Word Problems with Quadratic Equations

Word Problems with Quadratic Equations exercise

• Explain how to solve a quadratic equation by factoring.

  Hints

  The FOIL method a way of multiplying two binomials together. For example, $(x+9)(x+3)$:

  • multiply the First $x^2$
  • multiply the Outer $9x$
  • multiply the Inner $3x$
  • multiply the Lasr $27$

  Next, you add the terms together and combine like terms.

  Here is an example for solving an equation; subtraction is the opposite operation of addition.

  Use the zero factor property: if a product is equal to zero then one of its factors must also be equal to zero.

  Solution

  The servants have to calculate the total length of the carpet.

  To do this they first use the FOIL method for multiplying the two binomials:

  $(x+9)(x+3)=x^2+9x+3x+27=x^2+12x+27$

  This quadratic term is equal to the given total area. So we get

  $x^2+12x+27=72$

  which is equivalent to

  $\begin{array}{rcl} x^2+12x+27&=&~72\\ \color{#669900}{-72} & &\color{#669900}{-72}\\ x^2+12x-45&=&~0. \end{array}$

  To factor the left-hand side of the equation above we are looking for pairs of factors of $-45$ which sum to $12$:

  • $1\times(-45)=-45$ but $1-45=-44$
  • ...
  • $3\times (-15)=-45$ but $3-15=-12$
  • $-3\times 15=-45$ and $-3+15=12$ $~~~~~$✓

  Thus we have $(x-3)(x+15)=0$.

  To get the solutions we use opposite operations twice:

  $\begin{array}{rcl} x-3&=&~0\\ \color{#669900}{+3} & &\color{#669900}{+3}\\ x&=&~3 \end{array}$

  and

  $\begin{array}{rcl} x+15&=&~0\\ \color{#669900}{-15} & &\color{#669900}{-15}\\ x&=&~-15 \end{array}$

  For $x=3$ we get that the length is $3+9=12$ and the width is $3+3=6$. And for $x=-15$, the length is $-15+9=-6$ ... so we can stop because there don't exist any negative lengths.

  This means that both $x=3$ as well as $x=-15$ are solutions to the quadratic equation, but only $x=3$ is a possible solution in our given situation.

• Determine how to solve quadratic equations graphically.

  Hints

  The function above is a quadratic function.

  Any linear function is given by $f(x)=mx+b$, where $m$ is the slope and $b$ the $y$-intercept. The graph of such a function is a line.

  The graph of a quadratic function is a parabola. Each parabola has at most two $x$-intercepts.

  Solution

  There exist different ways to solve quadratic equations like $ax^2+bx+c=0$.

  For example, lets draw the parabola in a coordinate plane which corresponds to the graph of

  $f(x)=-0.25x^2+2x+5$

  1. Plug in a few values for $x$ and calculate the corresponding $y$-values.
  2. Plot the resulting ordered pairs $(x,y)$ in a coordinate plane.
  3. Connect those pairs to get the corresponding parabola.

  The solutions of the equation $-0.25x^2+2x+5=0$ are the $x$-intercepts of the parabola, which can be seen in the graph pictured to the right, namely $x=-2$ and $x=10$.

• Determine how to solve quadratic equations by completing the square.

  Hints

  The FOIL method is a way of multiplying two binomials; for example, $(x+2)(x-3)$:

  • multiply the First $x^2$
  • multiply the Outer $-3x$
  • multiply the Inner $2x$
  • multiply the Lasr $-6$

  Then add the terms together and combine the like terms.

  To complete a quadratic term $x^2+bx$ add $\left(\frac b2\right)^2$.

  Here is an example for solving an equation by taking a square root.

  Don't forget the $\pm$ sign in your calculation!

  Check the solutions. Do they makes sense when trying to determine the positive length and width of the moat?

  Solution

  First, we use the FOIL method for multiplying the two binomials

  $(x+2)(x-3)=x^2-3x+2x-6=x^2-x-6$

  This must be equal to the given total area. So we get $x^2-x-6=50$, which is equivalent to

  $\begin{array}{rcl} x^2-x-6&=&~50\\ \color{#669900}{+6} & &\color{#669900}{+6}\\ x^2-x&=&~56 \end{array}$

  Now we complete the quadratic term on the left-hand side to a square. For this we add $(\frac12)^2$:

  $\begin{array}{rcl} x^2-x&=&~56\\ \color{#669900}{+\left(\frac12\right)^2} & &\color{#669900}{+\left(\frac12\right)^2}\\ x^2-x+\left(\frac12\right)^2&=&~56+\left(\frac12\right)^2\\ \left(x-\frac12\right)^2&=&~56.25 \end{array}$

  Next, we take the square root of both sides to get

  $x-\frac12=\pm7.5$

  Lastly we add $\frac12=0.5$, which leads to the desired solutions

  $x=0.5+7.5=8$ or $x=0.5-7.5=-7$

  Let's check if both solutions work with the problem at hand, keeping in mind that we are trying to find the length and width of a moat.

  For $x=8$ we get the length $8+2=10$ and the width $8-3=5$. Both are positive, so this solution works.

  For $x=-7$ we get the length $-7+2=-5$ and the width $-7-3=-10$. A negative length and width doesn't make any sense, so this solution can't work in our given situation.

• Calculate where the castle ditches have to be built by using the quadratic formula.

