first put the equation in standard form.

How can you make factoring easier here?

Divide both sides of the equation by 4 to get an x 2 coefficient of 1.

The number 2 does not have factors that add up to 6. What can you do besides factoring?

You only have three terms with no perfect squares, so grouping and square rooting are out. Try completing the square instead.

Now that you have two possible values of x, what should you do next?

Identify the solution that is among the answer choices.





Step 3: Check that you answered the right question

One of your solutions is an exact match for (A), the correct answer.


CONNECTIONS BETWEEN QUADRATICS AND PARABOLAS

A quadratic function is simply a quadratic equation set equal to y or f (x) instead of 0. To solve one of these, you would follow the same procedure as before: substitute 0 for y, or f (x), then solve using one of the three methods demonstrated (algebra, graphing, quadratic formula). Consider the graphical connection: when you set y equal to 0, you’re really finding the x-intercepts.

The graph of every quadratic equation (or function) is a parabola, which is a symmetric U-shaped graph that opens either up or down. To determine whether a parabola will open up or down, examine the value of a in the equation. If a is positive, the parabola will open up; if a is negative, it will open down. Take a look at the examples below to see this graphically.

Like quadratic equations, quadratic functions will have zero, one, or two real solutions, corresponding to the number of times the parabola crosses the x-axis. As you saw with previous examples, graphing is a powerful way to determine the number of solutions a quadratic function has.

Two Real Solutions

One Real Solution

No Real Solutions

There are three algebraic forms that a quadratic equation can take: standard, factored, and vertex. Each is provided in the following table along with some features that are revealed by writing the equation in that particular form.

Standard

Factored

Vertex

y = ax2 + bx + c

y = a(xm)(xn)

y = a(xh)2 + k

y-intercept is c.

Solutions are m and n.

Vertex is (h, k).

In real-world contexts, starting quantity is c.

x-intercepts are m and n.

Minimum/maximum of function is k.

Format used to solve via quadratic formula.

Vertex is halfway between m and n.

Axis of symmetry is given by x = h.

You’ve already seen standard and factored forms earlier in this chapter, but vertex form might be new to you. In vertex form, a is the same as the a from standard form, and h and k are the coordinates of the vertex (h, k). If a quadratic function is not in vertex form, you can still find the x-coordinate of the vertex by plugging the appropriate values into the equation , which is also the equation for the axis of symmetry (see graph that follows). Once you determine h, plug this value into the quadratic function and solve for y to determine k, the y-coordinate of the vertex.

In addition to familiarity with the various forms a quadratic equation/function can take, you should have a foundational knowledge of the structure of a parabola. Some of the basic pieces you could be asked about on Test Day are shown above. You already know how to determine the solutions and the vertex, and finding the axis of symmetry is straightforward. The equation of the axis of symmetry of a parabola is x = h, where h is the x-coordinate of the vertex.

Note

The formula for a parabola’s axis of symmetry is easy to remember: it’s the quadratic formula without the radical component. If the x-intercepts are rational numbers, you can also determine the axis of symmetry by finding the midpoint, the point exactly halfway between.

Take some time to explore the questions on the next several pages to test your new wealth of quadratic knowledge.

A question like this next one could arise in either the calculator or the no-calculator section. Think critically about how you’d solve it in either case.

4. Will the graph of intersect the graph of ?

1. Yes, only at the vertex of the parabola

2. Yes, once on each side of the vertex

3. Yes, twice to the right of the vertex

4. No, the graphs will not intersect

Work through the Kaplan Method for Math step-by-step to solve this question. The following table shows Kaplan’s strategic thinking on the left, along with suggested math scratchwork on the right. Because this question could occur in either the calculator or the no-calculator section, graphical and algebraic solutions are included.

Strategic Thinking

Math Scratchwork

Step 1: Read the question, identifying and organizing important information as you go

You are given a linear function and a quadratic function and asked whether their graphs intersect.

Step 2: Choose the best strategy to answer the question

What should you find first, the vertex or the intersection points?

Because the equations may not even intersect, ignore the vertex for now. You can always find it later, if need be.

How can you discover whether these intersect without graphing?

Set the equations equal to each other and combine all of your terms on one side to get a quadratic in standard form. Factor to find the solutions.

Your quadratic has two solutions; what does this mean?

