Solving and Graphing Quadratic Equations

Consider the function f(x) = 2x² – 4x – 3.

1. Decide whether the function has a maximum or minimum value.
2. Evaluate the maximum or minimum value of the function.
3. State the domain and range of the function.

Solution

Here, a = 2, b = –4, and c = –3.

1. Since a > 0, the graph opens up and the function has a minimum value.

2. The minimum value of the function is the y-coordinate of the vertex.

The x-coordinate of the vertex is $-\frac{(-4)}{2\left(2\right)}=1$.

To find the y-coordinate of the vertex, we shall evaluate the function at x = 1.

$$\begin{gathered} f\left(1\right)=2{\left(1\right)}^2-4\left(1\right)-3 \\ \Rightarrow f\left(1\right)=-5 \end{gathered}$$

Hence, the minimum value of the function is –5.

3. The domain is the set of all real numbers. The range is the set of all real numbers greater than or equal to the minimum value, f(x) ≥ –5. In set notation, [–5, +∞ [.

The graph is shown below.

Example 1, Aishah Amri - StudySmarter Originals

1. Determine the y-intercept, the equation of the axis of symmetry, and the x-coordinate of the vertex.
2. Create a table of values that includes the vertex.
3. Graph the function based on the results from Questions 1 and 2.

Solution

Here, a = 1, b = –6 and c = 3.

1. To find the y-intercept we shall evaluate f at x = 0. In doing so we obtain, f(0) = c = 3. Thus the y-intercept is (0, 3). The equation of the axis of symmetry is found by using the formula

$$\begin{gathered} x=-\frac{b}{2a}=-\frac{(-6)}{2\left(1\right)} \\ \Rightarrow x=3 \end{gathered}$$

Therefore, the equation of the axis of symmetry is x = 3. It follows that the x-coordinate of the vertex is 3.

2. We shall now make our table of values. To do this, choose some values for x that are less than 3 and some that are greater than 3. This ensures that the points on each side of the axis of symmetry are graphed.

Let us apply the Location Principle to this example. Looking at the table above, there happens to be a change in sign between x = 0 and x = 1. This indicates that there is one real zero between these two values of x. Similarly, we observe a change in sign between x = 5 and x = 6 which again tells us that one real zero must exist between this pair of x values.

3. The graph is shown below.

Example 2, Aishah Amri - StudySmarter Originals

Solve the quadratic equation x² + 5x – 2 = 0 by graphing. If exact solutions cannot be found, state the consecutive integers between which the solutions are located.

Solution

Let f(x) = x² + 5x – 2.

We must now make a table of values for a domain x. We need to make a guess of the range of x-values we want to apply the Location Principle on. We shall choose the integer values between x = –6 and x = 2.

By the Location Principle, notice that the x-intercepts of the graph suggest that one solution is between –6 and –5, and the other is between 0 and 1.

Express a quadratic equation with $-\frac{1}{3}$ and 4 as its roots. Write the equation in standard form.

Solution

The intercept form of a quadratic equation is given by (x – p)(x – q) = 0 where p and q represent the roots of the expression. As there is no information on the coefficient of x² in this question, this infers that a = 1. Here, we are given

$p=-\frac{1}{3}andq=4$

Substituting this into the form above we obtain

$$\begin{gathered} \left(x-\left(-\frac{1}{3}\right)\right)(x-4)=0 \\ \Rightarrow \left(x+\frac{1}{3}\right)(x-4)=0 \end{gathered}$$

We now want to write this equation in the standard form of a quadratic equation. To do this, we must expand the right-hand side of the expression using the FOIL method.

$$\begin{gathered} \Rightarrow x^2-4x+\frac{1}{3}x-\frac{4}{3}=0 \\ \Rightarrow x^2-\frac{11}{3}x-\frac{4}{3}=0 \end{gathered}$$

Multiplying the entire equation by 3, we finally obtain

$\Rightarrow 3x^2-11x-4=0$

Factor the quadratic equation 3x² = 9x and graph the respective function.

Solution

We begin by rearranging the equation to the standard form of a quadratic equation

$3x^2-9x=0$

Notice that we can factor out 3x from the expression above as

$3x(x-3)=0$

By the Zero Product Property, the roots of this function is

$$\begin{gathered} 3x=0\Rightarrow x=0 \\ x-3=0\Rightarrow x=3 \end{gathered}$$

Since the coefficient of x² is a = 3 > 0, the parabola opens upward. The graph is shown below.

Example 3, Aishah Amri - StudySmarter Originals

Factor the quadratic equation 6x² = 1 – x and graph the respective function.

Solution

We begin by rearranging the equation to the standard form of a quadratic equation

$6x^2+x-1=0$

This is a general trinomial in which we can factorize by (ax + b)(cx + d). Note that the coefficient 6 can be factorized as $6\times 1or2\times 3$ while the constant -1 can be factorized as $-1\times 1$.

