Find the Roots (Zeros) 2x^2+x+2=0

Solution:

Step:1

Apply the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Step:2

Insert the coefficients $a = 2$, $b = 1$, and $c = 2$ into the formula: $x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot 2}}{2 \cdot 2}$.

Step:3

Begin simplification.

Step:3.1

Start with the numerator.

Step:3.1.1

Recognize that any number raised to the power of 2 is itself squared: $x = \frac{-1 \pm \sqrt{1 - 4 \cdot 2 \cdot 2}}{2 \cdot 2}$.

Step:3.1.2

Perform the multiplication inside the square root.

Step:3.1.2.1

Calculate $-4$ times $2$: $x = \frac{-1 \pm \sqrt{1 - 8 \cdot 2}}{2 \cdot 2}$.

Step:3.1.2.2

Next, multiply $-8$ by $2$: $x = \frac{-1 \pm \sqrt{1 - 16}}{2 \cdot 2}$.

Step:3.1.3

Subtract $16$ from $1$: $x = \frac{-1 \pm \sqrt{-15}}{2 \cdot 2}$.

Step:3.1.4

Express $-15$ as $-1 \cdot 15$: $x = \frac{-1 \pm \sqrt{-1 \cdot 15}}{2 \cdot 2}$.

Step:3.1.5

Split the square root of the product into the product of square roots: $x = \frac{-1 \pm \sqrt{-1} \cdot \sqrt{15}}{2 \cdot 2}$.

Step:3.1.6

Replace $\sqrt{-1}$ with $i$: $x = \frac{-1 \pm i \sqrt{15}}{2 \cdot 2}$.

Step:3.2

Simplify the denominator by multiplying $2$ by $2$: $x = \frac{-1 \pm i \sqrt{15}}{4}$.

Step:4

The roots are complex and given by $x = \frac{-1 \pm i \sqrt{15}}{4}$.

Knowledge Notes:

The problem involves finding the roots of a quadratic equation, which is an equation of the form $ax^2 + bx + c = 0$. The roots are the values of $x$ that satisfy the equation. There are various methods to find the roots, but one of the most common is using the quadratic formula:

$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

This formula provides the solutions to the quadratic equation, where $a$, $b$, and $c$ are coefficients from the equation, and $\sqrt{b^2 - 4ac}$ is called the discriminant. The discriminant determines the nature of the roots:

• If the discriminant is positive, there are two distinct real roots.

• If the discriminant is zero, there is one real root (also called a repeated or double root).

• If the discriminant is negative, there are two complex roots, which are conjugates of each other.

In the given problem, the discriminant is negative, indicating that the roots will be complex numbers. Complex numbers have a real part and an imaginary part, and are often represented as $a + bi$, where $i$ is the imaginary unit, defined by $i^2 = -1$. When dealing with square roots of negative numbers, we use the imaginary unit to express them as multiples of $i$. The process of simplification involves arithmetic operations and the application of the properties of square roots and complex numbers.