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Chapter 2: Problem 46

Use the discriminant to determine how many real roots each equation has. $$4 x^{2}-28 x+49=0$$

Short Answer

Expert verified
The equation has exactly one real root.

Step by step solution

01

Rewrite the Quadratic Equation

The given equation is already in standard form: \(4x^2 - 28x + 49 = 0\). In this form, \(a = 4\), \(b = -28\), and \(c = 49\).
02

Calculate the Discriminant

The discriminant \(D\) of a quadratic equation \(ax^2 + bx + c = 0\) is given by the formula \(D = b^2 - 4ac\). Substitute \(a = 4\), \(b = -28\), and \(c = 49\) into the formula to find \(D\): \[ D = (-28)^2 - 4(4)(49) \] \[ D = 784 - 784 = 0 \].
03

Interpret the Discriminant

The value of the discriminant \(D = 0\) indicates that the quadratic equation has exactly one real root. When \(D = 0\), the parabola touches the x-axis at a single point, which means there is one repeated root (double root).

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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 valuable tool in analyzing quadratic equations. It helps us understand how many real roots a quadratic equation has. The general formula for the discriminant \(D\) is \(D = b^2 - 4ac\), where \(a\), \(b\), and \(c\) are coefficients from the quadratic equation \(ax^2 + bx + c = 0\). This simple formula can tell us a lot:
  • When \(D > 0\), there are two distinct real roots.
  • When \(D = 0\), there is exactly one real root, indicating a repeated or double root.
  • When \(D < 0\), there are no real roots, but two complex roots instead.
By calculating the discriminant, we gain insight into the nature of the roots without having to explicitly solve the equation. This makes it an efficient first step in understanding any quadratic equation.
Real Roots
Real roots are the solutions of a quadratic equation that can be plotted on the number line. When we talk about real roots, we refer to those solutions of the equation which are not imaginary. In the context of quadratic equations:
  • Real roots exist if the discriminant \(D\) is greater than or equal to zero.
  • Two distinct real roots exist if \(D > 0\), meaning the parabola formed by the equation crosses the x-axis at two points.
  • One real root (or a double root) exists if \(D = 0\). This means the parabola touches the x-axis at a single point, known as the vertex.
Real roots are crucial because they represent actual intersections of the quadratic's parabolic graph with the x-axis. Understanding whether we have real roots helps in visualizing the behavior of quadratic equations and is central to solving them.
Quadratic Formula
The quadratic formula is a powerful method for solving quadratic equations. It's especially useful when factoring is complex or impossible. The formula is given by:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] This formula directly incorporates the discriminant, \(b^2 - 4ac\), under the square root term. By using the quadratic formula:
  • You can find the roots of any quadratic equation, whether they are real, double, or complex.
  • The '卤' symbol indicates that there are usually two solutions unless the discriminant is zero, in which case both rows give the same solution.
The quadratic formula is a fundamental tool in mathematics, enabling the solution of almost any quadratic equation and providing a clear pathway to understanding more about the properties of these equations.

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Most popular questions from this chapter

Solve the inequalities. Suggestion: A calculator may be useful for approximating key numbers. $$\frac{2-x}{3-2 x} \geq 0$$

Given two positive numbers \(a\) and \(b,\) we define the geometric mean (G.M.) and the arithmetic mean (A.M.) as follows: $$\text { G.M. }=\sqrt{a b} \quad \text { A.M. }=\frac{a+b}{2}$$ (a) Complete the table, using a calculator as necessary so that the entries in the third and fourth columns are in decimal form. $$\begin{array}{rrrr} & & \sqrt{a b} & (a+b) / 2 & \text { Which is larger, } \\ a & b & (G . M .) & (A . M .) & G . M . \text { or } A . M . ? \\ \hline 1 & 2 & & & \\ 1 & 3 & & & \\ 1 & 4 & & & \\ 2 & 3 & & & \\ 3 & 4 & & & \\ 5 & 10 & & & \\ 9 & 10 & & & \\ 99 & 100 & & & \\ 999 & 1000 & & & & \\ \hline \end{array}$$ (b) Prove that for all nonnegative numbers \(a\) and \(b\) we have $$\sqrt{a b} \leq \frac{a+b}{2}$$ Hint: Use the following property of inequalities: If \(x\) and \(y\) are nonnegative, then the inequality \(x \leq y\) is equivalent to \(x^{2} \leq y^{2}\) (c) Assuming that \(a=b\) (and that \(a\) and \(b\) are nonnegative), show that inequality (1) becomes an equality. (d) Assuming that \(a\) and \(b\) are nonnegative and that \(\sqrt{a b}=\frac{a+b}{2},\) show that \(a=b\) Remark: Parts (b) through (d) can be summarized as follows. For all nonnegative numbers \(a\) and \(b,\) we have \(\sqrt{a b} \leq \frac{a+b}{2},\) with equality holding if and only if \(a=b\) This result is known as the arithmetic- geometric mean inequality for two numbers. The mini project at the end of this section shows an application of this result.

Solve each equation. \(\sqrt{x^{2}+3 x-4}-\sqrt{x^{2}-5 x+4}=x-1,\) where \(x>4\) Hint: Factor the expressions beneath the radicals. Then note that \(\sqrt{x-1}\) is a factor of both sides of the equation.

Solve the inequalities Suggestion: A calculator may be useful for approximating key numbers. $$x^{2}\left(3 x^{2}+11\right) \geq 4$$

Find all real solutions of each equation. For Exercises \(31-36,\) give two forms for each answer: an exact answer (involving a radical) and a calculator approximation rounded to two decimal places. $$x^{-4}+x^{-2}+1=0$$
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