91Ó°ÊÓ

Quadratic equations can have complex roots when they do not intersect the x-axis. Complex roots always come in conjugate pairs like \(a + bi\) and \(a - bi\), where \(i\) is the imaginary unit representing \(\sqrt{-1}\). In the given exercise, the roots provided are \(\{-8i, 8i\}\). These are purely imaginary roots (since the real part \(a = 0\)) and they are also conjugates of each other.

When a quadratic equation has complex roots, it means you cannot graphically represent the solutions on the real-number line. Instead, they are located on the complex plane. Complex roots emerge from the fundamental theorem of algebra, which states that every non-constant single-variable polynomial with complex coefficients has at least one complex root.

Vieta’s Formulas

For any quadratic equation in the standard form \(ax^2 + bx + c = 0\), the sum of its roots can be found using Vieta's formulas. Specifically, the sum of the roots \(p + q\) is equal to \(\frac{-b}{a}\).

In our exercise, the sum of the complex roots \(\{-8i, 8i\}\) is \(0\). This aligns with Vieta’s Formula, as the middle term \(b\) of the standard form disappears, indicating \(b = 0\). It's interesting to see how even with complex numbers, this formula still applies.

Connection to the Equation's Constant

The product of the roots of a quadratic equation \(ax^2 + bx + c = 0\) is \(\frac{c}{a}\), another piece of Vieta’s formulas. This means that even without explicitly calculating the product of the roots, you can deduce it from the constant term in the equation.

In the given exercise, the product of the roots \(\{-8i, 8i\}\) is \(64\), which will be the constant term \(c\) in our standard form equation. The product here is positive because the roots are conjugates, and when multiplied, their imaginary parts cancel out, leaving a positive real number.

Quadratic equations are solved to find the values of \(x\) that make the equation true. Solutions to quadratic equations can be found by factoring, completing the square, using the quadratic formula, or graphing. The solutions may be real or complex and they signify the points where the graph of the equation intersects the x-axis (real roots) or fall on the complex plane (complex roots).

In our exercise, having complex roots suggests no real intersection point. Therefore, when we write our quadratic equation \(x^2 + 64 = 0\) in standard form, we show that the solutions are purely on the imaginary axis. To solve for \(x\), we would take the square root of \(64\), realizing that in this case, the solutions embody the initial roots given, \(\pm 8i\), confirming that the equation we formed is indeed correct.