Throughout this entire unit, the focus has been on solving quadratic equations. This topic introduces many different forms that a quadratic equation has. Once it has been written into standard form, any method discussed can help find solutions. Since there will be application problems, make sure to check if a solution makes sense.
Solve \(\large{\frac{2}{x}+\frac{1}{x-2}=\frac{5}{3}}\).
If you think this equation looks familiar, it is. This is from the Rational Things unit, and some problems there created quadratic equations, but those we solved by factoring. The difference this time is that the quadratic equation created may not be factorable and will need to be solved using one of the other methods: Complete the Square, Method of Square Roots or the Quadratic Formula.
Common Denominator: \(3x(x-2)\) \(\begin{array}{l} \implies\large{\frac{2}{x}}\normalsize{\cdot\color{red}{3x(x-2)}+}\large{\frac{1}{x-2}}\normalsize{\cdot\color{red}{3x(x-2)}=}\large{\frac{5}{3}}\normalsize{\cdot\color{red}{3x(x-2)}} \\ \implies2\cdot3(x-2)+1\cdot3x=5\cdot x(x-2) \\ \implies6x-12+3x=5x^2-10x \\ \implies0=5x^2-19x+12 \end{array}\) |
Quadratic Formula: \(x=\large{\frac{-(-19)}{2\cdot5}}\normalsize{\pm}\large{\frac{\sqrt{(-19)^2-4(5)(12)}}{2\cdot5}}\) \(\begin{array}{l} \implies x=\large{\frac{19}{10}}\normalsize{\pm}\large{\frac{\sqrt{361-240}}{10}} \\ \implies x=\large{\frac{19}{10}}\normalsize{\pm}\large{\frac{\sqrt{121}}{10}} \\ \implies x=\large{\frac{19}{10}}\normalsize{\pm}\large{\frac{11}{10}} \\ \implies x=\large{\frac{19}{10}}\normalsize{+}\large{\frac{11}{10}} \normalsize{\implies x=}\large{\frac{19+11}{10}}\normalsize{\implies x=}\large{\frac{30}{10}}\normalsize{\implies x=3\;\color{green}{\checkmark}} \\ \implies x=\large{\frac{19}{10}}\normalsize{-}\large{\frac{11}{10}} \normalsize{\implies x=}\large{\frac{19-11}{10}}\normalsize{\implies x=}\large{\frac{8}{10}}\normalsize{\implies x=}\large{\frac{4}{5}}\normalsize{\;\color{green}{\checkmark}} \end{array}\) |
In 4 hours, Kerrie can go 15 miles upriver and come back. If the speed of the current in the river is 5 mph, then find the speed of the boat in still water.
First, let's fill in a table with the information from the problem to help build an equation:
| \(d\) | \(r\) | \(t\) | |
|---|---|---|---|
| upstream | \(15\) miles | \(x-5\) mph | \(\large{\frac{15}{x-5}}\) hours |
| downstream | \(15\) miles | \(x+5\) mph | \(\large{\frac{15}{x+5}}\) hours |
To build an equation, it takes 4 hours to go upriver and come back:
$$\underbrace{\frac{15}{x-5}}_{\text{time upriver}}+\underbrace{\frac{15}{x+5}}_{\text{time downriver}}= \underbrace{4}_{\text{total time}}$$| Common Denominator = \((x-5)(x+5)\) |
