/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 64 \(8 x^{2}-4 x-3=0\)... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 64

\(8 x^{2}-4 x-3=0\)

Short Answer

Expert verified
The roots of the equation \(8x^{2}-4x-3=0\) are \(x = \frac{1 + \sqrt{7}}{4}\) and \(x = \frac{1 - \sqrt{7}}{4}\).

Step by step solution

01

Identify Coefficients

In the given equation \(8x^{2}-4x-3=0\), the coefficient a is 8, the coefficient b is -4, and the constant c is -3.
02

Substitute into the Quadratic Formula

Next, substitute the values of a, b and c into the Quadratic formula \(x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\). Plugging these values in gives the equation \(x = \frac{-(-4) \pm \sqrt{(-4)^{2}-4*8*-3}}{2*8}\).
03

Evaluate Inside the Square Root

The next step is to simplify the equation. This starts with calculating the value under the square root. This gives \(x = \frac{4 \pm \sqrt{16+96}}{16}\).
04

Simplify the Square Root and the Equation

Simplifying the square root gives \(x = \frac{4 \pm \sqrt{112}}{16}\). Also, \(x = \frac{4 \pm 4\sqrt{7}}{16}\), which can be simplified further to \(x = \frac{1 \pm \sqrt{7}}{4}\).
05

Final Solutions

So, the two roots (values for x) of the given equation are \(x = \frac{1 + \sqrt{7}}{4}\) and \(x = \frac{1 - \sqrt{7}}{4}\).

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Key Concepts

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

Quadratic Formula
The quadratic formula is a crucial tool in algebra for finding the roots of any quadratic equation. A quadratic equation is generally in the form \(ax^2 + bx + c = 0\) where the letters \(a\), \(b\), and \(c\) represent constants. The quadratic formula itself is given by: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
This formula allows us to find solutions for \(x\), which are also known as the "roots" of the equation. Here's how it works:
  • The term \(-b\) flips the sign of \(b\).
  • \(b^2 - 4ac\) is called the discriminant and it determines the nature of the roots.
  • The square root symbol \(\pm\) indicates there are generally two solutions.
You insert your \(a\), \(b\), and \(c\) values from the specific equation into this formula and follow through the steps to find your answers.
Roots of Equations
Roots of a quadratic equation allow you to find where the equation equals zero. For the quadratic equation, these are the values of \(x\) that satisfy the equation \(ax^2 + bx + c = 0\). Using the quadratic formula, the solutions are derived as: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
The nature of the roots depends heavily on the value of the discriminant \(b^2 - 4ac\):
  • If the discriminant is positive, there are two distinct real roots.
  • If it is zero, there is exactly one real root, which means the parabolic graph of the equation touches the x-axis only once.
  • If the discriminant is negative, the roots are complex, indicating the parabola does not intersect the x-axis at all.
Understanding roots is vital because they reveal where a quadratic function will cross the horizontal (x) axis, which is useful in various mathematical and real-world applications.
Simplification of Radicals
Simplification of radicals is often needed while solving quadratic equations. A radical is an expression that uses the square root symbol, and simplifying it can make equations easier to solve and interpret. For example, consider \(\sqrt{112}\) from our equation where we simplify eventually to \(4\sqrt{7}\).
Here's a simple guide to simplify radicals:
  • Find factors of the number under the square root. Look for perfect square factors.
  • Take the square root of the perfect square factor out of the radical.
  • Continue this process until the expression under the square root can't be factored further by perfect squares.
For our example, since \(112\) can be expressed as \(16 \times 7\), and \(16\) is a perfect square (\(4^2\)), we took \(4\) out of the square root, making it \(4\sqrt{7}\). This simplification step is essential as it leads directly to finding the exact roots of the equation, providing clearer insights into their value.

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Most popular questions from this chapter

In Exercises 55-64, verify the identity. $$ \sin (x+y)+\sin (x-y)=2 \sin x \cos y $$

Let \(x=\pi / 6\) in the identity in Exercise 91 and define the functions \(f\) and \(g\) as follows. $$ \begin{aligned} &f(h)=\frac{\cos (\pi / 6+h)-\cos (\pi / 6)}{h} \\ &g(h)=\cos \frac{\pi}{6}\left(\frac{\cos h-1}{h}\right)-\sin \frac{\pi}{6}\left(\frac{\sin h}{h}\right) \end{aligned} $$ (a) What are the domains of the functions \(f\) and \(g\) ? (b) Use a graphing utility to complete the table. $$ \begin{array}{|l|l|l|l|l|l|l|} \hline h & 0.01 & 0.02 & 0.05 & 0.1 & 0.2 & 0.5 \\ \hline f(h) & & & & & & \\ \hline g(h) & & & & & & \\ \hline \end{array} $$ (c) Use a graphing utility to graph the functions \(f\) and \(g\). (d) Use the table and the graphs to make a conjecture about the values of the functions \(f\) and \(g\) as \(h \rightarrow 0\).

In Exercises 133-136, find (if possible) the complement and supplement of each angle. (a) \(55^{\circ}\) (b) \(162^{\circ}\)

In Exercises 55-64, verify the identity. $$ \sin \left(\frac{\pi}{6}+x\right)=\frac{1}{2}(\cos x+\sqrt{3} \sin x) $$

In Exercises 45-50, find the exact value of the trigonometric function given that \(\sin u=-\frac{7}{25}\) and \(\cos v=-\frac{4}{5}\). (Both \(u\) and \(v\) are in Quadrant III.) $$ \cos (u+v) $$
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