Remember that you have to test all possible pairs and positions for these two products. Using trial and error, we can factorize this as

$(3x-1)(2x+1)=0$

By the Zero Product Property, the roots of this function are

$$\begin{gathered} 3x-1=0\Rightarrow x=\frac{1}{3} \\ 2x+1=0\Rightarrow x=-\frac{1}{2} \end{gathered}$$

Since the coefficient of x² is a = 6 > 0, the parabola opens upward. The graph is shown below.

Example 4, Aishah Amri - StudySmarter Originals

Factor the quadratic equation x² – 25 = 0 and graph the respective function.

Solution

Notice that $25=5^2$.

Thus, the equation takes the form of the difference of two squares. Therefore, we can factorize this as

$$\begin{gathered} x^2-25=0 \\ \Rightarrow (x-5)(x+5)=0 \end{gathered}$$

By the Zero Product Property, the roots of this function are

$$\begin{gathered} x-5=0\Rightarrow x=5 \\ x+5=0\Rightarrow x=-5 \end{gathered}$$

Since the coefficient of x² is a = 1 > 0, the parabola opens upward. The graph is shown below.

Example 5, Aishah Amri - StudySmarter Originals

Solve the quadratic equation x² – 12x + 36 = 25 using the Square Root Property.

Solution

Notice that the left-hand side of this expression takes the form of a perfect square trinomial. Factorizing this, we obtain

$$\begin{gathered} x^2-12x+36=25 \\ \Rightarrow {(x-6)}^2=25 \end{gathered}$$

Now using the Square Root Property to solve this, we obtain

$$\begin{gathered} \Rightarrow x-6=\pm \sqrt{25} \\ \Rightarrow x=\pm \sqrt{25}+6 \\ \Rightarrow x=\pm 5+6 \end{gathered}$$

Thus, the two roots are x = 1 and x = 11. The graph is shown below.

Example 6 (1), Aishah Amri - StudySmarter Originals

Zooming into this graph, we see that the roots are indeed x = 1 and x = 11.

Example 6 (2), Aishah Amri - StudySmarter Originals

Solve the quadratic equation x² + 8x + 16 = 20 using the Square Root Property.

Solution

Notice that the left-hand side of this expression takes the form of a perfect square trinomial. Factorizing this, we obtain

$$\begin{gathered} x^2+8x+16=20 \\ \Rightarrow {(x+4)}^2=20 \end{gathered}$$

Now using the Square Root Property to solve this, we obtain

$$\begin{gathered} \Rightarrow x+4=\pm \sqrt{20} \\ \Rightarrow x=\pm \sqrt{20}-4 \\ \Rightarrow x=\pm 2\sqrt{5}-4 \end{gathered}$$

Thus, the two roots are

$$\begin{gathered} x=-2\sqrt{5}-4\approx -8.47(correcttotwodecimalplaces) \\ x=2\sqrt{5}-4\approx 0.47(correcttotwodecimalplaces) \end{gathered}$$

A rough sketch of this graph is shown below.

Example 7 (1), Aishah Amri - StudySmarter Originals

Zooming into this graph, we see that the roots are approximately x ≈ -8,47 and x ≈ 0.47.

Example 7 (2), Aishah Amri - StudySmarter Originals

Solve the quadratic equation x² – 10x + 24 = 0 by completing the square.

Solution

Subtract 24 from both sides of the equation as

$$\begin{gathered} x^2-10x+24=0 \\ \Rightarrow x^2-10x=-24 \end{gathered}$$

To complete the square of this expression, we must find the missing values of

$x^2-10x+\square =-24+\square$

Since

${\left(-\frac{10}{2}\right)}^2={(-5)}^2=25$

we shall add 25 to each side to balance the equation

$$\begin{gathered} \Rightarrow x^2-10x+25=-24+25 \\ \Rightarrow x^2-10x+25=1 \end{gathered}$$

Rewrite the left-hand side as a perfect square by factoring

$\Rightarrow {(x-5)}^2=1$

We can now use the Square Root Property

$$\begin{gathered} \Rightarrow x-5=\pm \sqrt{1}=\pm 1 \\ \Rightarrow x=\pm 1+5 \end{gathered}$$

Thus, the two solutions are x = 4 and x = 6. The graph is shown below.

Example 8, Aishah Amri - StudySmarter Originals

Solve the quadratic equation 3x² + 10x – 8 = 0 by completing the square.

Solution

Divide the equation by the coefficient 3

$$\begin{gathered} 3x^2+10x-8=0 \\ \Rightarrow x^2+\frac{10}{3}x-\frac{8}{3}=0 \end{gathered}$$

Add $\frac{8}{3}$ on both sides

$\Rightarrow x^2+\frac{10}{3}x=\frac{8}{3}$

To complete the square of this expression, we must find the missing values of

$x^2+\frac{10}{3}x+\square =\frac{8}{3}+\square$

Since

${\left(\frac{10}{2\times 3}\right)}^2={\left(\frac{5}{3}\right)}^2=\frac{25}{9}$

we shall add $\frac{25}{9}$ to each side to balance the equation

$$\begin{gathered} \Rightarrow x^2+\frac{10}{3}x+\frac{25}{9}=\frac{8}{3}+\frac{25}{9} \\ \Rightarrow x^2+\frac{10}{3}x+\frac{25}{9}=\frac{8}{3}+\frac{25}{9}=\frac{49}{9} \end{gathered}$$

Rewrite the left-hand side as a perfect square by factoring

$\Rightarrow {\left(x+\frac{5}{3}\right)}^2=\frac{49}{9}$

We can now use the Square Root Property

$$\begin{gathered} \Rightarrow x+\frac{5}{3}=\pm \sqrt{\frac{49}{9}} \\ \Rightarrow x=\pm \frac{\sqrt{49}}{\sqrt{9}}-\frac{5}{3} \\ \Rightarrow x=\pm \frac{7}{3}-\frac{5}{3} \end{gathered}$$

Thus, the two solutions are $x=-4andx=\frac{2}{3}$. The graph is displayed below.

