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Chapter 7: Problem 10

solve the given quadratic equations, using the quadratic formula. Exercises \(5-8\) are the same as Exercises \(11-14\) of Section 7.2. $$x^{2}+10 x-4=0$$

Short Answer

Expert verified
The solutions are \(x = -5 + \sqrt{29}\) and \(x = -5 - \sqrt{29}\).

Step by step solution

01

Identify Coefficients

The given quadratic equation is in the form \(ax^2 + bx + c = 0\). Here, \(a = 1\), \(b = 10\), and \(c = -4\).
02

Write the Quadratic Formula

The quadratic formula to solve equations of the form \(ax^2 + bx + c = 0\) is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
03

Calculate the Discriminant

The discriminant is the part under the square root in the quadratic formula: \(b^2 - 4ac\). Calculate it as \(10^2 - 4 \times 1 \times (-4) = 100 + 16 = 116\).
04

Substitute Values into the Formula

Substitute \(a = 1\), \(b = 10\), and \(c = -4\) into the quadratic formula: \(x = \frac{-10 \pm \sqrt{116}}{2 \times 1}\).
05

Simplify the Expression

First, simplify \(\sqrt{116}\). Since \(116 = 4 \times 29\), \(\sqrt{116} = \sqrt{4 \times 29} = 2\sqrt{29}\). Then the expression becomes \(x = \frac{-10 \pm 2\sqrt{29}}{2}\).
06

Solve for x

Separate the expression into two terms: \(x = \frac{-10 + 2\sqrt{29}}{2}\) and \(x = \frac{-10 - 2\sqrt{29}}{2}\). These simplify to \(x = -5 + \sqrt{29}\) and \(x = -5 - \sqrt{29}\).
07

Conclude with Solutions

The solutions to the equation \(x^2 + 10x - 4 = 0\) are \(x = -5 + \sqrt{29}\) and \(x = -5 - \sqrt{29}\).

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Key Concepts

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

Quadratic Equations
Quadratic equations are a type of polynomial equation that are characterized by their highest degree, which is two. They are generally written in the standard form as \(ax^2 + bx + c = 0\). Here, \(a\), \(b\), and \(c\) are constants, where \(a\) cannot be zero because that would make it a linear equation, not quadratic.
Understanding these key components is essential, as they define the shape of the parabola when graphed:
  • The term \(ax^2\) determines the direction the parabola opens, upwards if \(a > 0\) and downwards if \(a < 0\).
  • The linear term \(bx\) influences the tilt and horizontal shift of the parabola.
  • The constant \(c\) affects the vertical position of the graph.
Understanding how these parts fit together can help make solving these equations more intuitive, as each part plays a crucial role in determining the nature and position of the solutions.
Discriminant
The discriminant is a part of the quadratic formula that provides valuable insights about the nature of the roots of a quadratic equation. It is represented by \(b^2 - 4ac\).
Here's how the discriminant affects the solutions:
  • If \(b^2 - 4ac > 0\), the quadratic equation has two distinct real roots. This means the parabola intersects the x-axis at two different points.
  • If \(b^2 - 4ac = 0\), there is exactly one real root, indicating the vertex of the parabola touches the x-axis.
  • If \(b^2 - 4ac < 0\), the equation has no real roots, implying the parabola does not intersect the x-axis at all.
In our example, we calculated the discriminant to be 116, which is greater than zero, confirming that there are two distinct real solutions.
Solving Quadratics
Solving quadratic equations using the quadratic formula involves substituting the constants \(a\), \(b\), and \(c\) into the expression \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). This method is particularly useful because it works for all types of quadratic equations, regardless of whether the solutions are real or complex.
Let's break this process down:
  • Substitute: Insert the values of \(a\), \(b\), and \(c\) from the equation into the formula.
  • Calculate: Use the discriminant to determine the nature of the roots, then compute inside the square root.
  • Simplify: Solve for \(x\) by performing basic arithmetic and algebraic operations, considering both the plus and minus parts of the \(\pm\) symbol.
This comprehensive approach, illustrated by the example \(x^2 + 10x - 4 = 0\), reveals the solutions \(x = -5 + \sqrt{29}\) and \(x = -5 - \sqrt{29}\), demonstrating how neatly the quadratic formula ties the components of a quadratic equation into clear solutions.

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Most popular questions from this chapter

Solve the given problems. All numbers are accurate to at least two significant digits. Solve \(6 x^{2}-x=15\) for \(x\) by (a) factoring, (b) completing the square, and (c) the quadratic formula. Which is (a) longest?, (b) shortest?

Solve the given applied problem. Under specified conditions, the pressure loss \(L\) (in \(\mathrm{lb} / \mathrm{in}\). \(^{2}\) per \(100 \mathrm{ft}),\) in the flow of water through a fire hose in which the flow is \(q\) gal/min, is given by \(L=0.0002 q^{2}+0.005 q .\) Sketch the graph of \(L\) as a function of \(q,\) for \(q<100\) gal/min.

Solve the given applied problem. The diagonal of a rectangular floor is \(3.00 \mathrm{ft}\) less than twice the length of one of the sides. If the other side is \(15.0 \mathrm{ft}\) long, what is the area of the floor?

Solve the given problems. All numbers are accurate to at least two significant digits. Solve the equation \(x^{4}-5 x^{2}+4=0\) for \(x\). (Hint: The equation can be written as \(\left(x^{2}\right)^{2}-5\left(x^{2}\right)+4=0 .\) First solve for \(x^{2}\).)

Use a calculator to solve the given equations. If there are no real roots, state this as the answer. \(3 x^{2}-25=20 x\)
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