The two graphs intersect at two points so you can eliminate A and D.

and

How can you determine where the vertex of the parabola is?

Use the formulas for h and k. Putting a parabola in standard form into vertex form is too time-consuming.

Where are the points of intersection with respect to the vertex?

Compare the two solutions you found earlier to the x-coordinate of the vertex. They are both to the right of the vertex. The correct answer is (C).

Vertex: (−2, −5)

How can your calculator make this question much easier?

You can simply graph the equations on the same graph and visually estimate the vertex and intersection points. Make sure you set an appropriate viewing window. You need to be able to see the vertex of the parabola and the points where the graphs intersect (if they do). Upon further investigation, it is clear that there are two intersection points to the right of the vertex, which matches (C).

image

Step 3: Check that you answered the right question

Using either method, you found that the line intersects the parabola at two locations to the right of the vertex, which is what the question asked for.


Expert Tip

Sometimes questions involving complex algebra or simplification will appear in the calculator section. Avoid autopilot. Use your calculator to cut out as much algebra as you can. Remember, you don’t get extra points for solving questions the hard way on the PSAT!

In one final type of quadratic-related question, you may be asked to match a function to a graph, or vice versa. An example of this follows; unfortunately, it is not likely to appear in the calculator section of the test:

A parabola graphed on a coordinate plane. The parabola opens downward, has vertex three comma two, and crosses the x-axis at two and at four.

5. Which of the following represents the function shown in the graph?

1.

2.

3.

4.

Use the Kaplan Method for Math to work through this question step-by-step. The following table shows Kaplan’s strategic thinking on the left, along with suggested math scratchwork on the right.

Strategic Thinking

Math Scratchwork

Step 1: Read the question, identifying and organizing important information as you go

Notice that the answer choices all represent quadratic functions written in vertex form. Your task is to determine which matches the graph.


Step 2: Choose the best strategy to answer the question

Use what you know about vertex form to systematically eliminate choices. What is the vertex of the function in the graph? Which choices can you eliminate?

vertex = (3, 2)

Eliminate C and D.

How might you determine which remaining function matches without using a calculator?

Check a few key points. The graph crosses the x-axis at x = 2 and x = 4. Check those points first.

key points: (2, 0) and (4, 0)

Choice A:

f  (2) = −(2 − 3)2 + 2 = −1 + 2 = 1

1 ≠ 0

Eliminate A.

Choice B:

f (2) = −(2−3)2 + 2 = —2 + 2 = 0

0 = 0

Step 3: Check that you answered the right question

Choice (B) must be the correct answer.


Note

Remember to check for differences in the a values of the answer choices. In the previous example, all the choices have a negative a, so a isn’t much help here. However, it might help you eliminate incorrect answers on Test Day, so always check it.

GRAPHING QUADRATIC EQUATIONS ON A CALCULATOR

At this point, you’ve become quite an expert at working with quadratics on paper. In this part, we’ll explore how you can use your calculator to efficiently graph quadratics. Calculators can be great time-savers when you’re allowed to use them.

Graphing

All quadratic equations can be solved by graphing, unless they happen to have imaginary solutions. That said, you might ask why you need to learn all the algebra techniques. There are a few reasons why graphing shouldn’t be the first option you turn to:

· Remember, there’s a no-calculator section on the PSAT; graphing isn’t an option here.

· Graphing is often slower because entering complex equations and then zooming in to find points of interest can be tedious.

· It is easy to accidentally mistype when you’re being timed—a misplaced parenthesis or negative sign will likely lead to a trap answer choice.

However, if you have complicated algebra ahead (e.g., fractional coefficients), decimals in the answer choices, or time-consuming obstacles to overcome, graphing can be a viable alternative to solving quadratic equations algebraically. Here is a set of straightforward steps for graphing on a calculator:

1. Manipulate the equation so that it equals 0.

2. Substitute y = for the 0.

3. Enter the equation into your calculator.

4. Trace the graph to approximate the x-intercepts (usually the answer choices will be sufficiently different to warrant an approximation over an exact value) or use your calculator’s built-in capability to find the x-intercepts exactly.

