What to do when you are stuck: Create a similar, but simpler problem.
- Page ID
- 159694
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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 similar problem that is simpler. We illustrate with the following example:
Solve for \(y\) in the equation \(y^2+4 y=x^2+2 x+3\)
Since we are solving for \(y\), we 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 \(R\) 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: \[ \notag \begin{aligned} & y^2+4 y=x^2+2 x+3 \\ & y^2+4 y-x^2-2 x-3=0 \end{aligned} \]
Recall the quadatric formula: If \(ay^2 + by + c = 0\), then \[y=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\] where \(a\) is the coefficient of \(y^2\), \(b\) is the coefficient of \(y\) and \(c\) is everything else, assuming you have the correct format where one side of the equality equals 0.
Using the quadratic formula: \[ \notag \begin{gathered} y=\frac{-4 \pm \sqrt{4^2-4(1)\left(-x^2-2 x-3\right)}}{2(1)}=\frac{-4 \pm \sqrt{16+4\left(x^2+2 x+3\right)}}{2(1)}=\frac{-4 \pm \sqrt{4\left(4+x^2+2 x+3\right)}}{2} \\ =\frac{-4 \pm 2 \sqrt{4+x^2+2 x+3}}{2}=\frac{2\left(-2 \pm \sqrt{4+x^2+2 x+3}\right)}{2}=-2 \pm \sqrt{4+x^2+2 x+3} \end{gathered} \]