Even though Math B is winding down, I didn’t expect a good exam. And New York State did not fail to disappoint. 13 had ridiculous context. 33 included math from outside of the curriculum. Some reasonable questions, and answers:

2 .The expression $\frac{5}{3+\sqrt{2}}$ is equivalent to…

Normal solution: multiply numerator and denominator by $3 - \sqrt{2}$ (the conjugate)

Cheat solution: plug the expression into a calculator. Equals about 1.1. Check the four answers. Only choice (3) is close.

Correct solution: $\frac{15 - 5\sqrt{2}}{7}$

4. What is the solution of the inequality $x^2 - x - 6 < 0$ ?

Cheat. Check a couple of numbers.  0 works, and eliminates 2 choices. 2 works and eliminates another choice.

Normal solution: Factor the quadratic: (x-3)(x+2) < 0. Sign changes at 3 and -2.

Correct solution: -2 < x < 3

20. The accompanying diagram [not here, just imagine] shows part of the architectural plans for a structural support of a building. PLAN is a rectangle and $\overline{AS} \perp \overline{LN}$

My comment: The context, as usual, is contrived, silly, and distracting. LN is a diagonal. AS is an altitude to LN at S.

Which equation can be used to find the length of $\overline{AS}$ ?

(1) $\frac{LS}{AS} = \frac{AS}{SN}$ (3) $\frac{AS}{SN} = \frac{AS}{LS}$

(2) $\frac{AN}{LN} = \frac{AS}{LS}$ (4) $\frac{AS}{LS} = \frac{LS}{SN}$

I’ve got no cheat here. \triangle ASL ~ \triangle NSA. Sides are in proportion, so choice (1).

21. Solve for x: $\sqrt{x+18} - 2 = 2$

Transpose: $\sqrt{x+18} = 4$

Square both sides: x + 18 = 16, and x = -2. Checks, too.

25. What is the solution of the inequality |2x – 5| ≤ 11/2 ?

Abnormal solution: Factor out a 2: |x – 5/2| ≤ 11/2 This is the interior of a circle with center 5/2 and radius 11/2, so -3 ≤ x ≤ 8

26. The volume of Earth can be calculated by using the formula $V = \frac{4}{3}\pi r^3$. Solve for r in terms of V.

Multiply both sides by 3/4pi, then take the cube root: $r = \root 3\of{\frac{3V}{4\pi}}$

28. Perform the indicated operation and express in simplest form:

$\frac{3x^2+12x-15}{x^2+2x-15} \div \frac{3x^2-3x}{3x-x^2}$

A former student of mine explained this one well: You factor, you cancel. She used to abbreviate it on the board, and her kids never forgot (UF, UC). I choose not to write that in class. Here goes:

$\frac{3(x+5)(x-1)}{(x+5)(x-3)} \div \frac{3x(x-1)}{x(3-x)}$

$\frac{3(x+5)(x-1)}{(x+5)(x-3)} \times \frac{x(3-x)}{3x(x-1)}$

Everything cancels. That x-3 over 3-x gives -1. The answer.

31. Find the roots of the equation $x^2 + 7 = 2x$

Cheat solution. Complete the square. $x^2 - 2x + 1 = -6$

$(x - 1)^2 = -6$

$x - 1 = \pm \sqrt{-6}$

$x = 1 \pm 6i$

34. $Given PROE is a rhombus, \overline{SEO}, \overline{PEV}, \angle SPR \cong \angle VOR$

$Prove \overline{SE} \cong \overline{EV}$

Ok, I am missing the diagram, but the description is sufficient.

Take those congruent angles, subtract the congruent opposite angles from the rhombus, that gives us a pair of congruent angles (SPE and VOE)

PE = OE (sides of a rhombus)

$\angle SEP \cong \angle VEO$ (opp angles)

Now the two little triangles are congruent by ASA, and what we want is congruent by corresponding parts.

January 28, 2009 am31 10:37 am 10:37 am

I’m curious. Why is completing the square a “cheat solution”?

2. January 28, 2009 pm31 5:12 pm 5:12 pm

I called it a cheat, but you are right, it’s not. Just, most kids will go straight to the formula. I get the sense that many think it is unfair that completing the square works… I used “cheat” on that one in jest.