Question

Find the zeros of the function. \(f(x)=3 x^2+6\)

Short answer

The function \(f(x)=3 x^2+6\) has no zeros in the real number set because the discriminant is negative.

Step-by-step solution

Write Down The Function

The given function is \(f(x)=3 x^2+6\).

Set the Function to Zero

To find the zeros of the function, the function has to be set to zero, which gives the equation \(3 x^2+6 = 0\).

Solve the Quadratic Equation

The quadratic equation is now simplified and solved. First, divide all terms by 3: \(x^2+2 = 0\), then subtract 2 from both sides: \(x^2 = -2\). We do not have real solutions for this equation because the square of x cannot be a negative number. Therefore, the function \(f(x)=3 x^2+6\) does not have any zeros within the real numbers.

Key concepts

Solving Quadratic Equations

When faced with a quadratic equation, the goal is often to find the values of x that make the equation true, commonly known as the 'roots' or 'zeros' of the equation. A quadratic equation is typically in the form of \(ax^2 + bx + c = 0\), where 'a', 'b', and 'c' are coefficients and 'x' represents the unknown variable.

To solve for x, you can use several methods, such as factoring, completing the square, using the quadratic formula, or graphically observing where the function crosses the x-axis. In our exercise, where we have \(3x^2+6=0\), there's no 'b' or 'c' term, so we attempt to isolate x by itself. After dividing both sides by 3 and moving 6 to the other side, we're left with \(x^2 = -2\), which hints at the need for complex numbers to find solutions, as no real number squared can give us a negative result.

Complex Numbers

Complex numbers come into play when we encounter equations that involve the square roots of negative numbers. In the real number system, there is no number that can be squared to give a negative result. That's where the imaginary unit 'i' comes in, which is defined as \(i = \sqrt{-1}\).

Using this definition, we can express the square root of any negative number as a multiple of i. For instance, in our exercise, we have \(x^2 = -2\), which can be rewritten as \(x = \pm\sqrt{-2}\). Breaking this down further, we get \(x = \pm\sqrt{2}i\). This is how the roots of the function are represented using complex numbers, extending our understanding of numbers beyond the real number line and allowing us to solve a broader set of quadratic equations.

No Real Solutions

The notion of 'no real solutions' in the context of quadratic equations means that the function does not cross the x-axis at any point. Such a scenario occurs when the quadratic equation yields a negative number inside the square root of the quadratic formula \(\frac{-b \pm \sqrt{b^2-4ac}}{2a}\).

If this part, known as the discriminant \(b^2-4ac\), is less than zero, there are no real numbers that can satisfy the equation. This is because, in the real number system, we cannot take the square root of a negative number to find a real value. Thus, when we say a quadratic equation has 'no real solutions,' it implies the existence of complex solutions. In our example, where no real numbers can square to equal -2, the quadratic equation does not factor into real roots, confining the zeros of the function to the realm of complex numbers.