Posts tagged with: OT Test 4 Section 3

College Board Test 4 Section 3 #9:
____
√x-a = x-4
If a=2 what is the solution set of the preceding equation?
A. {3, 6}
B. {2}
C. {3}
D. {6}

Is there another way to solve this quickly besides plugging in the answer choices?

Backsolve is far and away my preferred method on a question like this. The answer choices only show you 2, 3, and 6 as possible solutions, so you have to try, at most, three numbers.

However, of course there is an algebraic solution! If you go that way, you must be careful because you will generate extraneous solutions. To begin, square both sides, combine like terms, and solve the resulting quadratic:

    \begin{align*}\sqrt{x-2}&=x-4\\\left(\sqrt{x-2}\right)^2&=(x-4)^2\\x-2&=x^2-8x+16\\0&=x^2-9x+18\\0&=(x-6)(x-3)\end{align*}

From there, you must check that both solutions you generated actually work in the original equation, because when you square both sides of an equation you can cause extraneous solutions.

    \begin{align*}\sqrt{3-2}&=3-4\\1&\ne -1\end{align*}

    \begin{align*}\sqrt{6-2}&=6-4\\2&=2\end{align*}

As you can see, 3 doesn’t work in the original equation, so it’s an extraneous solution. 6 DOES work in the original equation, so it’s the one we go with.

Hopefully this reinforces the appeal of backsolving here: even if you do the algebra, you still need to try the answers you find in the original equation anyway. Why not just start by trying the answers provided!? 🙂

Test 4 Section 3 #20 please. Also Mike, I still didn’t know how to answer the question even after I went and reviewed it from the official test breakdown section of your book.

There are a lot of words here, but when you strip that all away, this is basically a question about slope. You know you have a change in distance from the earth’s surface of 50 km to 80 km, and a change in temperature from –5° C to –80° C.

The big trick is that we’re told we need to find the degrees by which the temperature changes every 10 km, so we should think about our distance values in tens. That is, instead of 50 km and 80 km, we should use 5 and 8 and our units will be tens of km.

So let’s calculate the change in temperature in degrees Celsius per change in 10 km distance.

    \begin{align*}&\dfrac{\text{Change in Temperature in degrees Celsius}}{\text{Change in Distance in 10 kilometer units}}\\\\=&\dfrac{-5-(-80)}{5-8}\\\\=&\dfrac{75}{-3}\\\\=&-25\end{align*}

So there you go. Every 10 km further you go from the Earth’s surface, the temperature drops by 25° C.

Test 4 Section 3 Number 17

Above I’ve recreated the figure on the test, only my version adds labeled side lengths ab, and c. We’re told in the question that the sine of x° is 0.6. From SOH-CAH-TOA, we know that the sine of an angle is the ratio of its opposite side in a right triangle over the hypotenuse. In other words, \sin x^\circ =\frac{b}{c}=0.6.

We’re asked for the cosine of y°. Again, from SOH-CAH-TOA we know that the cosine of an angle is the ratio of its adjacent side in a right triangle over the hypotenuse. In this triangle, that means \cos y^\circ =\frac{b}{c}. Wait a minute…don’t we already know something about \frac{b}{c}? We sure do—that’s the same ratio as \sin x^\circ!

Therefore, we know that the cosine of y° is 0.6.

Can you explain #15 in Section 3 of test 4? I understand how to do the problem but was wondering if there is a shortcut or trick to seeing the correct answer?

Great question! First, I don’t know if this is really a shortcut but for me this question becomes much easier to tackle when I multiply by 2 to eliminate the fraction. So the quadratic I’m working with after multiplying and rearranging is:

    \begin{align*}x^2-\dfrac{k}{2}x&=2p\\2x^2-kx&=4p\\2x^2-kx-4p&=0\end{align*}

So, for quadratic formula purposes, a=2, b=-k, and c=-4p.

The real shortcut, for this question and ones like it, is to focus on the differences in the answer choices rather than doing the whole quadratic formula process and trying to match a choice.

In this question, for example, you might notice that you only have two choices for the first term: \dfrac{k}{2} or \dfrac{k}{4}. If you start by focusing only on that term, you can eliminate half the choices right away!

The quadratic formula is x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}. That means the first term is \dfrac{-b}{2a}. In this question, then, the first term is \dfrac{-(-k)}{2(2)}=\dfrac{k}{4}. So we know the answer must be A or B.

From there, we know the bit under the radical must either be \sqrt{k^2+2p} or \sqrt{k^2+32p}.

Since the radical bit equals \sqrt{b^2-4ac}, we can plainly see that \sqrt{k^2+32p} is the way to go. Choice B is the answer.

