/*! 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 39 Let \(g(x)=4 x^{2}-2 x-3 .\) Fin... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 39

Let \(g(x)=4 x^{2}-2 x-3 .\) Find \(x\) such that \(g(x)=0\)

Short Answer

Expert verified
The solutions are \( x = \frac{1 + \sqrt{13}}{4} \) and \( x = \frac{1 - \sqrt{13}}{4} \).

Step by step solution

01

- Write down the equation

Start by writing down the given equation: \[ g(x) = 4x^2 - 2x - 3 \]Given: \( g(x) = 0 \)
02

- Set the equation to zero

Since we want to find \( x \) such that \( g(x) = 0 \), set the equation equal to zero: \[ 4x^2 - 2x - 3 = 0 \]
03

- Apply the quadratic formula

To solve for \( x \) in the quadratic equation \( 4x^2 - 2x - 3 = 0 \), use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]For our equation, \( a = 4 \), \( b = -2 \), and \( c = -3 \).
04

- Solve for the discriminant

Calculate the discriminant \( \Delta \) using the formula \[ \Delta = b^2 - 4ac \]: \[ \Delta = (-2)^2 - 4 \cdot 4 \cdot (-3) \]\[ \Delta = 4 + 48 \]\[ \Delta = 52 \]
05

- Compute the solutions

Substitute the discriminant \( \Delta \) and the constants \( a \) and \( b \) back into the quadratic formula to find the solutions: \[ x = \frac{-(-2) \pm \sqrt{52}}{2 \cdot 4} \]\[ x = \frac{2 \pm \sqrt{52}}{8} \]\[ x = \frac{2 \pm 2\sqrt{13}}{8} \]\[ x = \frac{1 \pm \sqrt{13}}{4} \]
06

- Present the final answers

Thus, the solutions for \( x \) are: \[ x = \frac{1 + \sqrt{13}}{4} \]and\[ x = \frac{1 - \sqrt{13}}{4} \]

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Key Concepts

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

quadratic formula
To solve quadratic equations like the one given (4x^2 - 2x - 3 = 0), the quadratic formula is an essential tool. The quadratic formula allows us to find the roots of any quadratic equation in the form ax^2 + bx + c = 0. The formula is:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
This formula might look complex initially, but it breaks down into manageable parts:
  • -b: The opposite of the coefficient of x
  • \pm: Plus or minus, indicating two potential solutions
  • \sqrt{b^2 - 4ac}: The square root of the discriminant, which we'll discuss next
  • 2a: Twice the coefficient of or the squared term
By substituting the values of a, b, and c from our equation into the quadratic formula, we can solve for x. This method works even when the equation doesn't factor easily, making it a powerful tool for solving quadratics.
discriminant
The discriminant is a crucial part of the quadratic formula, represented by \( b^2 - 4ac \), and it helps determine the nature of the roots without fully solving the equation. We compute the discriminant as follows:
\[ \Delta = b^2 - 4ac \]
Depending on the value of \( \Delta \):
  • If \( \Delta > 0 \), there are two distinct real roots
  • If \( \Delta = 0 \), there is exactly one real root or a repeated root
  • If \( \Delta < 0 \), there are no real roots, but two complex roots
In our exercise, \( \Delta \) was calculated as:
\( \Delta = 52 \)
Since \( 52 > 0 \), we can confidently say there are two distinct real roots for the equation 4x^2 - 2x - 3 = 0.
roots of quadratic equations
Finding the roots of a quadratic equation means identifying the values of x that make the equation (ax^2 + bx + c = 0) true. These roots can be found using the quadratic formula, factoring, or completing the square. In this exercise, we use the quadratic formula due to its general applicability.
Given:\[ x = \frac{1\pm \sqrt{13}}{4} \]
This breaks down to two solutions:
  • \( x = \frac{1 + \sqrt{13}}{4} \) - a positive root
  • \( x = \frac{1 - \sqrt{13}}{4} \) - a negative root
These solutions are the points where the quadratic function intersects the x-axis. Understanding the nature and location of these roots is essential in graphing quadratic functions and solving real-world problems involving quadratics.

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Most popular questions from this chapter

Solve by completing the square. Show your work. $$ x^{2}+12 x+32=0 $$

Can a fourth-degree equation have exactly three irrational solutions? Why or why not?

Find the \(x\) -intercepts of the function given by \(g(x)=2 x^{2}-3 x-1\).

A quadratic function has \((4,0)\) as one of its intercepts and \((-1,7)\) as its vertex. Find an equation for the function.

Review finding intercepts and completing the square (Sections \(3.3,6.2,6.3, \text { and } 11.1)\) Find the \(x\) -intercepts. $$ y=2 x^{2}-x-3 $$
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