Problem 51 - Algebra 2

Question

Find the exact solutions of \(2 i x^{2}-3 i x-5 i=0\) by using the Quadratic Formula.

Step-by-Step Solution

Answer

The solutions are \( x = \frac{3 - i\sqrt{31}}{4} \) and \( x = \frac{3 + i\sqrt{31}}{4} \).

Step 1: Identify Coefficients

The given quadratic equation is \( 2ix^2 - 3ix - 5i = 0 \). Here, the coefficients are \( a = 2i \), \( b = -3i \), and \( c = -5i \).

Step 2: Write the Quadratic Formula

The quadratic formula is given by \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). We will substitute the values of \( a \), \( b \), and \( c \) into this formula.

Step 3: Calculate the Discriminant

Calculate the discriminant \( b^2 - 4ac \). This is \((-3i)^2 - 4(2i)(-5i)\). Calculate \((-3i)^2 = -9\), and \(4(2i)(-5i) = -40i^2 = 40\) (since \(i^2 = -1\)). Thus, the discriminant becomes \(-9 + 40 = 31\).

Step 4: Compute the Square Root of Discriminant

The square root of the discriminant \( \sqrt{31} \) is a real number since 31 is positive. Thus, \( \sqrt{31} \) simplifies to \( \sqrt{31} \).

Step 5: Plug Values into the Quadratic Formula

Substitute back into the formula: \( x = \frac{-(-3i) \pm \sqrt{31}}{2(2i)} \) which simplifies to \( x = \frac{3i \pm \sqrt{31}}{4i} \).

Step 6: Simplify the Expression

Simplify \( x = \frac{3i \pm \sqrt{31}}{4i} \) by multiplying both numerator and denominator by \(-i\) to remove the imaginary unit from the denominator: \( x = \frac{(3i \pm \sqrt{31})(-i)}{4(-i^2)} = \frac{3 - i\sqrt{31}}{4} \) or \( x = \frac{3 + i\sqrt{31}}{4} \).

Key Concepts

Complex Numbers

Complex numbers are like regular numbers, but they have a special component known as the imaginary part. They're written in the form of \( a + bi \), where \( a \) is a real number, and \( bi \) is the imaginary part. The imaginary unit \( i \) is defined as the square root of \( -1 \). A complex number essentially allows for solutions beyond the normal number line, reaching into a two-dimensional plane. This plane is known as the complex plane.

In the equation provided, each coefficient of the quadratic equation \( 2ix^2 - 3ix - 5i = 0 \) is a complex number, specifically imaginary numbers because they don't have a real part."

They allow us to handle equations that cannot be solved using only real numbers, unlocking a whole new world of mathematical possibilities.

Quadratic Equations

A quadratic equation is any equation that can be rearranged in the standard form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \).
These equations form a parabola when plotted in the Cartesian plane, and they can have zero, one, or two solutions.
Quadratic equations are often used in various fields like physics, biology, and engineering to model parabolic trends or curved structures.

For our exercise, the quadratic equation is \( 2ix^2 - 3ix - 5i = 0 \), where the coefficients are complex numbers.

Discriminant

The discriminant is a special part of the quadratic formula. It helps us determine the nature of the roots of the quadratic equation.

In the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), the expression \( b^2 - 4ac \) is the discriminant.
The value of this discriminant can tell us the number and type of solutions:

• If the discriminant is positive, the roots are real and distinct.
• If it's zero, the roots are real and identical.
• If it's negative, the roots are complex and conjugate pairs.

In this exercise, the discriminant \( b^2 - 4ac = 31 \) is positive, implying that our solutions will be complex numbers with no real numbers involved in their imaginary parts themselves. Even though the discriminant is positive, since we deal with imaginary coefficients (\( a, b, c \) are imaginary thus rooted in complex form), the solutions are complex.

Complex Solutions

Complex solutions occur when the roots of a quadratic equation are not real numbers and must include an imaginary component. This happens when the discriminant is negative or when dealing with quadratic equations in the complex plane.

In the current problem, despite having a positive discriminant, we began with imaginary coefficients \( 2i, -3i, -5i \). Hence, our roots will be complex. In simple terms, while solving for \( x \) using the quadratic formula, the given quadratic coefficients led to solutions involving complex numbers.

The solutions we obtained are \( x = \frac{3 - i\sqrt{31}}{4} \) and \( x = \frac{3 + i\sqrt{31}}{4} \). These are complex because they include an imaginary part \( i\sqrt{31} \), and they highlight how quadratic equations in the complex domain can have solutions that lie beyond the real number axis.