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Use the function \(f(x)=a x^{2}+2 x+1\) where a is a real number. For what value(s) of \(a\) will \(f\) have no real zeros?

Short Answer

Expert verified
The quadratic function \(f(x)=ax^{2}+2x+1\) has no real zeros when \(a > 1\)

Step by step solution

01

Write down the formula for the discriminant

The formula for the discriminant is \(D = b^2 - 4ac\). Here in the given quadratic equation, \(b = 2\) and \(c = 1\). Therefore, I need to plug these values into the formula in order to isolate \(a\).
02

Substitute the values of \(b\) and \(c\) into the formula

The equation after substituting these values will become: \(D = 2^2 - 4*a*1 = 4 - 4a\)
03

Set the Discriminant D < 0

The task is to find when there are no real roots, and this occurs when \(D < 0\). Therefore, the equation that needs to be solved is \(4 - 4a< 0\).
04

Solve the inequality for \(a\)

Solving this simple linear inequality, we divide both sides by \(-4\) and we get \(a > 1\). Please remember that when dividing or multiplying both sides of an inequality by a negative number, the direction of the inequality flips!

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Discriminant
The discriminant is a key component in understanding the nature of the roots of a quadratic equation. It comes from the formula \(D = b^2 - 4ac\) where \(a\), \(b\), and \(c\) are coefficients from the quadratic equation \(ax^2 + bx + c = 0\). The value of the discriminant tells us whether the equation's solutions are real or complex.

  • If \(D > 0\), the quadratic equation has two distinct real roots.
  • If \(D = 0\), the equation has exactly one real root, also called a repeated or double root.
  • If \(D < 0\), the roots are not real numbers; instead, they are complex conjugates.

In this exercise, we are dealing with the function \(f(x)=a x^{2}+2 x+1\) and want to find when it has no real zeros. This means we need \(D < 0\), thus focusing on solving the inequality \(4 - 4a < 0\) to discover when there will be no intersection with the x-axis.
Quadratic Equation
A quadratic equation is a second-degree polynomial equation in the form of \(ax^2 + bx + c = 0\). Here, \(a\), \(b\), and \(c\) are constants with \(a eq 0\). The solutions to a quadratic equation can be understood through the parabola shape that represents it. The vertex form or the standard form provides different insights into the properties of the functional graph.

Finding solutions to quadratic equations is an essential step to understand their properties. These solutions are also known as the roots or zeros of the quadratic function, and they can be found using the quadratic formula:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.\]
Depending on the discriminant \(D\), the square root term in the quadratic formula will determine whether the results are real numbers or complex numbers.

In our context, the quadratic equation is derived from the function \(f(x) = ax^2 + 2x + 1 = 0\), and by altering \(a\), we can determine the nature of its solutions.
Real Zeros
Real zeros of a polynomial function, such as a quadratic, are the values of \(x\) where the function intersects the x-axis, meaning \(f(x) = 0\). These are the solutions where the output of the function becomes zero. A function with no real zeros implies that its graph does not touch or cross the x-axis.

To determine the presence of real zeros in a quadratic function, we use the discriminant \(D = b^2 - 4ac\). When \(D < 0\), it indicates the absence of real zeros because the parabola representing the quadratic equation lies entirely above or below the x-axis. Complex solutions arise, hinting that the equation doesn't end with a real number solution.

For the problem \(f(x)=a x^{2}+2 x+1\), calculating \(4 - 4a < 0\) and solving for \(a\) determined that for \(a > 1\), there are no real zeros. Thus, the parabola will always be either above or below the x-axis depending on the nature and direction of its opening.

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