Equations

Algebra: Equations

Read the following SAT test question and then select the correct answer. 

You should read every math problem on the SAT carefully.  Identify your bottom line, assess your options for reaching it, and then select the most efficient method to attack the problem.  Once you have an answer, loop back to make sure that it matches your bottom line.

y = x² - 4x + c

In the quadratic equation above, c is a constant. The graph of the equation in the xy-plane contains the points (-2, 0) and (6,0). What is the value of c?

Bottom Line: c = ?

Assess your Options:  For this problem, you are given two points (x, y).  That means that you could plug in the x and y values for either point and solve for c (this is the method that most students will use):

(-2, 0)	(6, 0)
0 = (-2)² - 4(-2) + c	0 = (6)² - 4(6) + c
0 = 4 – (-8) + c	0 = 36 – 24 + c
0 = 12 + c	0 = 12 + c
-12 = c	-12 = c

Isolating the variable in either of these two equations will get you the correct value of c.  However, notice how many steps there are.  Can you just look at the two points and know the answer?  Yes!  Think about you find the roots of an equation.

Attack the Problem:  Your original equation is already set equal to zero.  You know this because both of the points have a y value of 0.  In order to factor a polynomial, you need two binomials.  Here you already know that x = -2 or 6.  That means that your two binomials are (x + 2) and (x – 6).  Now c is the last number that you would get in your polynomial if you multiplied (x + 2) by (x – 6).  What number is that? -12!  

All you had to do was multiply the last two numbers (2 × -6) because every other combination would have an x.  If you don’t see how that works, multiply out (x + 2)(x – 6):

Use FOIL (First, Inner, Outer, Last)

x² + 2x – 6x – 12 (combine like terms)

x² - 4x – 12

When you compare this equation to the original equation, you will see that in place of the c you now have a -12.

Loop Back:  During a test, you would never work through a problem three times (time waster!), so this is where you would check to make sure that you solved for the correct variable.  Look down at your answer choices.

(A) -12

(B) -6

(C) 4

(D) 6

(E) 12

The correct answer is (A).

On sat.collegeboard.org, 51% of the responses were correct.

For more help with SAT math, visit www.myknowsys.com!

Quadratic Equations

Read the following question and select the correct answer choice.

As always, remember to follow the Knowsys method for math. Read the problem carefully and identify the bottom line (what you are looking for). Then, consider your options. How could you solve it? How should you solve it? Next, attack the problem using the method that you selected. Finally, loop back and verify that your answer matches the bottom line.

If , which of the following statements must also be true?

. 

.

. 

This problem is going to be a tricky one. On the actual SAT, this would probably be one of the last problems in a math section (and since problems go in order from easiest to hardest, this gives you a clue that this problem is a difficult one). This means that you need to take your time solving this problem. If you think you have found the answer in 20 seconds, you have probably fallen for a trap (a common wrong answer that the test makers put in the answer choices to trick you). Take your time and follow the Knowsys method to avoid traps.

Start by reading the problem carefully and identifying the bottom line. You are looking for the statements that must be true. That means you will need to evaluate each step carefully to find out if it must be true. A good way to test if a statement must bet true is to try and prove it false. If you can't prove it false, then it must be true.

Now consider your options. Because there are variables in both the problem and the answer choices, you could pick numbers for the variables and test the answer choices. However, since the formula given to you could be expanded, it's probably a better idea to expand the formula first and see what you can deduce from that.







Notice that the only way for this formula to be true is if 2xy = 0. In other words, either x or y must be zero. Now, take a look at the three statements and try to prove them false, given that either x or y must be zero. If you can prove that it is false, eliminate it.

. 

I could be true, since either x or y must be zero. However, it does not need to be true. x could have some other value as long as y is zero. Since it is possible for this statement to be false, you can eliminate it.

.

II is a little bit more difficult to evaluate. However, if you look closely, you should notice that it looks a lot like the original equation you were given. If you expand the equation, you get the following:







Once again, notice that as long as 2xy = 0, this equation is true. In other words, since you already know that x or y must be zero, this equation must be true.

. 

III also has to be true because you already know that either x or y must be zero.

You now know that (II) and (III) must be true. Choose the answer choice that matches your prediction.

(A) None
(B) I only
(C) II only
(D) III only
(E) II and III

The correct Answer Choice is (E).

On sat.collegeboard.org, 33% of the responses were correct.

For more help with math, visit www.myknowsys.com.