Solutions to the two challenge problems posed during the CEER 2012 Math review last 21 July 2012.

1. Answers to
Math Challenge Problems

2. Challenge Question
1
The "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.

3. CQ1 Solution
The equation will have distinct RATIONAL roots if
the discriminant D = b2 4ac is a PERFECT
SQUARE.
Compute the discriminant. In this case, b = 2:
b2 4ac 22 4ac
4 4ac
Note that 4 4ac 4 1 ac

4. CQ1 Solution
Since 4 is a perfect square, 1 – ac MUST also be a
perfect square so that 4 1 ac 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).

5. 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! 

6. 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. 

7. Challenge Question
2
CHALLENGE: What is the value of ?
5 4 2
a b c d 2
3 3 3
HINT: What must be the power of 8 to have 16?

8. CQ 2 Solution
Using the definition of the logarithm (see the
solution to Q11 in the CEER presentation):
y
y log8 16 8 16
At this point, you can SUBSTITUTE the choices to
the exponent y. The answer is 4/3 since
4
3
84/ 3 8 24 16

9. CQ 2 Solution
Here’s the formal 8 y
16
solution. The idea is 3
y
4
to make each side 2 2
have the same 2 3y
2 4
smallest possible 3y 4
base, and equate
the exponents to 4
y
solve for y. 3

10. Dakal a
Salamat!