Graphing on the TI-83/84

While on the home screen, press [Y=]. Then enter the function to be graphed. Press [GRAPH] and allow the function to plot. If you can’t see everything or want to make sure there isn’t something hiding, consider pressing [WINDOW] to set your own manual parameters or hitting [ZOOM] to quickly zoom in and out. If you want to simply investigate your graph, press [TRACE] and use the right and left arrow keys to move around on the graph. If you type in any x-value and press [ENTER], the y-value will be returned on screen.

Determining Solutions on the TI-83/84

Once you have your graph on screen, you’re ready to find solutions. Press [2ND] [TRACE] to pull up the CALC menu, which has options for finding points of interest. Select option 2:ZERO by highlighting and pressing [ENTER]. You will be taken back to the graph. Use the arrow keys to move to the left of the x-intercept (zero) that you want to calculate. Once you are just to the left of only the zero you are interested in, press [ENTER]—this is called the Left Bound. Next, move to the right of that zero only, careful not to go past any others, and press [ENTER]—this is called the Right Bound. Finally, the calculator will ask you to “Guess,” so move left or right to approximate this zero, and press [ENTER].

Because you’ve already set the quadratic equation equal to zero, you know the zeros that your calculator returns will be the solutions to the overall equation.

Note

Take the time to get comfortable with your calculator functions regardless of what calculator you have. You can find great instructions and even video demonstrations on the Internet.

Next, you’ll get to try a sample test-like question that could be solved via graphing or the quadratic formula. Choose wisely. In almost every case, graphing will be faster, but familiarize yourself with the quadratic formula approach in case you encounter a question like this in the no-calculator section:

6. Which of the following are the real values of x that satisfy ?

1. 3 and −2

2. and

3. 0

4. No real solutions

Work through the Kaplan Method for Math step-by-step to solve this question. The following table shows Kaplan’s strategic thinking on the left, along with suggested math scratchwork on the right.

Strategic Thinking

Math Scratchwork

Step 1: Read the question, identifying and organizing important information as you go

You need to identify the values of x that satisfy the given equation.


Step 2: Choose the best strategy to answer the question

With quadratics, you have a few options: solve algebraically, graphically, or via the quadratic formula.

What is the first step?

Set the quadratic equal to 0 by subtracting 5x from each side of the equation. Once there, you’ll see that solving algebraically is not wise, as the equation doesn’t look easy to factor. Graphing and the quadratic formula will be quicker.

Graphical Approach

After plugging into your calculator, you get the graph on the right. The graph does not cross the x-axis, so there are no real solutions.



image

Quadratic Formula (No-Calculator) Approach

Plug in the coefficients and the constant carefully. It helps to jot down the numbers you’ll be using: a = 3, b = —3, and c = 4. You’ll notice that the discriminant will be negative, meaning there are no real solutions.

Don’t worry about simplifying—that’s beyond the scope of this question.

Step 3: Check that you answered the right question

This equation has no real solutions, which is (D).


Note

Don’t worry if graphing is still somewhat foreign to you; the next section has more examples to get you comfortable with this route.

Nicely done! Take a look at another example:

7. The equation is satisfied when x = 4 and when x = −3. What is the value of 2k ?

1. —6

2. —3

3.

4. 3

Although this question seems more complicated than others you’ve seen in this chapter, if you use the Kaplan Method for Math, you’ll arrive at the correct answer. The following table shows Kaplan’s strategic thinking on the left, along with suggested math scratchwork on the right.

Strategic Thinking

Math Scratchwork

Step 1: Read the question, identifying and organizing important information as you go

You’re asked to find the value of 2k.


Step 2: Choose the best strategy to answer the question

Notice that the equation in the question stem is not in standard form. Distributing the won’t result in unmanageable fractions, so doing so won’t cost you a lot of time. After distributing, set the equation equal to 0.

The “normal” routes to the solutions (factoring, etc.) would be difficult to take here because of the presence of k. Instead, use the solutions to construct and FOIL two binomials.

The quadratic expressions must be equal because they share the same solutions. Set them equal to each other, and then use algebra to solve for k.







Step 3: Check that you answered the right question

Be careful! Many students will bubble in choice B.

You’re asked for 2k. Multiply your previous result by 2, and select (A), the correct response.

As demonstrated, even the most daunting quadratic equation questions are made more straightforward by using the Kaplan Method for Math.

