Download - CEER 2012 mathematics challenge questions
Answers to
Math Challenge Problems
Challenge Question 1The "UPCAT-level" problem that can be solved using
the discriminant:
The equation ax2 + 2x + c = 0 has two distinct
RATIONAL roots, where a and c are both rational.
Which of the following is a possible value for ac?
The condition “where a and c are both rational” is added to avoid ambiguity.
CQ1 Solution
Compute the discriminant. In this case, b = 2:
2 24 2 4
4 4
b ac ac
ac
Note that
The equation will have distinct RATIONAL roots if
the discriminant D = b2 4ac is a PERFECT
SQUARE.
4 4 4 1ac ac
CQ1 Solution
Since 4 is a perfect square, 1 – ac MUST also be a
perfect square so that is a perfect
square.
Among the choices, ac = 3 is a possible value
since 1 – (– 3) = 4 is a perfect square.
Therefore, the answer is (b).
4 1 ac
CQ1 Solution
QUESTION: If ac = 1, 1 – 1 = 0 is also a perfect
square. Why is this NOT allowed to be a value of
ac?
CLUE: Read the question again!
CQ1 Solution
If ac = 1, 1 – 1 = 0 is also a perfect square. BUT, it
will make D = 0, which means that the equation
will have EQUAL ROOTS.
Now, read the question again. The given equation
must have DISTINCT rational roots, which means
D > 0. Kaya dapat, ang conditions ay:
1. D is a perfect square
2. D > 0.
CHALLENGE: What is the value of ?
Challenge Question 2
5 4 2a b c d 2
3 3 3
HINT: What must be the power of 8 to have 16?
Using the definition of the logarithm (see the
solution to Q11 in the CEER presentation):
CQ 2 Solution
8log 16 8 16yy
At this point, you can SUBSTITUTE the choices to
the exponent y. The answer is 4/3 since 4
34/ 3 48 8 2 16
Here’s the formal
solution. The idea is
to make each side
have the same
smallest possible
base, and equate
the exponents to
solve for y.
CQ 2 Solution
3 4
3 4
8 16
2 2
2 2
3 4
4
3
y
y
y
y
y
Dakal a Salamat!