The full algebra for anyone who came here looking for it:

    \begin{align*}2x^2-kx-4p&=0\\\dfrac{-(-k)\pm\sqrt{k^2-4(2)(-4p)}}{2(2)}&=x\\\dfrac{k\pm\sqrt{k^2+32p}}{4}&=x\\\dfrac{k}{4}\pm\dfrac{\sqrt{k^2+32p}}{4}&=x\end{align*}

Practice Test 4, Section 3, Number 11 (No Calc)

I love this question because the fastest way to go involves almost no math. You just have to know a little bit about the shapes of lines and parabolas.

First, think about the parabola. Its equation is y=(2x-3)(x+9). From that we know it’s a parabola that opens up (the x^2 term will be positive) and has x-intercepts at \dfrac{3}{2} and -9. You should figure out its y-intercept, too, by plugging in zero for x:

    \begin{align*}y&=(2(0)-3)(0+9)\\y&=(-3)(9)\\y&=-27\end{align*}

Do a very rough drawing of that on your paper (forgive my MS Paint skillz, but your drawing can be sloppier than mine and still be plenty good enough):

Now do a rough drawing of the line. To do that, put it in slope-intercept form:

    \begin{align*}x&=2y+5\\x-5&=2y\\y&=\dfrac{1}{2}x-\dfrac{5}{2}\end{align*}

The important detail there is that the y-intercept is -\dfrac{5}{2}, which is above the parabola’s y-intercept of -27, so you know the line will intersect the parabola twice. Like so:

Dear Mike,

Can you please do #16 in the Official Practice Test #4 NO CALCULATOR section
(aka Section 3)?

You bet!

This is a question about similar triangles. Here’s a simplified figure we can use to work through it.

Note that I’ve taken the step of drawing a vertical line from the top vertex to the base. Because the shelves are parallel, we know the angles they form with the vertical segment I drew and with the sides of the triangle will all be the same. That tells us that we have angle-angle similarity, so we know that the 2:3:1 ratio we have on the left side will also apply down the middle, like so:

Since we know that the height of the unit is 18 inches, we can solve for y:

y + 3y + 2y = 18
6y = 18
y = 3

From there, we know that the shampoo needs to fit on the middle shelf, so it needs to fit into a space that’s 3y = 3(3) = 9 inches tall.

Test 4 Section 3 #14

When you’re dividing complex numbers, you have to multiply the top and bottom of the fraction by the complex conjugate of the bottom. This creates a real number in the bottom of the fraction, which is awesome.

The bottom of this fraction is 3-2i, so its complex conjugate is 3+2i. To get started, then, we write the following:

    \begin{align*}\dfrac{8-i}{3-2i}\times\dfrac{3+2i}{3+2i}\end{align*}

Then we simplify as much as we can. First we FOIL:

    \begin{align*}&\dfrac{(8-i)(3+2i)}{(3-2i)(3+2i)}\\=&\dfrac{24+16i-3i-2i^2}{9+6i-6i-4i^2}\\=&\dfrac{24+13i-2i^2}{9-4i^2}\end{align*}

Now, remembering that i=\sqrt{-1} and therfore that i^2=-1, we make that substitution and simplify further:

    \begin{align*}&\dfrac{24+13i-2(-1)}{9-4(-1)}\\=&\dfrac{24+13i+2}{9+4}\\=&\dfrac{26+13i}{13}\\=&2+i\end{align*}

So there you have it. Once we simplify that complex fraction into a+bi form, we see that it’s just 2+i, which means a=2.

Test 4 Section 3 #13

You’re told that the given parabolas intersect at (k,0) and (-k,0). Important insight: when two graphs intersect at a point, that point is on both graphs. Therefore, we don’t need to use both equations to solve this question, we can just pick one and find the values of x that will result in a function value of zero.

f(x)=8x^2-2

f(k)=8k^2-2

0=8k^2-2

2=8k^2

\dfrac{1}{4}=k^2

\pm\dfrac{1}{2}=k

The given figure tells you that k is positive (look where k and -k are on the x-axis) so you know the answer is \dfrac{1}{2}.

Note: You can also backsolve this one. Put the answer choices into one of the functions until you get a result of 0. Even without a calculator, that won’t take long (but will probably take longer than the algebra unless you’re stuck).

Test 4 Section 3 Number 19

This one is a great opportunity to practice solving a system of linear equations by elimination. Here are the equations you’re given:

    \begin{align*}-3x+4y&=20\\6x+3y&=15\end{align*}

Multiply both sides of the first equation by 2:

    \begin{align*}-6x+8y&=40\end{align*}

Now add that to the original second equation and you’re left with only y terms!

    \begin{align*}-6x+8y&=40\\+(6x+3y&=15)\\11y&=55\end{align*}

From that, you know that y = 5. The question asked you for x, though, so plug in 5 for y in one of the original equations to finish up:

    \begin{align*}6x+3(5)&=15\\6x+15&=15\\6x&=0\\x&=0\end{align*}