Practice

Now you’ll have a chance to try a few test-like questions in a scaffolded way. We’ve provided some guidance, but you’ll need to fill in the missing parts of explanations or the step-by-step math in order to get to the correct answer. Don’t worry—after going through the worked examples at the beginning of this section, these questions should be completely doable.

8. A projectile is launched from a cannon on top of a building. Its height in feet can be described as a function of elapsed time according to the following quadratic equation: f (t) = —16t 2 + 128t + 320. What is the product of the maximum height of the projectile and the time it takes the projectile to hit the ground?

1. 2,048

2. 3,842

3. 4,096

4. 5,760

Use the scaffolding in the table that follows as your map through the question. If you aren’t sure where to start, fill in the blanks in the table as you work from top to bottom.

Strategic Thinking

Math Scratchwork

Step 1: Read the question, identifying and organizing important information as you go

You’re asked to find the product of the maximum height and the time the object takes to hit the ground.


Step 2: Choose the best strategy to answer the question

How can you approach this systematically? What component of the parabola would correspond to the maximum height?

Hint: this parabola opens down. You know the maximum height would be at the vertex.

Which coordinate would give height, h or k?

k = f(h)

k = —16(_)2 + 128(_) + 320

k = _____

Are you finished? Make sure you calculate the time it takes to hit the ground. What height will the projectile be at when it hits the ground?

The projectile will be at a height of zero. Set up an equation and solve for the time. Hint: you can easily factor this quadratic.

What are the possible solutions for t? Which of these solutions doesn’t make sense in the context of the question? Is there anything left to do here?

0 = __________

0 = (________)(________)

t = ________

t = ________

Step 3: Check that you answered the right question

Did you calculate the product of the maximum height and the time the projectile spent in the air?

If you answered (D), you’re absolutely right.

product = _____ x _____ = _____

If you’re up for a challenge, try the next question.

9. Which of the following functions has x-intercepts at x = —2 and x = 5 ?

1.

2.

3.

4.

Use the scaffolding that follows as your map through the question. If you aren’t sure where to start, fill in the blanks in the table as you work from top to bottom.

Strategic Thinking

Math Scratchwork

Step 1: Read the question, identifying and organizing important information as you go

You’re asked to find the function that crosses the x-axis at the x-values of —2 and 5.


Step 2: Choose the best strategy to answer the question

What will the y-coordinate of your function be at the x-intercept?

The y-coordinate will be 0 when the function crosses the x-axis.

Can you eliminate any answer choices quickly by plugging in x values? Can you immediately tell that some choices will not have zeroes at these x values?

Yes! Start with the ones that are easy to calculate. As soon as you get a nonzero result, go ahead and eliminate the answer choice.

f (—2) = _______

f (5) = _______

A: f (—2) = ____________

C: f (5) = ____________

Now that you’ve got it down to two choices, look critically at them. What is the same in both choices?

Notice that in each case, you’ll be subtracting .

What is a decimal approximation for ? Don’t worry about accuracy . . . what two integers is it between?

Now use number properties to narrow it down. If you’re looking for f(x) = 0, you’ll need to subtract from something pretty close to that decimal value. Are you looking for a number greater or less than 52? Should you add or subtract when you plug in x = 5?

Do you need to go any further?

No. There are no other possible matching answer choices.


52 = _______

which is much _______ than .

Therefore, you should



(add / subtract) .

Step 3: Check that you answered the right question

If your answer is (B), you’re correct! Choice D would result in a squared quantity that is much greater than 25, such that subtracting would never give you a zero.


Expert Tip

Notice that we didn’t do a lot of calculating in the previous question. Remember, the PSAT tests math and critical thinking, not computation. Don’t calculate a quantity unless you absolutely have to. Calculations take time and are often unnecessary to eliminate answer choices. Instead, think about the relationships between numbers.

Perform

Now that you’ve seen the variety of ways in which the PSAT can test you on quadratics, try the following questions to check your understanding. Give yourself 4 minutes to tackle the following four questions. Make sure you use the Kaplan Method for Math as often as you can. Remember, you want to emphasize speed and efficiency in addition to simply getting the correct answer.

10.What positive value(s) of z satisfy the equation    ?

