Solving Quadratic Equations by Completing the Square

Solving Quadratic Equations by Completing the Square exercise

• Determine the vertex of the parabola.

  Hints

  What is the format of a perfect square trinomial? You will need to add a term to the expression inside the parentheses on the left hand side of the equation in order to turn it into a perfect square trinomial.

  For an expression in the form $ax^2 + bx$, the third term that must be added to make it a perfect square trinomial is of the form: $(\frac{b}{2})^2$.

  Remember that the vertex form of the equation is $a (x-h)^2 + k = 0$. Pay close attention to the signs when you are determining the values of $h$ and $k$.

  Solution

  Kata Ana knows that if she rearranges the function into vertex form, she will be able to easily identify the coordinates of the vertex. Vertex form is $a(x-h)^2+k = 0$, and gives the vertex as $(h,k)$.

  If the given equation was in standard form, she would begin by moving the constant term to the right hand side of the equation. Then she would factor out the coefficient on the $x^2$ term to give an expression in parentheses on the left hand side. She notices that this has already been done to the given equation, $\frac{1}{8}(x^2+20x)=-\frac{9}{2}$.

  She knows that she now needs to get the left hand side of the given function into a form that includes a perfect square trinomial. She knows that this trinomial is the sum of three terms: a term with $x^2$, another with $x$, and a constant. The expression on the left hand side of the given function is missing the constant term. Kata Ana knows that she must determine what constant term should be added to result in an expression that is a perfect square. She can find this constant by taking the coefficient of the $x$ term, $20$, dividing it by two, and then squaring it. She finds $(\frac{20}{2})^2 = 10^2$.

  She wants to add this term to the expression that is inside the parentheses on the left hand side. However, she knows that this entire expression in parentheses is being multiplied by $\frac{1}{8}$. So she really needs to add $\frac{1}{8}\times10^2$ to both sides, in order to add $10^2$ inside the parentheses on the left hand side.

  $\begin{array}{rclcl} \frac{1}{8}(x^2+20x+10^2) & = & -\frac{9}{2}+\frac{10^2}{8} & | & \text{simplifying} \\ \frac{1}{8}(x^2+20x+10^2) & = & 8 \\ \end{array}$

  Now she has a perfect square trionomial on the left hand side, $(x^2+20x+10^2)$. She can factor this trinomial to result in the following function:

  $\frac{1}{8}(x+10)^2 = 8$.

  Now, she can use opposite operations to get the function into vertex form:

  $\frac{1}{8}(x+10)^2 -8 = 0$.

  This shows her that $h = -10$, and $k=-8$. Therefore the vertex coordinates are $(-10,-8)$.

• Find the zeroes of the quadratic equation.

  Hints

  Remember that a positive or negative number can result from taking the square root of a given number.

  You must pay close attention to signs.

  Solution

  Kata Ana knows that she must find the zeroes, or roots, of the quadratic function. To do this, she must isolate the variable $x$.

  She starts by observing that the given equation is in vertex form, which is $a (x-h)^2 + k = 0$. Next she uses opposite operations to move $k$ over to the the right hand side, and then divides both sides by $a$.

  $ \begin{array}{rclcl} -\frac{1}{10}(x-8)^2 + 10 &=& 0&|&\text{subtract 10 from both sides} \\ -\frac{1}{10}(x-8)^2 &=&-10 &|&\text{divide by -1/10} \\ (x-8)^2 &=&100&|&\text{divide by -1/10} \\ \end{array} $

  To continue to isolate $x$, she must take the square root of both sides of the equation. When taking the square root of a number, it is important to remember that the answer can be positive or negative. For example $\sqrt9 = \pm 3$ because $9=3^3=(-3)^2$. In this case:

  $x-8=\pm 10$.

  She then finishes isolating $x$, and arrives at:

  $x = 8 \pm 10$.

  She can now find the zeroes of the function to be:

  $8 - 10 = -2$

  and

  $8 + 10 = 18$.

• Solve for the zeroes of each quadratic equation.

  Hints

  You can find the zeroes by isolating $x$.

  The given functions are in the form $a (x-h)^2 + k = 0$. How can you move $k$ and $a$ to the right hand side?

  Remember that when you take the square root of a number, there are two possible answers. For example $\sqrt 81 = \pm9$.

  Solution

  These functions have already been set equal to zero. The pair of roots for each function can therefore be found by isolating $x$.

  For the first function:

  $2(x-5)^2-8=0$.

  We move the constant to the right hand side:

  $2(x-5)^2=8$.

  Then we can divide through by the term that is outside the parentheses on the left hand side:

  $(x-5)^2=\dfrac{8}{2}=4$.

  Then we can take the square root of each side:

  $x-5=\sqrt4$.

  Remember that the square root can be a positive or negative value:

  $x-5=\pm 2$.

  Finally, we can move the constant to the right hand side, and determine the roots:

  $x=5 \pm 2$

  $\begin{array}{rcl} x=5+2 & or & x=5-2 \\ x=7 & or & x=3 \end{array}$

  So we find that the roots are $3$ and $7$.

