Complex Solutions of Quadratic Equations: Videos & Practice Problems

Solving quadratic equations using the square root property involves isolating the variable and applying the square root to both sides. When encountering a negative number under the square root, the solution includes imaginary numbers, represented by $i$, where $\sqrt{-1}=i$. For example, solving $x^2=\; -16$ yields $x=\; \pm 4i$. This approach integrates understanding of polynomials, exponents, and complex numbers, essential for mastering quadratic equations and their solutions.

Imaginary Solutions

When solving quadratic equations using the square root property, encountering a negative number under the square root indicates the presence of imaginary or complex solutions. This occurs because the square root of a negative number is not defined within the real numbers but can be expressed using the imaginary unit i, where i is defined as the square root of -1, or \(i = \sqrt{-1}\).

For example, consider the quadratic equation \$2x^2 + 32 = 0\(. To solve for x, first isolate the squared term by subtracting 32 from both sides:

\[2x^2 = -32\]

Next, divide both sides by 2 to simplify:

\[x^2 = -16\]

Applying the square root property, take the square root of both sides:

\[x = \pm \sqrt{-16}\]

Since the radicand (the number under the square root) is negative, rewrite the square root using the product property:

\[\sqrt{-16} = \sqrt{16} \times \sqrt{-1}\]

Recognizing that \)\sqrt{16} = 4\( and \)\sqrt{-1} = i\(, the expression simplifies to:

\[x = \pm 4i\]

Thus, the solutions are complex numbers: \)x = 4i\( and \)x = -4i$. This method demonstrates how the square root property extends to complex solutions by incorporating the imaginary unit i. Understanding this process is essential for solving quadratic equations that do not have real roots, expanding the scope of solutions to include the complex number system.

Solve each quadratic equation. If roots are not real, use $i$.
$2x^2+18=0$

• A. $x=\pm3$
• B. $x=\pm\frac92$
• C. $x=\pm3i$
• D. $x=\pm\frac92i$

Answer: (C)

Solve each quadratic equation. If roots are not real, use $i$.
$\frac{w^2}{4}+8=0$

• A. $w=\pm4\sqrt2i$
• B. $w=\pm4i$
• C. $w=\pm4$
• D. $w=\pm4\sqrt2$

Answer: (A)

Solve each quadratic equation. If roots are not real, use $i$.
${(3z-1)}^2+5=0$

• A. $z=\frac{1\pm2i}{3}$
• B. $z=\frac{1\pm\sqrt5i}{3}$
• C. $z=\pm3i$
• D. $z=\pm\frac13i$

Answer: (B)

Complete the Square Example 1

To solve a quadratic equation by completing the square, start by visualizing the quadratic expression as a geometric square diagram. Represent the x² term as the area of a square with side lengths x. Next, split the linear term, such as −2x, into two equal parts, each corresponding to rectangles attached to the square. For example, splitting −2x into −x and −x means the smaller side lengths of these rectangles are −1. Completing the square involves adding the area of the small square formed by these side lengths, which is (−1) × (−1) = 1. This transforms the quadratic expression into a perfect square trinomial: (x − 1)² = x² − 2x + 1.

However, if the original quadratic is x² − 2x + 3, the constant term differs from the perfect square trinomial by 2. To balance the equation, add 2 to both sides, rewriting the quadratic as (x − 1)² + 2. Setting this equal to zero allows solving for x:

\[(x - 1)^2 + 2 = 0\]

Subtract 2 from both sides:

\[(x - 1)^2 = -2\]

Applying the square root property, take the square root of both sides:

\[x - 1 = \pm \sqrt{-2}\]

Since the square root of a negative number involves imaginary numbers, express \sqrt{-2} as \sqrt{2} \times \sqrt{-1}. Recognizing that \sqrt{-1} = i, the imaginary unit, the solutions become:

\[x = 1 \pm \sqrt{2}i\]

This method highlights how completing the square not only simplifies quadratic expressions but also reveals complex solutions when the quadratic does not intersect the x-axis. Understanding the geometric interpretation and algebraic manipulation involved in completing the square is essential for solving quadratic equations, especially when dealing with complex numbers.

Here’s what students ask on this topic:

How do you solve quadratic equations with complex solutions using the square root property?

To solve quadratic equations with complex solutions using the square root property, first isolate the squared term. For example, consider the equation $x^2=-16$. Since the right side is negative, taking the square root involves imaginary numbers. Apply the square root to both sides: $x=\pm \sqrt{-16}$. Rewrite the square root of a negative number as the product of the square root of the positive number and $i$, where $i^2=-1$. So, $\sqrt{-16}=\sqrt{16}\sqrt{-1}=\; 4i$. Therefore, the solutions are $x=\pm 4i$. This method applies to any quadratic where the term under the square root is negative, leading to complex solutions.

What does it mean when a quadratic equation has a negative number under the square root?

When a quadratic equation has a negative number under the square root, it means the solutions are not real numbers but complex or imaginary numbers. This occurs because the square root of a negative number is not defined in the set of real numbers. Instead, we use the imaginary unit $i$, where $i^2=-1$. For example, if you have $\sqrt{-16}$, it can be rewritten as $4i$. This indicates the solutions are complex numbers, combining real and imaginary parts. Understanding this concept is essential for solving quadratic equations that do not have real roots.

How do you simplify the square root of a negative number in quadratic equations?

To simplify the square root of a negative number in quadratic equations, separate the negative sign from the number under the root. For example, $\sqrt{-16}$ can be rewritten as $\sqrt{-1}\sqrt{16}$. Since $\sqrt{-1}=i$ and $\sqrt{16}=\; 4$, the expression simplifies to $4i$. This process allows you to express complex solutions clearly when solving quadratic equations with negative values under the square root.

Can you explain the role of the imaginary unit $i$ in solving quadratic equations?

The imaginary unit $i$ plays a crucial role in solving quadratic equations when the discriminant (the value under the square root) is negative. Since the square root of a negative number is not a real number, $i$ is defined such that $i^2=-1$. This allows us to express solutions involving the square root of negative numbers. For example, if $x^2=-16$, then $x=\pm 4i$. Using $i$ extends the set of solutions to include complex numbers, ensuring all quadratic equations have solutions.

What steps should I follow to solve $2x^2\; +\; 32\; =\; 0$ and find complex solutions?

To solve $2x^2\; +\; 32\; =\; 0$ for complex solutions, follow these steps: First, isolate the squared term by subtracting 32 from both sides: $2x^2\; =\; -32$. Next, divide both sides by 2 to get $x^2\; =\; -16$. Now, apply the square root property: $x\; =\pm \sqrt{-16}$. Since the square root of a negative number involves $i$, rewrite it as $x\; =\pm 4i$. These are the complex solutions to the equation.

How do complex solutions relate to the graph of a quadratic function?

Complex solutions of a quadratic function occur when the graph does not intersect the x-axis. This happens when the discriminant (the expression under the square root in the quadratic formula) is negative. In such cases, the quadratic equation has no real roots but two complex conjugate solutions involving the imaginary unit $i$. For example, if the equation is $2x^2\; +\; 32\; =\; 0$, the solutions are $x\; =\pm 4i$, indicating the parabola lies entirely above or below the x-axis without crossing it. Understanding this helps connect algebraic solutions to their graphical representations.