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 xyplane 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

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!