  Similarly, for the other equations, we find the roots:

  $\begin{array}{rcl} -3(x+2)^2+27=0 & | & \text{roots are -5 and 1} \\ 4(x-2)^2-64=0 & | & \text{roots are -2 and 6} \\ -3(x-9)^2+12=0 & | & \text{roots are 7 and 11} \\ \end{array}$

• Calculate the vertex and zeroes of the quadratic equation.

  Hints

  It will be helpful if you always reduce rational numbers to their simplest form during your solution.

  At every step of the way, every term in this problem (both coefficients and constants) can be simplified to an integer. If you have a rational number that you can't immediately simplify to an integer in your solution, you should double check your calculations.

  Remember, you can convert the standard form equation to vertex form by completing the square. The vertex will then be at the coordinates $(h,k)$.

  Once you have the function in vertex form, you can solve for the zeroes by setting the function equal to zero, and isolating $x$.

  There are two roots for this function.

  Solution

  We need to find the vertex, and the roots of this function. We can find the vertex by converting the function into vertex form, $a (x-h)^2 + k = 0$.

  We have the standard form function:

  $-2x^2+24x-64=0$.

  We start by moving the constant term to the right hand side:

  $-2x^2+24x=64$.

  Next, we factor out the coefficient on the $x^2$ term from the left hand side:

  $-2(x^2-12x)=64$.

  Then we take the coefficient on the $x$ term, divide it by two, and square the result:

  $(\frac{12}{2})^2 = 6^2 = 36$.

  Adding this constant to the expression inside the parentheses on the left hand side of the equation will give us a perfect square trinomial, which we know how to factor. In order to add the constant inside the parentheses on the left hand side, we have to add the constant times negative two on the right hand side. This is because all terms inside the parentheses on the left hand side are multiplied by negative two. So really, we add $36\times-2$ to both sides.

  $ \begin{array}{rclcl} -2(x^2-12x+36) & = & 64+(36\times-2) &|&\text{Simplifying} \\ -2(x^2-12x+36) & = & 64-72 \\ -2(x^2-12x+36) & = & -8 \end{array} $

  Now we can factor the perfect square trinomial on the left hand side, to get:

  $-2(x-6)^2 = -8$.

  And finally, we can move the constant to the left hand side, to get the equation in vertex form:

  $-2(x-6)^2+8=0$.

  This tells us that the vertex is at $(6,8)$.

  Now, we can isolate for $x$ to find the roots.

  The first step is to move the constant to the right hand side, which we have already done in an earlier step: $-2(x-6)^2 = -8$.

  Then we can divide through by the term that is outside the parentheses on the left hand side:

  $ \begin{array}{rclcl} (x-6)^2 & = & \frac{-8}{-2} &|&\text{simplifying} \\ (x-6)^2 & = & 4 \end{array} $

  Taking the sqaure root of both sides is the next step to isolate $x$:

  $x-6 = \sqrt4$.

  This gives us two possible solutions, a positive one and a negative one:

  $x-6 = \pm 2$.

  We can then find the two roots of the function:

  $\begin{array}{rcl} x=6+2 & or & x=6-2 \\ x=8 & or & x=4 \end{array}$

• Identify the vertex form of a quadratic function.

  Hints

  Remember that the standard form is the most simplified form. This is not the vertex form.

  Remember that the vertex takes its name from the fact that it includes the coordinates of the parabola's vertex.

  The vertex form gives the coordinates of the vertex as $(h,k)$.

  Solution

  The standard form of a quadratic functon is $ax^2+bx+c=0$. This is the most simplified format, and is a trinomial.

  The vertex form is $a (x-h)^2 + k = 0$. It reveals the coordinates of the vertex, which are $(h,k)$.

  The factored form of a quadratic function is $a(x-p)(x-q)=0$. It shows the intercepts, which are $p$ and $q$.

  $ax^2+bx=-c$ is a rearranged version of the standard form. It shows the first step in converting a standard form function into the vertex form. But it is not yet in the vertex form.

• Identify the graphs of the functions.

  Hints

  The equations are given in vertex form.

  Pay close attention to the signs.

  How can you determine the coordinates of the vertex from the values $h$ and $k$ in a vertex form equation?

  It may help to start by finding the correct vertices of each given equation. Then you can compare those vertices to those on the graph.

  Solution

  We are given the vertices of three parabolas, and many quadratic functions. The quadratic functions are in vertex form.

  We know that in general, the vertex form is:

  $a (x-h)^2 + k = 0$,

  Where the vertex is $(h,k)$.

  Let's find the values of $h$ and $k$ that correspond to the vertex that is given for each parabola.

  For the purple parabola, the vertex is given as $(2,-4)$. Therefore $h = 2$ and $k = -4$. Substituting these values into the vertex form of a quadratic function, we get:

  $a (x-2)^2 -4 = 0$.

  We don't yet know the value of $a$. The only function that is listed that matches the values that we found for $h$ and $k$ is:

  $2(x-2)^2-4=0$.

  Therefore this is the function that describes the purple parabola.

  Similarly, for the blue parabola, we know the vertex is at $(4,12)$, which means that $h = 4$ and $k = -12$. The only given function that matches is:

  $-3(x-4)^2+12=0$.

  For the gold parabola, we know the vertex is at $(-1,-4)$, which means that $h = -1$ and $k = -4$. The only given function that matches is $(x+1)^2-4=0$.