Example 9 (1), Aishah Amri - StudySmarter Originals

Zooming into this graph, we see that the roots are true as required.

Example 9 (2), Aishah Amri - StudySmarter Originals

Identify the number and type of roots for the quadratic equation 3x² + 5x + 1 = 0 . Determine the roots of this expression using the Quadratic Formula.

Solution

Here, $a=3,b=5andc=1.$

We begin by finding the value of the discriminant as

$$\begin{gathered} D=b^2-4ac={\left(5\right)}^2-4\left(3\right)\left(1\right) \\ \Rightarrow D=25-12 \\ \Rightarrow D=13 \end{gathered}$$

Since D > 0, we must have two real roots. Also, note that D is not a perfect square so the roots for this quadratic equation must be irrational. We shall now use the Quadratic Formula to determine these roots.

$$\begin{gathered} x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-\left(5\right)\pm \sqrt{13}}{2\left(3\right)} \\ \Rightarrow x=\frac{-5\pm \sqrt{13}}{6} \end{gathered}$$

Thus, the two roots are

$$\begin{gathered} x=\frac{-5-\sqrt{13}}{6}\approx -1.43(correcttotwodecimalplaces) \\ x=\frac{-5+\sqrt{13}}{6}\approx -0.23(correcttotwodecimalplaces) \end{gathered}$$

The graph is shown below.

Example 10, Aishah Amri - StudySmarter Originals

Determine the number and type of roots for the quadratic equation x² – 6x + 10 = 0 . Determine the roots of this expression using the Quadratic Formula.

Solution

Here, $a=1,b=-6andc=10.$

We begin by finding the value of the discriminant as

$$\begin{gathered} D={(-6)}^2-4\left(1\right)\left(10\right) \\ \Rightarrow D=36-40 \\ \Rightarrow D=-4 \end{gathered}$$

Since D < 0, we must have two complex conjugate roots. We shall now use the Quadratic Formula to determine these roots.

$$\begin{gathered} x=\frac{-(-6)\pm \sqrt{-4}}{2\left(1\right)} \\ \Rightarrow x=\frac{6\pm i\sqrt{4}}{2} \\ \Rightarrow x=\frac{6\pm 2i}{2} \\ \Rightarrow x=3\pm i \end{gathered}$$

Thus, the two complex conjugate roots are $x=3-iandx=3+i.$

Example 11, Aishah Amri - StudySmarter Originals

Express the equation y = x² + 4x + 6 in vertex form and plot the graph of the function.

Solution

Notice that y = x² + 4x + 6 is not a perfect square. We begin by completing the square of this expression.

$y=(x^2+4x+\square )+6-\square$

To find the missing values, we add ${\left(\frac{4}{2}\right)}^2=4$ and balance the equation by subtracting 4 as

$y=(x^2+4x+\mathbf{4})+6-\mathbf{4}$

Now writing this as a perfect square, we obtain

$\Rightarrow y={(x+2)}^2+2$

Since h = –2 and k = 2, the vertex is at (–2, 2). The axis of symmetry is x = –2. Since a = 1, the graph opens up and has the same shape as the graph of y = x². Furthermore, the graph is translated 2 units left and 2 units up.

The graph is shown below.

Example 12, Aishah Amri - StudySmarter Originals

Write the equation y = –2x² – 12x + 17 in vertex form and plot the graph of the function.

Solution

Notice that y = –2x² – 12x + 17 is not a perfect square. We begin by factoring the first two terms of the right-hand side as

$y=-2(x^2+6x)+17$

Next, we shall complete the square of this expression.

$y=-2(x^2+6x+\square )+17-(-2)\square$

To find the missing values, we add ${\left(\frac{6}{2}\right)}^2=9$ and balance the equation by subtracting –2(9) as

$y=\mathbf{-}\mathbf{2}(x^2+6x+\mathbf{9})+17-\mathbf{(}\mathbf{-}\mathbf{2}\mathbf{)}\mathbf{\left(}\mathbf{9}\mathbf{\right)}$

Now writing this as a perfect square, we obtain

$\Rightarrow y={-2(x+3)}^2+35$

The vertex is at (–3, 35), and the axis of symmetry is x = –3. Since a = –2, the graph opens downward and is narrower than the graph of y = x². Additionally, the graph is translated 1 unit right and 2 units up.

The graph is shown below.

Example 13, Aishah Amri - StudySmarter Originals