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Chapter 0: Problem 39

Find all real zeros of the polynomial. $$ x^{2}-3 $$

Short Answer

Expert verified
The real zeros of the polynomial \(x^{2}-3\) are \(x = \sqrt{3}\) and \(x = -\sqrt{3}\).

Step by step solution

01

Set up the equation

First, the given polynomial, \(x^{2}-3\), needs to be set equal to zero, because the real zeros of the polynomial are the values that result in the polynomial being zero. This gives the equation \(x^{2}-3 = 0\).
02

Rearrange the equation

In order to isolate \(x^{2}\), add 3 to both sides of the equation, which leads to: \(x^{2} = 3\).
03

Solve for x

Taking the square root of both sides to solve for \(x\), it needs to be noted that this will result in both a positive and negative solution, as when a number is squared, it becomes positive regardless of whether it was initially positive or negative. This calculates to \(x = \sqrt{3}\) and \(x = -\sqrt{3}\).

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Key Concepts

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

Real Numbers
Real numbers include all the numbers we encounter in everyday life. They consist of integers like -1, 0, and 5, fractions like 1/2, irrational numbers such as the square root of 2, and all the decimals in between. Real numbers are crucial because they make up the continuous number line we use in mathematics.
  • Understanding real numbers helps in solving equations since it allows us to categorize solutions as real or non-real.
  • Real zeros, for instance, are the x-values for which a polynomial equals zero, indicating where a graph intersects the x-axis.
In our exercise, we are dealing with a polynomial whose real zeros we need to find. This means we are looking for the values of x in the real number system that satisfy the equation, resulting in zero when substituted back into the polynomial.
Quadratic Equation
Quadratic equations are polynomial equations of degree two. The general form of a quadratic equation is:\[ ax^2 + bx + c = 0 \]where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\).
These equations form parabolic graphs on a plane, and solving them involves finding the values of x that make the equation true (meaning the polynomial equals zero).
  • When given in standard form, replacing \(b\) and \(c\) with zero makes it easier, as in the exercise \(x^2 - 3 = 0\), where \(b = 0\) and \(c = -3\).
  • Quadratic equations can have two solutions, one solution, or no real solution at all, depending on the discriminant (the value inside the square root in the quadratic formula).
Square Root Method
The square root method is a straightforward technique used to solve quadratic equations, especially when there is no linear term (\(bx\)) in the equation. It involves isolating the term with \(x^2\) on one side and taking the square root on both sides to solve for x.
In the exercise given:
  • After setting up \(x^2 - 3 = 0\), we isolated \(x^2\) to get \(x^2 = 3\).
  • Taking the square root of \(x^2 = 3\), we find two solutions, \(x = \sqrt{3}\) and \(x = -\sqrt{3}\).
Remember that when taking square roots, there are always two potential solutions: positive and negative. This is because squaring either a positive or a negative number returns a positive result. This method provides an efficient way to find real zeros in polynomials when we don't have any term with \(x\).

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Most popular questions from this chapter

a certificate of deposit has a principal of P and an annual percentage rate of r (expressed as a decimal) compounded n times per year. Enter the compound interest formula $$ A=P\left(1+\frac{r}{n}\right)^{N} $$ into a graphing utility and use it to find the balance after \(N\) compoundings. $$ P=\$ 5000, \quad r=5.5 \%, \quad n=4, \quad N=60 $$

find the domain of the given expression. $$ \frac{1}{\sqrt[3]{x+4}} $$

Perform the indicated operations and rationalize as needed. $$ \frac{\frac{\sqrt{4-x^{2}}}{x^{4}}-\frac{2}{x^{2} \sqrt{4-x^{2}}}}{4-x^{2}} $$

Rationalize the numerator or denominator and simplify. $$ \frac{13}{6+\sqrt{10}} $$

simplify the expression. $$ \frac{7 x^{2}}{x^{-3}} $$
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