algebra problems with multiple solutions

Question

$x^2+6x+8=0$

you can factorise this to $(x+4) \cdot (x+2)=0$ and it's quite obvious that there are two solutions and why they work. you can also solve it like this:

$x^2+6x+8=0$

$(x+3)^2+8-9=0$

$(x+3)^2=1$

$x+3=±1$

$x=-3±1$

now i understand every step, but why taking the square root gives the same two solutions feels kinda magical to me. it's even less intuetiv for me when there aren't any exponents in the first place.

1) $\tan x = 2$, find $\cos x$

2) $\tan x = 2$

3) $\frac{\sin x}{\cos x} = 2$

4) $\sin x = 2\cos x$

5) $\sin^2 x = 4\cos^2 x$

6) $\sin^2 x - 4\cos^2 x = 0$

7) $\sin^2 x + \cos^2 x = 5\cos^2 x$

8) $1 = 5\cos^2 x$

9) $\frac{1}{5} = \cos^2 x$

10) ±$\frac{1}{\sqrt{5}} = \cos x$

again, i know why this problem should have two solution, because i know the unit circle, but that taking the square root can find that other solution, i dont get at all. especially since the original expression didnt even have powers of two before i introduced them in step 5.

in other cases, if you square a number you can make a solution less accurate.

$x=-1$

$x^2=1$

$x=±1$

and then there was an instance where i dont have an example. i just remember a algebra problem where the teacher said that we shouldn't divide by sinx, because if we did that we would loose one of the two solutions, and i think i understood it in that particular instance, but i had no idea how i could see it with out being told, and what the general rule was.

so i have a few question.

what method are there to find more solution to problems? (e.g taking the square root)

in what ways can we loose solutions? (e.g squaring numbers, or dividing by certain numbers)

how do we know we have found all the solutions? (to e.g more difficult problems than those i mentioned)

also, is there an intuitive link between those examples i mentioned? like how factorising and taking the square root could lead you to the same two answers, even tough the methods semms different to me. and how taking the square root also is linked with the unit circle, even tough the expression "tan x" or "sinx/cosx" wasnt squared in the first place.

Answer 1

There's quite a few questions here so I'll do my best to answer all of them.

---

Square roots giving two solutions:

This is based off the fact that $$(-x)^2\equiv x^2$$

Therefore if you squareroot an equation like:

$$(x+3)^2=1$$ we get: $$x+3=\sqrt{1}\tag 1$$ AND $$-(x+3)=\sqrt{1}\tag2$$

Multiply $(2)$ by $-1$ and you get $x+3=-1$, so combining $(1)$ and the edited $(2)$ we get $x=-3\pm 1$

Each of these may have solutions, and so the results both count. However as you notice, it may create solutions which don't actually work. These are called extraneous solutions. You can spot them by plugging them back into your original problem and seeing if they fit.

__

Dividing by $\sin (x)$ removes solutions

Let's take the random equation:

$$(x+1)\sin(x)=0$$ You can see when $x=-1$ we get $(-1+1)\sin(-1)=0\sin(-1)=0$

Are there any other solutions?

Well, since $\sin(k\pi)=0$ for any integer $k$, if we plug $x=k\pi$ into the equation we get:

$$(k\pi+1)\sin(k\pi)=(k\pi+1)(0)=0$$ another set of infinite solutions. However, if we had simply divided by $\sin(x)$ the equation would reduce to: $$x+1=\frac{0}{\sin(x)}\to x+1=0$$ and we lose that entire set of solutions. That's why we avoid doing this, it's a legitimate mathematical step but with undesirable consequences.

__

If $\tan x = 2$, find $\cos x$

The two most important rules in trig are without question:

$$\frac{\sin x}{\cos x}=\tan x$$ $$\sin^2 x+\cos^2 x =1$$

Applying these, do you see how when $\tan x =2$; $$\pm\sqrt{1-\cos^2x}=2\cos x$$ using both those identities. From there, squaring gives:

$$4\cos^2 x =1-\cos^2 x\to \cos x =\pm\sqrt\frac15$$

Answer 2

The problem of finding $\cos x$ given $\tan x$ is underdetermined, because there are two angles in $(-\pi,\pi]$ which satisfy $\tan x=2$ and they have different cosine.

The relation is $$ \frac{\sin x}{\cos x}=2 $$ that you can indeed square, but with some care afterwards. You get $$ 1-\cos^2x=4\cos^2x $$ and therefore $$ \cos^2x=\frac{1}{5} $$ Both $\cos x=1/\sqrt{5}$ and $\cos x=-1/\sqrt{5}$ are acceptable as answers, because the first corresponds to the angle $x$ in the interval $(0,\pi)$ having $\tan x=2$, whereas the second corresponds to the angle $x$ in the interval $(-\pi,0)$.

If $x_1$ is the angle satisfying $0<x_1<\pi$ and $\tan x=2$, then $\cos x_1=1/\sqrt{5}$, because $\sin x_1>0$ and so $\cos x_1>0$ as well.

The angle $x_2=x_1+\pi$ satisfies $\sin x_2=-\sin x_1$ and $\cos x_2=-\cos x_1=-1/\sqrt{5}$, so $\tan x_2=2$.

