Posts filed under: Q and A

Whenever you have to square both sides to solve, you have to check for extraneous solutions.


That tells you m could be 2 or –10, but because part of the solution was squaring both sides, you need to run both possible solutions through the original equation.Try 2 first:


That works, now how about –10?


Nope. Remember that the square root function √ returns only positive results, so –10 is an extraneous solution that doesn’t work in the original equation. The sum of all solutions is just 2.

One more note here: I often think it’s worth graphing questions like this.

It should be obvious from that graph that there’s only one intersection of the two functions, at x = 2.

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Basically, the question is: how many seconds is h greater than 21? (This tennis ball is being thrown on a planet other than Earth, by the way. I challenge anyone to throw a tennis ball that stays in the air anywhere near as long as this one does.)

To figure it out, solve for the two times the equation equals 21. The first time will be when the ball crosses into view, and the second one will be the time the ball falls out of view. The time in between is the answer you want.

So there you go–the ball is at height 21 at 1 second and at 21 seconds. There are 20 seconds between there, so that’s the amount of time the ball is visible to the kids on the roof.

The other way some folks might choose to solve this is by graphing. If you graph the given function and also a horizontal line at y = 21, you get a nice visual of the ball’s flight, which helps make the 20-second window the ball is visible more intuitive.

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Think of a 5-12-13 triangle (that’s one of the Pythagorean triples you should know).


Say angle A measures x°, which would make angle C measure (90 – x)°. (I’m choosing those based on the fact that I already know that the sine of angle C will be 12/13.)


Now that we’ve got it set up, all we need to do is SOH-CAH-TOA it: sin x° = 5/13.

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A function will only have that property if it’s a line that passes through the origin. For example, f(x) = 5x has that property:

You can try the same with other linear functions to see why they won’t work. For example, if f(x) = 5x + 2:

Nonlinear functions also won’t work. For example, if f(x) = x^2:

Anyway, now that we know we’re dealing with a linear function through the origin, we can figure out that if f(6) = 12, then the function we’re dealing with must be f(x) = 2x. Therefore, f(2) = 4.

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First, a circle inscribed in a square looks like this:


If that square has an area of 2, that means each of its sides has a length of √2. And THAT means that the radius of the circle is √2/2.


The area of a circle is πr^2. Plug √2/2 in for r and you’ve got your answer:


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1/2 x = a

x + y = 5a

In the system of equations above, a is a constant such that 0 <a <1/3. If (x,y) is a solution to the system of equations, what is one possible value of y?

(Answer: 0 < x < 1)

The easiest way to go here is to plug in! Let’s say a = 0.1, which is certainly between 0 and 1/3.

The first equation tells us:


From there, we can solve for y in the second equation:


So there you go—grid in 0.3 as a possible value of y and move along.


To see the whole range of possible answers, plug in the endpoints you’re given for a.

If a = 0, then the first equation tells us that x = 0, too.

If x and a both equal zero, then the second equation becomes 0 + y = 5(0), which of course means y = 0, too!

OK, now plug in 1/3 for a. The first equation tells us:


Easy enough to solve for y in the second equation:


So if a = 0, y = 0, and if a = 1/3, y = 1. That’s why the full range of possible answers is 0 < y < 1.

Is it necessary for the math calculator section to have a graphing calculator like, say, a T1-84? I have CASIO fx-300ES plus, which is a scientific calculator. Are there any tricks which can help in the graph questions on SAT with this calculator.

It’s certainly not necessary—a scientific calculator in the right hands is more than enough to answer any question on the SAT. I do believe that there are times when using the graphing function of a graphing calculator is useful (e.g., to quickly see the intercepts of a function) and I have tried to point out those times in my book. However, there are other ways to achieve the same ends without graphing (e.g., manipulating function equations into forms that reveal intercepts).

I can’t think of any graph question tricks that apply to non-graphing calculators—if you don’t have the graphing option you just have to do the problem a different way.

Hi, I was wondering if you could explain to me how x + y= π/2, sin x = cos y. Thank you!

To understand this rule (which obviously works in degrees just as it does in radians), turn to right triangles.

Below, you’ve got a right triangle with acute angles measuring x° and y°. Because the third angle in the triangle is a right angle, and because the angles in any triangle add up to 180°, we know that no matter what x and y are,
= 90°.

So far so good?

Now let’s list out some trigonometric ratios using out old pal SOH-CAH-TOA (well, really just the SOH and CAH parts for this demonstration).