  Hints

  You can use the quadratic formula with

  • $a=-0.5$
  • $b=-1.5$
  • $c=5$

  You can also multiply the equation $-0.5x^2-1.5x+5=0$ by $-2$ to get $x^2+3x-10=0$.

  Both solutions are whole numbers. One is negative and the other one is positive.

  Solution

  You can solve each quadratic equation $ax^2+bx+c=0$ by using the quadratic formula,

  $x=\frac{-b \pm\sqrt{b^2-4ac}}{2a}$.

  First, determine $a$, $b$ and $c$ in the equation and then plug those values into the quadratic formula.

  For $-0.5x^2-1.5x+5=0$, we have that $a=-0.5$, $b=-1.5$, and $c=5$.

  So we calculate

  $\begin{array}{rcl} x&=&\frac{1.5\pm\sqrt{(-1.5)^2-4(-0.5)(5)}}{2(-0.5)}\\ &=&~\frac{1.5\pm\sqrt{2.25+10}}{-1}\\ &=&~-1.5\pm\sqrt{12.25}\\ x_1&=&~-1.5+3.5=2\\ x_2&=&~-1.5-3.5=-5 \end{array}$

  The desired solutions are then $x=2$ or $x=-5$.

  Perhaps you'd like to multiply the quadratic equation $-0.5x^2-1.5x+5=0$ with $-2$ to get $x^2+3x-10=0$. You don't have to do it but it makes the calculations a little bit less complicated.

• Name the methods for solving quadratic equations.

  Hints

  The solutions of the corresponding quadratic equation are $x=10$ and $x=90$.

  This is the quadratic formula.

  The zero factor property states that if a product is equal to zero, then one of its factors must also equal zero.

  It doesn't matter at all which method you choose to find the solutions. If the solutions exist, then they are always the same.

  Solution

  If you have to solve a quadratic equation like $ax^2+bx+c=0$, or with $a=1$, $x^2+bx+c=0$, which you can derive by dividing the equation $ax^2+bx+c=0$ by $a$, then you can decide from several methods:

  1. You could factor the left-hand side of the equation $x^2+bx+c=0$ to $(x+d)(x+e)=0$ and get the solutions $x=-d$ and $x=-e$.
  2. You can complete the square to get $(x+e)^2=d$. You can solve this equation by taking the square root of both sides.
  3. You can use the quadratic formula for solving $ax^2+bx+c=0$.
  4. You also could draw the parabola corresponding to $f(x)=ax^2+bx+c$. The $x$-intercepts of this parabola are the desired solutions.

  It doesn't matter which method you choose, as the solutions, if they exist, are always the same.

• Solve the following quadratic equations.

  Hints

  Decide which method you you like to use. No matter which method you choose, your results will always be the same.

  Here you see the quadratic formula for solving quadratic equations like $ax^2+bx+c=0$.

  For solving a quadratic equation $x^2+bx+c=0$ by factoring, find the factors of $c$ which sum to $b$.

  Here is an example of complete the square of a quadratic equation.

  Solution

  Let's solve the various quadratic equations using the different methods we learned:

  First, lets use the method of factoring for the quadratic equation $x^2-2x-8=0$:

  1. Look for the factors of $8$ which sum to $-2$, namely $2\times (-4)=-8$ and $2-4=-2$ $~~~~~$✓.
  2. Thus we get the equivalent equation $(x+2)(x-4)=0$ which we solve by using opposite operations:

  Either

  $\begin{array}{rcl} x+2&=&~0\\ \color{#669900}{-2} & &\color{#669900}{-2}\\ x&=&~-2 \end{array}$

  or

  $\begin{array}{rcl} x-4&=&~0\\ \color{#669900}{+4} & &\color{#669900}{+4}\\ x&=&~4 \end{array}$

  This method (better!) works for whole solutions.

  $~$

  Next, let's solve $x^2-8x+7=0$ by completing the square:

  1. Subtract $7$ from both sides to get $x^2-8x=-7$.
  2. Adding $4^2$ on both sides leads to $x^2-8x+4^2=-7+4^2$.
  3. $(x-4)^2=9$ is the resulting equation.
  4. Take the square root on both sides and don't forget the $\pm$ sign: $x-4=\pm3$.
  5. Almost done: add $4$ to get $x=4+3=7$ or $x=4-3=1$.

  $~$

  You can also use the quadratic formula, which we will use to solve $2x^2+4x-6=0$. Nothing that $a=2$, $b=4$, and $c=-6$, we have that

  $\begin{array}{rcl} x&=&~\frac{-4\pm\sqrt{(-4)^2-4(2)(-6)}}{2(2)}\\ &=&~\frac{-4\pm\sqrt{16+48}}{4}\\ &=&~\frac{-4\pm\sqrt{64}}{4}\\ x_1&=&~\frac{-4+8}{4}=1\\ x_2&=&~\frac{-4-8}{4}=-3 \end{array}$

  $~$

  For $0.2x^2+1.2x+1=0$, first multiply both sides of the equation by $5$ to get $x^2+6x+5=0$, and thus $a=1$, $b=6$ and $c=5$. This simplifies the following calculation:

  $\begin{array}{rcl} x&=&~\frac{-6\pm\sqrt{6^2-4(1)(5)}}{2(1)}\\ &=&~\frac{-6\pm\sqrt{36-20}}{2}\\ &=&~\frac{-6\pm\sqrt{16}}{2}\\ x_1&=&~\frac{-6+4}{2}=-1\\ x_2&=&~\frac{-6-4}{2}=-5 \end{array}$