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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