| $$\begin{array}{l} \implies \large{\frac{15}{x-5}}\normalsize{\cdot\color{red}{(x-5)(x+5)} +}\large{\frac{15}{x+5}}\normalsize{\cdot\color{red}{(x-5)(x+5)}=4\cdot\color{red}{(x-5)(x+5)}} \\ \implies 15\cdot(x+5)+15\cdot(x-5)=4(x-5)(x+5) \\ \implies 15x+75+15x-75=4(x^2-5x+5x-25) \\ \implies 30x=4x^2-100 \\ \implies 4x^2-30x-100=0 \\ \implies x=\large{\frac{30}{8}}\normalsize{\pm}\large{\frac{\sqrt{900-4(-400)}}{8}} \\ \normalsize{\implies x=}\large{\frac{15}{4}}\normalsize{\pm}\large{\frac{\sqrt{2500}}{8}} \\ \normalsize{\implies x=}\large{\frac{15}{4}}\normalsize{\pm}\large{\frac{50}{8}}\normalsize{\Rightarrow}\large{\frac{25}{4}} \\ \normalsize{\implies x=}\large{\frac{15+25}{4}}\normalsize{\Rightarrow x=}\normalsize{\frac{40}{4}}\normalsize{\Rightarrow x=10\;\color{green}{\checkmark}} \\ \implies x=\large{\frac{15-25}{4}}\normalsize{\Rightarrow x=}\large{\frac{-10}{4}}\normalsize{\Rightarrow x=-2.5\;\color{red}{\chi}} \end{array}$$ |
| The only solution is \(x=10\) since rate cannot be negative. The boat is moving at 10 mph. |
Sometimes equations will have three terms and appear to be quadratic. In fact, these equations are quadratic, but not in the single variable, it may be in another variable form. For example: $$2(4m-3)^2+7(4m-3)+5=0;\;\;\;x+\sqrt{x}=6;\;\;\;x^4-13x^2+36=0$$ certainly do not look like the quadratic equations we have studied so far. Using a substitution of variables, generates a new equation that is quadratic. Normally, the variable \(u\) is selected to act as the substitution variable. By setting \(u\) equal to the "middle term" we can rewrite the original equation into a quadratic one:
Use \(u\)-substitution to rewrite \(2(4m-3)^2+7(4m-3)+5=0\) into a quadratic equation.
Let \(u=4m-3\). Then substitute into the equation: $$2(\color{blue}{4m-3})^2+7(\color{blue}{4m-3})+5=0\iff2(\color{blue}{u})^2+7(\color{blue}{u})+5=0\implies2u^2+7u+5=0$$ This substitution has created a quadratic equation in the variable \(u\) and we can solve it. Once it has been solved, the solutions are for this substitution variable \(u\). By substituting back \(4m-3\) we can find the solutions in terms of the original variable \(m\).
| \(2(4m-3)^2+7(4m-3)+5=0\) | \(\implies2u^2+7u+5=0\) |
| This one happens to factor: | \(\begin{array}{l} \implies2*5=10\;\text{and}\;2u+5u=7u \\ \implies2u^2+2u+5u+5=0 \\ \implies2u(u+1)+5(u+1)=0 \\ \implies(u+1)(2u+5)=0 \end{array}\) |
| Solutions in terms of \(u\): | \(\implies u=-1\) and \(u=-\large{\frac{5}{2}}\) |
| Now back substitute in terms of \(m\): | \(u=4m-3\implies4m-3=-1\) and \(4m-3=-\large{\frac{5}{2}}\) |
| Solve in terms of \(m\): | \(\begin{array}{l} \implies4m=2\;\text{and}\;4m=3-\large{\frac{5}{2}} \\ \normalsize{\implies m=}\large{\frac{2}{4}} \normalsize{\;\text{and}\;4m=}\large{\frac{6}{2}-\frac{5}{2}} \\ \normalsize{\implies m=}\large{\frac{1}{2}}\normalsize{\;\text{and}\;4m=}\large{\frac{1}{2}} \\ \normalsize{\implies m=}\large{\frac{1}{2}}\normalsize{\;\color{green}{\checkmark}\;\text{and}\;m=}\large{\frac{1}{8}}\normalsize{\;\color{green}{\checkmark}} \end{array}\) |
To me the hardest part is making the substitution to create the new equation. Just for more practice with that, let's substitute and rewrite the other two equations from above:
| \(x+\sqrt{x}=6\) | \(x^4-13x^2+36=0\) |
|---|---|
| \(u=\sqrt{x}\) so \(u^2=\left(\sqrt{x}\right)^2\Rightarrow u^2=x\) | \(u=x^2\) so \(u^2=\left(x^2\right)^2\Rightarrow u^2=x^4\) |
| New equation: \(u^2+u-6=0\) | New equation: \(u^2-13u+36=0\) |