Solving problems and common errors
- Page ID
- 157556
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Sometimes when you are stuck on a problem, it can be helpful to create a simpler problem in order to determine how to solve the harder problem. We will illustrate this in the first example..
Solve for \(y\) in the equation \(y^2+4 y=x^2+2 x+3\)
Solution
Suppose we do not know how to solve the equation above. We will create a simpler problem that we know how to solve in order to determine how to solve this harder problem. Since you are solving for \(y\), I would leave the left-hand side (LHS) alone, and focus on creating a simpler problem by modifying the right-hand side (RHS):
Suppose we let the RHS be an integer such as 2 . Note 2 was chosen at random to be some simple integer to create a simpler problem.
Then we have the problem: \(y^2+4 y=2\)
You may now recognize that we have a quadratic equation, and thus we want to get 0 on one side. Moving the 2 to the LHS:
\[ \notag
\begin{gathered}
y^2+4 y=2 \\
y^2+4 y-2=0
\end{gathered}
\]
We can then use the quadratic formula:
\[ \notag
\begin{aligned}
& y=\frac{-4 \pm \sqrt{4^2-4(1)(-2)}}{2(1)}= \frac{-4 \pm \sqrt{16+8}}{2(1)}=\frac{-4 \pm \sqrt{24}}{2}=\frac{-4 \pm \sqrt{(4)(6)}}{2}=\frac{-4 \pm 2 \sqrt{6}}{2} \\
&=\frac{2(-2 \pm \sqrt{6})}{2}=-2 \pm \sqrt{6}
\end{aligned}
\]
If you can figure out how to do the original problem, \(y^2+4 y=x^2+2 x+3\), go ahead and do it. If not, make up a simpler problem, but instead of using an integer such as 2 , use a variable instead. For example, we can let the RHS \(=R=x^2+2 x+3\)
So we are now solving \(y^2+4 y=R\)
We do the same steps as before. Moving the \(\mathrm{R}\) to the LHS:
\[ \notag
\begin{gathered}
y^2+4 y=R \\
y^2+4 y-R=0
\end{gathered}
\]
Using the quadratic formula:
\[ \notag
\begin{gathered}
y=\frac{-4 \pm \sqrt{4^2-4(1)(-R)}}{2(1)}=\frac{-4 \pm \sqrt{16+4 R}}{2(1)}=\frac{-4 \pm \sqrt{4(4+R)}}{2}=\frac{-4 \pm 2 \sqrt{4+R}}{2} \\
=\frac{2(-2 \pm \sqrt{4+R})}{2}=-2 \pm \sqrt{4+R}
\end{gathered}
\]
Since you know that \(R=x^2+2 x+3\), you can plug this in for \(R\) to find that the solution to \(y^2+4 y=x^2+2 x+3\) is
\[ \notag
y=-2 \pm \sqrt{4+x^2+2 x+3}
\]
You can also solve this quadratic equation directly by following the steps above.
Solve \(y^2+4 y=x^2+2 x+3\)
We now recognize that we have a quadratic equation, and thus we want to get 0 on one side. Since we are solving for \(\mathrm{y}\), we will move all terms to the LHS:
\[ \notag
\begin{aligned}
& y^2+4 y=x^2+2 x+3 \\
& y^2+4 y-x^2-2 x-3 = 0
\end{aligned}
\]
Using the quadratic formula:
\[ \notag
\begin{gathered}
y=\frac{-4 \pm \sqrt{4^2-4(1)\left(-x^2-2 x-3\right)}}{2(1)}=\frac{-4 \pm \sqrt{16+4\left(x^2+2 x+3\right)}}{2(1)}=\frac{-4 \pm \sqrt{4\left(4+x^2+2 x+3\right)}}{2} \\
=\frac{-4 \pm 2 \sqrt{4+x^2+2 x+3}}{2}=\frac{2\left(-2 \pm \sqrt{4+x^2+2 x+3}\right)}{2}=-2 \pm \sqrt{4+x^2+2 x+3}
\end{gathered}
\]
Sometimes, especially on exams, one might wonder if a certain algebraic operation is correct.
For example, does \(\ln (x+y)=\ln (x)+\ln (y)\) ? NO!
An easy mistake, but upsetting to many instructors. You can prove the equality does not hold by plugging in numbers for \(x\) and \(y\). Sometimes you might choose the wrong numbers where the equality holds; but this doesn't mean the equality holds for all possible values \(x\) and \(y\). But in many cases, you can come up with a particular (counter-)example to show that the equality does not hold.
For example, let \(x=y=1\) :
\[ \notag
\ln (1+1)=2, \text { but } \ln (1)+\ln (1)=0+0
\]
Since \( \ln (1+1)=2 \neq 0 = \ln (1)+\ln (1)\), we know that \(\ln (x+y) \neq \ln (x)+\ln (y)\) for arbitrary \(x\) and \(y\).
You can often check if an equality holds by plugging in small numbers such as 0 and 1; but these are sometimes the exceptions, so if equality holds for such cases, you might try another example where you choose larger numbers and where \(x \neq y\). Checking several cases will NOT prove an equality, but can be used to determine if an equality does not hold in general.
- \( (a+b)^2 \neq a^2+b^2\)
- \( \sqrt{a+b} \neq \sqrt{a}+\sqrt{b}\)
In each of these cases, you can prove that equality does not hold by letting \(a=b=1\).
For the second example we see that \(\sqrt{1+1} = \sqrt{2} \neq 2 = 1+1 = \sqrt{1}+\sqrt{1}\). This is called proof by counter-example.