1. 2

2. 2 and —10

3. 4 and 2

4. None of the above

11.If a quadratic function f (x) has solutions a and b such that a < 0, b > 0, and | b | > | a |, which of the following could be f (x) ?

1.

2.

3.

4.

12.The positive difference between the zeros of is equal to what value?

image

13.Which equation represents the axis of symmetry of the graph of the quadratic function ?

1. x = 2

2. x = 1

3. x = −1

4. x = 0

On your own

1. Which of the following are solutions to the quadratic equation ?

1. x = −6, x = 4

2.

3. ,

4. ,

2. The factored form of a quadratic equation is y = (x + 2)(x — 7), and the standard form is y = x 2 — 5x — 14. Which of the following statements accurately describes the graph of y ?

1. The x-intercepts are —7 and 2, and the y-intercept is —14.

2. The x-intercepts are —2 and 7, and the y-intercept is —14.

3. The x-intercepts are —7 and 2, and the y-intercept is 14.

4. The x-intercepts are —2 and 7, and the y-intercept is 14.

3. image

The following quadratic equations are all representations of the graph shown. Which equation best reveals the values of the x-intercepts of the graph?

1.

2.

3.

4.

4. If x 2 + 8x = 48 and x > 0, what is the value of x — 5 ?

1. —9

2. —1

3. 4

4. 7

5. Which of the following equations has the same solutions as x 2 + 6x +17 = 0 ?

1. y = (x — 3)2 — 26

2. y = (x — 3)2 + 8

3. y = (x + 3)2 + 8

4. y = (x + 3)2 + 17

6.

If (x, y) is a solution to the system of equations shown here, what is the value of x 2 ?

1. 6

2. 36

3. 144

4. 1,296

7. Which of the following are the roots of the equation x 2 + 8x — 3 = 0 ?

1.

2.

3.

4.

8. If a catapult is used to throw a lead ball, the path of the ball can be modeled by a quadratic equation, y = ax 2 + bx + c, where x is the horizontal distance that the ball travels and y is the height of the ball. If one of these catapult-launched lead balls travels 150 feet before hitting the ground and reaches a maximum height of 45 feet, which of the following equations represents its path?

1. y = −0.008x 2 + 1.2x

2. y = −0.008x 2 — 150x

3. y = 45x 2 + 150x

4. y = 125x 2 + 25x

9. image

If the distance from —a to b in the figure shown is 10, which of the following could be the factored form of the graph’s equation?

1. y = (x — 7)(x — 3)

2. y = (x — 7)(x + 3)

3. y = (x — 8)(x — 2)

4. y = (x — 1)(x + 10)

10.image

If y = ax 2 + bx + c represents the equation of the graph shown in the figure, which of the following statements is not true?

1. The value of a is a negative number.

2. The value of c is a negative number.

3. The graph is increasing for x < 3 and decreasing for x > 3.

4. The zeros of the equation are x = −2 and x = 8.

11.Which of the following equations could represent a parabola that has a minimum value of —3 and whose axis of symmetry is the line x = 2 ?

1. y = (x — 3)2 + 2

2. y = (x + 3)2 + 2

3. y = (x — 2)2 — 3

4. y = (x + 2)2 — 3

12.If the axis of symmetry of the parabola given by the equation y = —(x + 5)2 + 1 is x = p, then what is the value of p ?

1. —5

2. —1

3. 1

4. 5

13.image

Shawna throws a baseball into the air. The equation h = −12t 2 + 36t + 4 represents the height of the ball in feet, t seconds after it was thrown. The graph of part of the equation is shown in the previous figure. Which of the following equations could represent the height of a second ball that was thrown by Meagan, if Meagan’s ball did not go as high as Shawna’s ball?

1. h = −12t 2 + 35t + 7

2. h = −12t 2 + 37t + 3

3. h = −3(2t — 1)2 + 35

4. h = −3(2t — 2)2 + 28

14.Given the equation y = −2(x — 6)2 + 5, which of the following statements is not true?

1. The y-intercept is (0, 5).

2. The axis of symmetry is x = 6.

3. The vertex is (6, 5).

4. The parabola opens downward.

15.The x-coordinates of the solutions to a system of equations are —8 and —3. Which of the following could be the system?

1.

2.

3.

4.

16.

If (a, b) is a solution to the system of equations shown, what is the value of a, given that a > 0 ?

image