\sin x^\circ = \frac{b}{c}

\sin y^\circ = \frac{a}{c}

\cos x^\circ = \frac{a}{c}

\cos y^\circ = \frac{b}{c}

There you have it. \sin x^\circ =\cos y^\circ and \sin y^\circ =\cos x^\circ, just by nature of x° and y° being the measures of the two acute angles in the same right triangle.

The trick is, for any two angles that add up to 90°, you can draw a right triangle that contains them. Therefore, for any two angles that add up to 90° (or π/2 radians) x and y, sin x = cos y and sin y = cos x.

I’ve been stuck at a math section score of 670. In addition to reviewing the topics related to the questions I got wrong on practice SATs in the PWN book, is there any other thing that would help, or should I just review the concepts of the questions I missed in the PWN book and I should be fine?

Without knowing a lot more about your situation it’s hard for me to be specific, but here are some questions to guide your thinking about other things you could/should be doing.

First, what do you mean by stuck? Like you got 670 twice? Or like you got it 4 times in a row? Improvements are rarely linear, so try to be patient with yourself as you deploy new techniques and solidify your content review. Sometimes it takes more than one test for those things to sink in and become second nature.

How are your mistakes distributed? Are you missing only hard questions, or are there some you think you really should have gotten (i.e., the ubiquitous “careless errors”)? Are you making more mistakes in no-calc than in calc sections? In grid-ins? Understanding where you’re losing points along multiple dimensions can help you figure out how to lose fewer of them.

How are you doing on time? Are you rushing to finish and losing points at the ends of sections? Are you finishing early? If so, how are you spending the time you have left?

What is your goal score? How many can you safely miss to hit it? Can you therefore ignore a less common topic that really drives you nuts and just plan to guess on that one or two questions if they come up?


Can you work out problem #11 of section four in Official SAT #5?

Sure. The math concept you need to know here is that perpendicular lines have negative reciprocal slopes. For example, if one line has a slope of \frac{7}{2}, then a line perpendicular to it must have a slope of -\frac{2}{7}.

In this case, none of the lines are given to you in slope-intercept form, so your first step should be to put at least the first line you’re given into that form.


That tells you that the slope of the given line is \frac{2}{3}, so the slope you need to find in the answer choices is -\frac{3}{2}.

They do you a real favor in the answer choices by only using 3 and 2 as coefficients together in one choice. So just from the get-go you should be leaning choice A. Let’s just put it in slope-intercept form to be sure.


Yep, that’s what we wanted. We’re all done!

Thank you, Mike, for your ever-awesome explanations! Here’s a question from April 2017 SAT, Section 4.22:

The graphs in the xy-plane of the following quadratic equations each have x-intercepts of -2 and 4. The graph of which equation has its vertex farthest from the x-axis?

A) y = -7 (x + 2)(x – 4)

B) y = 1/10 (x + 2)(x – 4)

C) y = -1/2 (x + 2)(x – 4)

D) y = 5 (x + 2)(x – 4)

It’s useful to note that all these equations are the same after the leading coefficient. So, for example, we could just say f(x)=(x+2)(x-4) and then the choices would just be:

A) y=-7f(x)

B) y=\frac{1}{10}f(x)

C) y=-\frac{1}{2}f(x)

D) y=5f(x)

The reason I point that out is that the effect of multiplying a function by a constant, in general, is that the larger the absolute value of the constant is, the more you’ll be stretching that function away from the x-axis. If you know that little fact, then you can immediately pick the answer choice with the coefficient with the largest absolute value, which is choice A.

If you don’t know that little fact, but you have a graphing calculator, you’re still in good shape. Just graph each one (on the same screen, if possible) to see which one has its vertex farthest from the x-axis.

An online candle business open five days a week (Monday through Friday) sells fifteen candles a day its first week. If each week, it sells three more candles a day, how many candles IN TOTAL does that business sell in Week 5?

A) 24 candles
B) 27 candles
C) 120 candles
D) 135 candles

You’d be forgiven (by me, anyway) for missing the fact that this online business is only open 5 days a week. It’s an important fact, though.

If they sold 15 candles per day in Week 1, then they’ll sell 18 candles per day in Week 2, 21 candles per day in Week 3, 24 candles per day in Week 4, and 27 candles per day in Week 5.

27 candles per day times 5 days in the work week equals 135 candles.

If the actual width is W, then the actual length must be 2W. We know this because in the scale drawing, the length is twice as many inches as the width.

The area of the living room floor, therefore will be length times width:


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If you remember that the square of i is –1, you can reduce these to purely real numbers before you even FOIL.


In a + bi form, 5 is just 5 + 0i, so a + b will be 5 + 0 = 5.

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None. The square root of something will never be negative, so if k < 0, it can’t be all alone on the right side of that equation.

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