Hence the data is insufficient to determine $\cos x$.

Note that squaring may introduce spurious solutions. Indeed, $\cos x=1/\sqrt{5}$ is satisfied by both $x_1$ and $x_3=2\pi-x_1$, but $\tan x_3=-2$. Similarly for $x_4=2\pi-x_2$, which satisfies $\cos x_4=-1/\sqrt{5}$, but $\tan x_4=-2$. But here the problem is not to determine $x$ after determining what's $\cos x$. Indeed, the condition $\tan x=2$ already determines $x_1$ and $x_2$.

Final comment: don't always expect automatic machinery that gives you the solutions without thinking. You should analyze carefully what you get and decide whether it is a solution or not.

Answer 3

If

$$f(x)=g(x)\tag{1}$$

and you square both sides

$$f^2(x)=g^2(x)\tag{2}$$

you may find that equation (2) has more solutions than equation (1) because equation (2) can be re-written

\begin{eqnarray} f^2(x)-g^2(x)&=&0\\(f(x)-g(x))(f(x)+g(x))&=&0 \end{eqnarray}

which leads to the conclusion that either

$$ f(x)-g(x)=0\tag{3} $$ or $$ f(x)+g(x)=0\tag{4} $$

Equation (3) is equivalent to equation (1) but equation (4) is a new equation with possibly different solutions called extraneous solutions which were not solutions to the original equation. This is why one must check for extraneous solutions when squaring both sides of an equation.

A similar situation can happen whenever you multiply both sides of an equation by an expression containing the variable.

Suppose we multiply equation (1) by $h(x)$ to obtain

\begin{eqnarray}h(x)f(x)&=&h(x)g(x)\\h(x)f(x)-h(x)g(x)&=&0\\h(x)(f(x)-g(x))&=&0\end{eqnarray}

So either

$$h(x)=0\tag{5}$$ or $$f(x)-g(x)=0\tag{6}$$

Equation (6) is equivalent to equation (1) but equation (5) can introduce extraneous solutions.

Now let us look at how dividing an equation by an expression containing the variable can lose solutions.

Suppose the original equation is (perhaps without even realizing it, because it involves a trigonometric identity or some non-obvious relation) of the form

$$u(x)v(x)=u(x)w(x)\tag{7}$$

This can be solved by

\begin{eqnarray} u(x)v(x)-u(x)w(x)&=&0\\u(x)(v(x)-w(x))&=&0 \end{eqnarray}

So either

$$u(x)=0\tag{8} $$

or

$$v(x)-w(x)=0$$

But if we had originally began by dividing equation (7) by $u(x)$ we would have lost the solutions given by equation (8).

Answer 4

There are several points in the question that could be answered separately (and several times have been answered separately), but you have managed to tie them together with a general theme, so let’s try to address that theme.

First, there is the general question of what to do when a problem involves a function that is not one to one. That is, how do we deal with a function that can produce the same “output” value from two or more “input” values.

The function $f(x)=x^2$ is one such function. For example, $2^2=(-2)^2=4$; there are two ways to solve $x^2=4,$ namely, $x=2$ or $x=-2.$ Similarly, if $(x+3)^2=1$ then also $(-(x+3))^2=1.$

That’s how squaring tends to produce multiple solutions. The multiple solutions do not happen because you took a square root; they happen because you had something that was squared. You can “undo” squaring by applying the square root function, but that only tells you one of the ways that the squared output value could have been produced. The other way is by squaring the negative of the square root.

The basic trig functions actually are not just like squaring, because they can produce the same “output” values in infinitely many ways. For example, $\sin(\pi/6)=\frac12$ (measuring the angle in radians), but also $\frac12= \sin(5\pi/6)= \sin(13\pi/6)= \sin(17\pi/6)=\cdots.$ Generally, if $\sin(x)=y$ then also $\sin(2k\pi+x)=y$ and $\sin((2k+1)\pi-x)=y$ for every integer $k.$ Often, people will put conditions on a trig problem, such as that the answer must be between $0$ and $2\pi,$ in order to avoid having to indicate an infinite number of solutions.

As for whether you can square each side of an equation in a problem or divide by a function of something, I have a slightly different attitude toward the question than your teacher or some others who might answer. My opinion is:

Yes, you can do that, but ...

Yes, you can square both sides of an equation, but you must remember that this can introduce incorrect solutions (as in your simple example with $x=1$) and that you must therefore check each “solution” at the end to make sure it was not an incorrect answer introduced by squaring.

Yes, you can divide by anything you like, but only when the thing you’re dividing by is not zero. So you must either prove that the thing isn’t zero, or you must do two solutions, one for the case where the thing isn’t zero and one for the case where it is. (Of course in the zero case you cannot perform the division, but you usually gain a simpler equation, for example if you divide by $\sin x$ in one case then in the other you have $\sin x=0,$ which tells you that $x$ is an integer multiple of $\pi$.)

In many cases it is simpler to solve the problem without introducing these complications. But sometimes it actually is easier to do such things, so I would not say you should never do them; just do them only when you need to, and be careful.