Sunday, September 05, 2021

Algebra 2 Problems of the Day (Algebra 2/Trigonometry Regents, January 2014)



Now that I'm caught up with the current New York State Regents exams, I'm revisiting some older ones.

More Regents problems.

Algebra 2/Trigonometry Regents, January 2014

Part I: Each correct answer will receive 2 credits.


11. The roots of the equation 2x2 + 4 = 9x are

1) real, rational, and equal
2) real, rational, and unequal
3) real, irrational, and unequal
4) imaginary

Answer: 2) real, rational, and unequal


Set the equation equal to zero and find the discriminant, b2 - 4ac.

If 2x2 + 4 = 9x
then 2x2 -9x + 4 = 0
and a = 2, b = -9, and c = 4

The discriminat is b2 - 4ac = (-9)2 - 4(2)(4) = 49

Since the discriminant is positive, there are two distinct solutions. Since 49 is a perfect square, those solutions will be ratioanl.

Those solutions will be (9 + 7) / (2*2) = (9+7)/4 or (9-7)/4, which is 4 or 1/2.





12. If d varies inversely as t, and d = 20 when t = 2, what is the value of t when d = -5?

1) 8
2) 2
3) -8
4) -2

Answer: 3) -8


If d varies inversely as t then d1t1 = d2t2

(20)(2) = (-5)t2

40/-5 = t

-8 = t





13. If sin A = -7/25 and ∠A terminates in Quadrant IV, tan A equals

1) -7/25
2) -7/24
3) -24/7
4) -24/25

Answer: 2) -7/24


If sin A = -7/25, then cos A = 24/25, using the Pythagorean Theorem, and the fact that we're in Quadrant IV, where x values are positive.

Tan A = sin A / cos A = (-7/25) / (24/25) = -7/24





14. Which expression is equivalent to Summation, n=1 to 4 (a - n)2?
1) 2a2 + 17
2) 4a2 + 30
3) 2a2 - 10a + 17
4) 4a2 - 20a + 30

Answer: 4) 4a2 - 20a + 30


You don't actually have to do all the math. Write out the terms and the correct answer will become obvious.

(a - 1)2 + (a - 2)2 + (a - 3)2 + (a - 4)2

Each binomial when expanded will have an a2 term. That sums to 4a2.

Each binomial when expanded will have a middle term that is -2n. Eliminate Choice (2).

Only Choice (4) remains.

Proof: a2 - 2a + 1 + a2 - 4a + 4 + a2 - 6a + 9 + a2 - 8a + 16 = 4a2 - 20a + 30





15. What are the coordinates of the center of a circle whose equation is x2 + y2 - 16x + 6y + 53 = 0?

1) (-8,-3)
2) (-8,3)
3) (8,-3)
4) (8,3)

Answer: 3) (8,-3)


Regroup the variables and complete the squares. That will put the equation in standard form.

x2 + y2 - 16x + 6y + 53 = 0
x2 - 16x + y2 + 6y = - 53
x2 - 16x + 82 + y2 + 6y + 32 = - 53 + 82 + 32
x2 - 16x + 64 + y2 + 6y + 9 = - 53 + 64 + 9
(x - 8)2 + (y + 3)2 = 20

The center of the circle is (8, -3).

Note that we didn't need the radius, so the last couple of steps weren't entirely necessary. Once you had half of -16x and half of 6y, you had enough information. But it doesn't hurt to be thorough.




More to come. Comments and questions welcome.

More Regents problems.

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